Trigonometric functions

Class 11 Mathematics, unit 9, book pages 216–249

Practise 95 MCQs Take the unit quiz Review 29 cards

The six trigonometric functions are studied as functions of a real number: where each is defined (domain), which values it takes (range), how often it repeats (period) and whether it is even or odd. The unit finds the greatest and least values of expressions such as $a + b sin (c θ + d)$ , draws the graphs of sine, cosine, tangent and the reciprocal functions from tables of values, and uses the transformed sine and cosine $y = a f [b (x - c)] + d$ to model periodic situations such as a Ferris wheel and alternating current.

Learning outcomes (6)

Find the domain and range of the trigonometric functions.
Discuss even and odd functions, and the periodicity of trigonometric functions.
Find the maximum and minimum value of a given function of the type $a + b sin θ$ , $a + b cos θ$ , $a + b sin (c θ + d)$ , $a + b cos (c θ + d)$ , and the reciprocal of the above, where $a, b, c$ and $d$ are real numbers.
Graph and analyse the trigonometric functions sine, cosine and tangent to solve problems.
Explain the properties of the graphs of $sin θ$ , $cos θ$ and $tan θ$ .
Apply the concepts of trigonometric functions, identities, graphs, periodicity, even and odd functions, and extreme values to real-world problems such as distance, elevation and direction of tall structures, navigation and mapping, lengths of irregular shapes, graphs to visualise and predict patterns in data, frequency and periodic length of a Ferris wheel, forces on a see-saw or lever, and the ideal angle for solar panel placement.

9.1Domain and range of trigonometric functionsOften examined

The domain of a function $f (x)$ is the set of all values of $x$ for which $f (x)$ is defined. The range is the set of all values that $f (x)$ takes when $x$ runs through the domain. This section finds the domain and range of the six trigonometric functions straight from the definition of the ratios, and then of functions built from them.

Domain of $f$: The set of all possible values of $x$ for which $f (x)$ is defined.
Range of $f$: The set of all possible values that $f (x)$ can take when $x$ is any number from the domain.

Let $θ$ be any real number. Construct the angle of $θ$ radian with its vertex at the origin of a rectangular coordinate system and its initial side along the $x$ -axis. Let $P (x, y)$ be any point on the terminal side $O A$ , and let $O P = r$ . Draw $P M$ perpendicular to the $x$ -axis, so that $O M P$ is a right-angled triangle with sides $x$ , $y$ and hypotenuse $r$ .

The angle

θ

in standard position, with

P (x, y)

on its terminal side

O A

and

P M

drawn perpendicular to the

x

-axis. In the right-angled triangle

O M P

sin θ = \frac{y}{r}

and

cos θ = \frac{x}{r}

Derivation Domain and range of $y = sin θ$

Aim. Find the domain and range of $sin θ$ .

From the right-angled triangle $O M P$ , the sine is the ratio of the side opposite $θ$ to the hypotenuse.
$sin θ = \frac{y}{r}$
For any $θ$ the ordinate $y$ cannot be longer than the distance $r$ from the origin to $P$ , so $y$ lies between $- r$ and $r$ .
$- r \leq y \leq r$
Divide every part by $r > 0$ . The direction of the inequalities does not change.
$- 1 \leq \frac{y}{r} \leq 1$
Replace $\frac{y}{r}$ by $sin θ$ . Nothing stops us from using any real $θ$ , so every real number is allowed as input.
$- 1 \leq sin θ \leq 1$

Result

Domain of sin θ = R = (- \infty, \infty), Range of sin θ = [- 1, 1]

The sine is defined for every real angle and never leaves the band from $- 1$ to $1$ .

Derivation Domain and range of $y = cos θ$

Aim. Find the domain and range of $cos θ$ .

From the right-angled triangle $O M P$ , the cosine is the ratio of the adjacent side to the hypotenuse.
$cos θ = \frac{x}{r}$
The abscissa $x$ also satisfies $- r \leq x \leq r$ for any $θ$ .
$- r \leq x \leq r$
Divide by $r > 0$ .
$- 1 \leq \frac{x}{r} \leq 1$
Replace $\frac{x}{r}$ by $cos θ$ .
$- 1 \leq cos θ \leq 1$

Result

Domain of cos θ = R = (- \infty, \infty), Range of cos θ = [- 1, 1]

Derivation Domain and range of $y = tan θ$

Aim. Find the domain and range of $tan θ$ .

In the triangle $O M P$ the tangent is the ratio of the opposite side to the adjacent side.
$tan θ = \frac{y}{x}$
This ratio is not defined when $x = 0$ , that is when the terminal side lies along the $y$ -axis. That happens exactly when $θ$ is an odd multiple of $\frac{π}{2}$ .
$θ \neq = (2 n + 1) \frac{π}{2}, n \in Z$
For every other value of $θ$ the ratio $\frac{y}{x}$ can be any real number: as the terminal side turns, $y$ and $x$ change and the ratio takes every real value.
$\frac{y}{x} \in R$

Result

Domain of tan θ = R - {(2 n + 1) \frac{π}{2}, n \in Z}, Range of tan θ = R

Derivation Domain and range of $y = cot θ$

Aim. Find the domain and range of $cot θ$ .

The cotangent is the ratio of the adjacent side to the opposite side.
$cot θ = \frac{x}{y}$
It is not defined when $y = 0$ , that is when the terminal side lies along the $x$ -axis. That happens when $θ$ is an even multiple of $\frac{π}{2}$ , in other words a multiple of $π$ .
$θ \neq = nπ, n \in Z$
For every other $θ$ the ratio $\frac{x}{y}$ can be any real number.
$\frac{x}{y} \in R$

Result

Domain of cot θ = R - {nπ, n \in Z}, Range of cot θ = R

Derivation Domain and range of $y = sec θ$

Aim. Find the domain and range of $sec θ$ .

The secant is the ratio of the hypotenuse to the adjacent side, so it is not defined when $x = 0$ , that is when $θ$ is an odd multiple of $\frac{π}{2}$ .
$sec θ = \frac{r}{x}, θ \neq = (2 n + 1) \frac{π}{2}$
For any other $θ$ , we know $- 1 \leq \frac{x}{r} \leq 1$ , that is $\frac{x}{r} \leq 1$ .
$\frac{x}{r} \leq 1$
Take reciprocals. For a number whose size is at most 1, the reciprocal has size at least 1.
$\frac{r}{x} \geq 1 ⟹ ∣ sec θ ∣ \geq 1$

Result

Domain of sec θ Range of sec θ = R - {(2 n + 1) \frac{π}{2}, n \in Z} = R - (- 1, 1)

$∣ sec θ ∣ \geq 1$ means $sec θ \geq 1$ or $sec θ \leq - 1$ . The values strictly between $- 1$ and $1$ are never taken.

Derivation Domain and range of $y = cosec θ$

Aim. Find the domain and range of $cosec θ$ .

The cosecant is the ratio of the hypotenuse to the opposite side, so it is not defined when $y = 0$ , that is when $θ$ is a multiple of $π$ .
$cosec θ = \frac{r}{y}, θ \neq = nπ$
For any other $θ$ , $- 1 \leq \frac{y}{r} \leq 1$ , that is $\frac{y}{r} \leq 1$ .
$\frac{y}{r} \leq 1$
Take reciprocals.
$\frac{r}{y} \geq 1 ⟹ ∣ cosec θ ∣ \geq 1$

Result

Domain of cosec θ = R - {nπ, n \in Z}, Range of cosec θ = R - (- 1, 1)

Remarks

Domain and range of the six trigonometric functions.
Function	Domain	Range
$y = sin x$	$R = (- \infty, \infty)$	$- 1 \leq y \leq 1$
$y = cos x$	$R = (- \infty, \infty)$	$- 1 \leq y \leq 1$
$y = tan x$	${x : x \in R, x \neq = (2 n + 1) \frac{π}{2}, n an integer}$	$R = (- \infty, \infty)$
$y = cot x$	${x : x \in R, x \neq = nπ, n an integer}$	$R = (- \infty, \infty)$
$y = sec x$	${x : x \in R, x \neq = (2 n + 1) \frac{π}{2}, n an integer}$	$y \geq 1$ or $y \leq - 1$
$y = cosec x$	${x : x \in R, x \neq = nπ, n an integer}$	$y \geq 1$ or $y \leq - 1$

Example 1

Find the domain and range of the following: (i)

y = 4 sin 3 x

(ii)

y = \frac{1}{2 cos x - 1}

(iii)

y = \frac{1}{1 + 2 sin θ}

(iv)

y = \frac{1}{2 - sin 3 x}

(i) $y = 4 sin 3 x$

$sin 3 x$ is defined for every real $x$ , so the function is too.
$Domain = D_{y} = (- \infty, \infty) = R$
For the range, put $3 x = θ$ . As $x$ runs through all real numbers, so does $θ$ , and $y = 4 sin θ$ .
$y = 4 sin θ$
Use the range of sine, $- 1 \leq sin θ \leq 1$ , and multiply by 4 (positive, so the inequalities keep their direction).
$- 4 \leq 4 sin θ \leq 4 ⟹ - 4 \leq 4 sin 3 x \leq 4$
So $- 4 \leq y \leq 4$ .
$Range = R_{y} = [- 4, 4]$

(ii) $y = \frac{1}{2 cos x - 1}$

The function is not defined where the denominator is zero.
$2 cos x - 1 = 0 ⟹ cos x = \frac{1}{2}$
$cos x = \frac{1}{2}$ at $x = \frac{π}{3}$ and $x = \frac{5 π}{3}$ in one turn, and then every $2 π$ .
$D_{y} = R - {x : x = 2 nπ + \frac{π}{3} or x = 2 nπ + \frac{5 π}{3}, n \in Z}$
For the range, start from $- 1 \leq cos x \leq 1$ and multiply by 2.
$- 2 \leq 2 cos x \leq 2$
Subtract 1 from every part.
$- 3 \leq 2 cos x - 1 \leq 1$
The denominator $2 cos x - 1$ lies in $[- 3, 1]$ but is never $0$ . Take reciprocals separately for the two parts. If $0 < 2 cos x - 1 \leq 1$ the reciprocal is at least $1$ . If $- 3 \leq 2 cos x - 1 < 0$ the reciprocal is at most $- \frac{1}{3}$ .
$y \geq 1 or y \leq - \frac{1}{3}$
Write the two pieces as intervals.
$R_{y} = (- \infty, - \frac{1}{3}] \cup [1, \infty)$

(iii) $y = \frac{1}{1 + 2 sin θ}$

The book solves this part in the variable $x$ . The denominator must not be zero.
$1 + 2 sin x = 0 ⟹ sin x = - \frac{1}{2}$
$sin x = - \frac{1}{2}$ at $x = \frac{7 π}{6}$ and $x = \frac{11 π}{6}$ in one turn, then every $2 π$ .
$D_{y} = R - {x : x = 2 nπ + \frac{7 π}{6} or x = 2 nπ + \frac{11 π}{6}, n \in Z}$
For the range, start from $- 1 \leq sin x \leq 1$ and multiply by 2.
$- 2 \leq 2 sin x \leq 2$
Add 1 to every part.
$- 1 \leq 1 + 2 sin x \leq 3$
The denominator lies in $[- 1, 3]$ and is never $0$ . If $0 < 1 + 2 sin x \leq 3$ the reciprocal is at least $\frac{1}{3}$ . If $- 1 \leq 1 + 2 sin x < 0$ the reciprocal is at most $- 1$ .
$y \geq \frac{1}{3} or y \leq - 1$
Write the two pieces as intervals.
$R_{y} = (- \infty, - 1] \cup [\frac{1}{3}, \infty)$

(iv) $y = \frac{1}{2 - sin 3 x}$

The denominator would be zero only if $sin 3 x = 2$ .
$2 - sin 3 x \neq = 0 ⟺ sin 3 x \neq = 2$
This never happens, because $- 1 \leq sin 3 x \leq 1$ . So the function is defined for every real $x$ .
$D_{y} = (- \infty, \infty) = R$
For the range, start from $- 1 \leq sin 3 x \leq 1$ and multiply by $- 1$ , which reverses the inequalities.
$1 \geq - sin 3 x \geq - 1$
Add 2 to every part. The denominator is positive and lies between 1 and 3.
$3 \geq 2 - sin 3 x \geq 1$
Take reciprocals of positive numbers: the order reverses again.
$\frac{1}{3} \leq \frac{1}{2 - sin 3 x} \leq 1$
So $\frac{1}{3} \leq y \leq 1$ .
$R_{y} = [\frac{1}{3}, 1]$

Worked example: domain and range of

y = sec 2 x

Find the domain, range and period of

y = sec 2 x

- π

- \frac{π}{2}

\frac{π}{2}

π

−5

−4

−3

−2

−1

x

y

y = sec 2 x

Graph of

y = sec 2 x

. It has no point at

x = \pm \frac{π}{4}, \pm \frac{3 π}{4}

(dashed lines) and never enters the strip

- 1 < y < 1

Write secant as a reciprocal. It is defined only where the denominator is not zero.
$sec 2 x = \frac{1}{cos 2 x}, cos 2 x \neq = 0$
Cosine is zero at odd multiples of $\frac{π}{2}$ . Put the angle $2 x$ equal to those and divide by 2.
$2 x = \frac{π}{2} + k π ⟹ x = \frac{π}{4} + \frac{k π}{2}, k \in Z$
So the domain is all real $x$ except $\pm \frac{π}{4}, \pm \frac{3 π}{4}, \dots$
$Domain = {x \in R : x \neq = \frac{π}{4} + \frac{k π}{2}}$
For the range, $- 1 \leq cos 2 x \leq 1$ and $cos 2 x \neq = 0$ . A number between $- 1$ and $1$ (not 0) has a reciprocal of size at least 1.
$∣ sec 2 x ∣ \geq 1$
Hence the range excludes $- 1 < y < 1$ . The period of $sec b x$ is $\frac{2 π}{b}$ (the same as cosine).
$Range = (- \infty, - 1] \cup [1, \infty), period = \frac{2 π}{2} = π$

Domain: x \neq = \frac{π}{4} + \frac{k π}{2}; Range: ∣ y ∣ \geq 1; Period: π

9.2Periodicity of trigonometric functionsOften examined

We meet periodic phenomena all around us in nature, technology and society: the 24-hour day and night cycle, or the tides caused by the moon revolving round the earth. A periodic function is a function whose value repeats after a fixed interval. A periodic function is written as

f(x+p)=f(x)

where p is the period of the function. The sine wave, the triangular wave, the square wave and the saw tooth wave are periodic in nature.

Four periodic waves. In each, one pattern repeats again and again after a fixed interval.

All trigonometric functions repeat themselves at regular intervals, or periods. The values of a trigonometric function at \theta and at 2n\pi\pm\theta, where \theta\in\mathbb{R} and n\in\mathbb{Z}, are the same. This repeating behaviour is called periodicity.

Periodic function: A function f for which there is a number p>0 with f(p+x)=f(x) for every x in the domain.
Period: The number p of a periodic function f(x). The period of a trigonometric function is the smallest positive value that, when added to the angle, gives the same value of the function.

Periodicity of the sine function (9.2.1)

Derivation Period of \sin\theta

Aim. Show that the period of \sin\theta is 2\pi.

Suppose p is the period of the sine function. Then, for every real \theta,
\sin(\theta+p)=\sin\theta\qquad(\text{i})
Put \theta=0 in (i). The candidates for p are the angles whose sine is 0.
\sin p=0\;\Longrightarrow\;p=\sin^{-1}(0)=0,\pm\pi,\pm2\pi,\pm3\pi,\dots
Case 1: try p=\pi, the smallest positive candidate. Using \sin(\pi+\theta)=-\sin\theta, relation (i) would say -\sin\theta=\sin\theta, which is not true for every \theta (for \theta=\dfrac{\pi}{2} it says -1=1). So \pi is not the period of \sin\theta.
\sin(\theta+\pi)=-\sin\theta\neq\sin\theta
Case 2: try p=2\pi, the next positive candidate. Adding 2\pi is one complete turn, so the terminal side returns to the same position and the sine has the same value.
\sin(\theta+2\pi)=\sin\left(4\cdot\dfrac{\pi}{2}+\theta\right)=\sin\theta\qquad(\text{ii})
Relation (ii) is true for every \theta, and 2\pi is the smallest positive number for which it is true (the only smaller positive candidate, \pi, failed).
\text{smallest }p>0\text{ with }\sin(\theta+p)=\sin\theta\text{ is }2\pi

Result\text{Period of }\sin\theta=2\pi

The sine curve repeats its shape every 2\pi radians (360^\circ).

Periodicity of the tangent function (9.2.2)

Derivation Period of \tan\theta

Aim. Show that the period of \tan\theta is \pi.

Suppose p is the period of the tangent function. Then, for every real \theta where it is defined,
\tan(\theta+p)=\tan\theta\qquad(\text{i})
Put \theta=0 in (i). The candidates for p are the angles whose tangent is 0.
\tan p=0\;\Longrightarrow\;p=\tan^{-1}(0)=0,\pi,2\pi,3\pi,\dots
Try p=\pi, the smallest positive candidate. Adding \pi turns the terminal side to the opposite ray, so x and y both change sign and the ratio \dfrac yx is unchanged.
\tan(\theta+\pi)=\tan\left(2\cdot\dfrac{\pi}{2}+\theta\right)=\tan\theta\qquad(\text{ii})
Relation (ii) holds for every \theta, and \pi is the smallest positive candidate, so no smaller positive number works.
\pi\text{ is the least positive number with }\tan(\theta+p)=\tan\theta

Result\text{Period of }\tan\theta=\pi

Periods of the trigonometric functions.
Function	Period	Function	Period
f(x)=\sin x	2\pi	f(x)=\sin ax or \sin(ax+b)	\dfrac{2\pi}{\|a\|}
f(x)=\cos x	2\pi	f(x)=\cos ax or \cos(ax+b)	\dfrac{2\pi}{\|a\|}
f(x)=\tan x	\pi	f(x)=\tan ax or \tan(ax+b)	\dfrac{\pi}{\|a\|}
f(x)=\cot x	\pi	f(x)=\cot ax or \cot(ax+b)	\dfrac{\pi}{\|a\|}
f(x)=\sec x	2\pi	f(x)=\sec ax or \sec(ax+b)	\dfrac{2\pi}{\|a\|}
f(x)=\operatorname{cosec}x	2\pi	f(x)=\operatorname{cosec}ax or \operatorname{cosec}(ax+b)	\dfrac{2\pi}{\|a\|}

Example 2

Find the periods of (i) f(x)=\sin3x (ii) f(x)=\cos\dfrac{2x}{5} (iii) f(x)=\tan\dfrac{5x}{7}.

(i)

The period of \sin x is 2\pi. For \sin(ax+b) the period is \dfrac{2\pi}{|a|}. Here a=3.
\text{Period of }\sin3x=\dfrac{2\pi}{3}

(ii)

The period of \cos x is 2\pi, and here a=\dfrac25.
\text{Period of }\cos\dfrac{2x}{5}=\dfrac{2\pi}{\frac25}=2\pi\cdot\dfrac52=5\pi
Check: adding 5\pi to x adds 2\pi to the angle \dfrac{2x}{5}.
\cos\left(\dfrac{2x}{5}+2\pi\right)=\cos\dfrac25\left(x+5\pi\right)=\cos\dfrac{2x}{5}

(iii)

The period of \tan x is \pi, and here a=\dfrac57.
\text{Period of }\tan\dfrac{5x}{7}=\dfrac{\pi}{\frac57}=\pi\cdot\dfrac75=\dfrac{7\pi}{5}

Function	Period	Function	Period
f(x)=a\sin x+b	2\pi	f(x)=a\operatorname{cosec}x+b	2\pi
f(x)=a\cos x+b	2\pi	f(x)=a\sec x+b	2\pi
f(x)=a\tan x+b	\pi	f(x)=a\cot x+b	\pi

Example 3

Find the period of f(x)=\cot3x+\sin\dfrac{2x}{3}.

Find the period of each term separately. For \cot3x use \dfrac{\pi}{|a|} with a=3.
\text{Period of }\cot3x=\dfrac{\pi}{3}
For \sin\dfrac{2x}{3} use \dfrac{2\pi}{|a|} with a=\dfrac23.
\text{Period of }\sin\dfrac{2x}{3}=\dfrac{2\pi}{\frac23}=3\pi
The period of the sum is the smallest number that is a whole number of periods of both terms. With the periods written as fractions \dfrac{\pi}{3} and \dfrac{3\pi}{1}, use the rule of the book: the period of f(x) is the L.C.M. of the numerators divided by the H.C.F. of the denominators.
\text{Period of }f(x)=\dfrac{\text{L.C.M. of }\pi\text{ and }3\pi}{\text{H.C.F. of }3\text{ and }1}=\dfrac{3\pi}{1}=3\pi
Check: 3\pi is 9 periods of \cot3x and 1 period of \sin\dfrac{2x}{3}.
3\pi=9\cdot\dfrac{\pi}{3}=1\cdot3\pi

3\pi

Example 4

Find the period of f(x)=7\sin(3x+5).

The period of \sin x is 2\pi. For \sin(ax+b) the period is \dfrac{2\pi}{|a|}, with a=3 here.
\text{Period of }\sin(3x+5)=\dfrac{2\pi}{3}
The factor 7 changes the height of the wave, not its period.
\text{Period of }7\sin(3x+5)=\dfrac{2\pi}{3}

\dfrac{2\pi}{3}

Worked example: the period of a sum of two functions

Find the period of f(x)=\sin2x+\cos3x, and show that \pi is not a period.

Find the period of each term using period =\dfrac{2\pi}{b} for \sin bx and \cos bx.
\sin2x:\ \dfrac{2\pi}{2}=\pi,\qquad \cos3x:\ \dfrac{2\pi}{3}
The sum repeats only when both terms repeat together, so the period is the smallest positive number that is a whole number of periods of each. Write p=m\pi=n\cdot\dfrac{2\pi}{3}, so 3m=2n.
3m=2n\ \Longrightarrow\ m=2,\ n=3\ \text{(smallest whole numbers)}
Then p=2\pi. Verify that adding 2\pi to x leaves f unchanged.
f(x+2\pi)=\sin(2x+4\pi)+\cos(3x+6\pi)=\sin2x+\cos3x=f(x)
Test \pi (the period of the first term only). The second term changes sign, so f does not repeat.
f(x+\pi)=\sin(2x+2\pi)+\cos(3x+3\pi)=\sin2x-\cos3x\neq f(x)

\text{period}=2\pi

9.3Maximum and minimum values of trigonometric functionsOften examined

This section finds the greatest (maximum) and least (minimum) values of functions of the types

(i) a+b\sin\theta
(ii) a+b\cos\theta
(iii) a+b\sin(c\theta+d)
(iv) a+b\cos(c\theta+d)
and the reciprocals of the above, where a,b,c,d are real numbers.

Everything rests on one fact: \sin u and \cos u stay between -1 and 1 for every angle u, and they reach both values.

Derivation Maximum and minimum of a+b\sin\theta and a+b\cos\theta

Aim. Find the greatest and least values M and m of y=a+b\sin\theta and of y=a+b\cos\theta.

Start from the range of sine, -1\le\sin\theta\le1. Multiply by |b|, which is positive or zero, so the inequalities keep their direction.
-|b|\le|b|\sin\theta\le|b|
The term b\sin\theta takes exactly the same set of values as |b|\sin\theta (if b<0 the two only swap their roles), so it is largest when |b|\sin\theta is largest. That is when \sin\theta is at its maximum, \sin\theta=1.
M=a+|b|(1)=a+|b|\quad\text{when }\sin\theta=1
Likewise a+b\sin\theta is least when b\sin\theta is least, which is when \sin\theta is at its minimum, \sin\theta=-1.
m=a+|b|(-1)=a-|b|\quad\text{when }\sin\theta=-1
The cosine has the same range [-1,1], so the argument is the same with \cos\theta in place of \sin\theta: the maximum occurs when \cos\theta=1 and the minimum when \cos\theta=-1.
a+b\cos\theta:\quad M=a+|b|\ (\cos\theta=1),\qquad m=a-|b|\ (\cos\theta=-1)

ResultM=a+|b|,\qquad m=a-|b|

The graph of a+b\sin\theta or a+b\cos\theta lies between a-|b| and a+|b|.

Derivation Maximum and minimum of a+b\sin(c\theta+d) and a+b\cos(c\theta+d)

Aim. Find the greatest and least values of y=a+b\sin(c\theta+d) and of y=a+b\cos(c\theta+d).

Write u=c\theta+d. As \theta runs through the real numbers, so does u (if c\neq0), and \sin u takes every value from -1 to 1.
y=a+b\sin u,\qquad-1\le\sin u\le1
a+b\sin(c\theta+d) is greatest when b\sin(c\theta+d) is greatest, that is when \sin(c\theta+d) is greatest.
M=a+|b|(1)=a+|b|\quad\text{when }\sin(c\theta+d)=1
It is least when \sin(c\theta+d) is least.
m=a+|b|(-1)=a-|b|\quad\text{when }\sin(c\theta+d)=-1
The same reasoning for the cosine gives the maximum when \cos(c\theta+d)=1 and the minimum when \cos(c\theta+d)=-1.
a+b\cos(c\theta+d):\quad M=a+|b|,\qquad m=a-|b|

ResultM=a+|b|,\qquad m=a-|b|

The numbers c and d decide only where the maximum and minimum occur. They do not change their values.

Maximum and minimum values.
Function	Maximum M	Occurs when	Minimum m	Occurs when
a+b\sin\theta	a+\|b\|	\sin\theta=1	a-\|b\|	\sin\theta=-1
a+b\cos\theta	a+\|b\|	\cos\theta=1	a-\|b\|	\cos\theta=-1
a+b\sin(c\theta+d)	a+\|b\|	\sin(c\theta+d)=1	a-\|b\|	\sin(c\theta+d)=-1
a+b\cos(c\theta+d)	a+\|b\|	\cos(c\theta+d)=1	a-\|b\|	\cos(c\theta+d)=-1

Example 5

Find the maximum and minimum values of y=3+4\sin\theta.

Compare y=3+4\sin\theta with y=a+b\sin\theta.
a=3,\qquad b=4
Maximum value.
M=a+|b|=3+4=7
Minimum value.
m=a-|b|=3-4=-1

M=7,\quad m=-1

Example 6

Find the maximum and minimum values of y=\dfrac13-5\cos\theta.

Compare with y=a+b\cos\theta.
a=\dfrac13,\qquad b=-5
Maximum value. Use |b|=|-5|=5.
M=a+|b|=\dfrac13+|-5|=\dfrac13+5=\dfrac{16}{3}
Minimum value.
m=a-|b|=\dfrac13-|-5|=\dfrac13-5=-\dfrac{14}{3}

M=\dfrac{16}{3},\quad m=-\dfrac{14}{3}

Example 7

Find the maximum and minimum values of the following trigonometric functions: (i) y=1+2\sin\theta (ii) y=3+2\cos(3\theta-2) (iii) y=\dfrac{1}{1+3\sin(2\theta-15)}.

(i) y=1+2\sin\theta

Compare with y=a+b\sin\theta.
a=1,\qquad b=2
Maximum value.
M=a+|b|=1+2=3
Minimum value.
m=a-|b|=1-2=-1

(ii) y=3+2\cos(3\theta-2)

Compare with y=a+b\cos(c\theta+d).
a=3,\qquad b=2
Maximum value.
M=a+|b|=3+2=5
Minimum value.
m=a-|b|=3-2=1

(iii) y=\dfrac{1}{1+3\sin(2\theta-15)}

The corresponding function of the reciprocal type is the denominator. Call it y_1 and compare with y=a+b\sin(c\theta+d).
y_1=1+3\sin(2\theta-15),\qquad a=1,\ b=3
Maximum value of y_1.
M=a+|b|=1+3=4>0
Minimum value of y_1.
m=a-|b|=1-3=-2<0
The book now forms the reciprocals of these two numbers for the original function.
M'=\dfrac1M=\dfrac14,\qquad m'=\dfrac1m=\dfrac{1}{-2}=-\dfrac12

Worked example: greatest and least values of a reciprocal function

Find the maximum and minimum values of y=\dfrac{3}{5+4\cos(2\theta-1)}.

Let u=\cos(2\theta-1). Whatever \theta is, -1\le u\le1.
Find the range of the denominator D=5+4u.
1\le5+4u\le9
The denominator is always positive, so it is never zero and the function is defined for every \theta.
A reciprocal reverses the order: the largest y comes from the smallest denominator, the smallest y from the largest. Do not pair maximum with maximum.
y_{\max}=\dfrac{3}{1}=3\ \ (\cos(2\theta-1)=-1),\qquad y_{\min}=\dfrac{3}{9}=\dfrac13\ \ (\cos(2\theta-1)=1)

y_{\max}=3,\quad y_{\min}=\dfrac13

9.4Graphs of trigonometric functionsOften examined

All the trigonometric functions are periodic, so their graphs can be drawn over an interval whose length equals the period. Once the graph is drawn on one such interval (one cycle), it is repeated over every interval of the same length to complete the graph.

Procedure for sketching the graphs of sine and cosine functions

To graph y=a\sin bx or y=a\cos bx with b>0, follow these steps.

Steps

Find the period \dfrac{2\pi}{b}. Start from 0 on the x-axis and lay off a distance \dfrac{2\pi}{b}.
Divide the interval into four equal parts. (a) Find the midpoint by adding the end points and dividing the sum by 2. (b) Find the quarter points, that is the midpoints of the two halves obtained in (a). (c) Continue in the same way until the required number of equal parts is obtained.
Evaluate the function at each of the five, nine or thirteen x-values from Step 2. The resulting points are the maximum points, minimum points and x-intercepts.
Plot the points of Step 3 and join them with a sinusoidal curve of amplitude |a|.
Draw the graph over additional periods if required.

The angles used in the tables, in radians and degrees:

Angles in degrees and radians.
Radians	0	\dfrac{\pi}{6}	\dfrac{\pi}{3}	\dfrac{\pi}{2}	\dfrac{2\pi}{3}	\dfrac{5\pi}{6}	\pi	\dfrac{7\pi}{6}	\dfrac{4\pi}{3}	\dfrac{3\pi}{2}	\dfrac{5\pi}{3}	\dfrac{11\pi}{6}	2\pi
Degrees	0^\circ	30^\circ	60^\circ	90^\circ	120^\circ	150^\circ	180^\circ	210^\circ	240^\circ	270^\circ	300^\circ	330^\circ	360^\circ

Angles in degrees and radians.
Radians	Degrees
0	0^\circ
\dfrac{\pi}{6}	30^\circ
\dfrac{\pi}{3}	60^\circ
\dfrac{\pi}{2}	90^\circ
\dfrac{2\pi}{3}	120^\circ
\dfrac{5\pi}{6}	150^\circ
\pi	180^\circ
\dfrac{7\pi}{6}	210^\circ
\dfrac{4\pi}{3}	240^\circ
\dfrac{3\pi}{2}	270^\circ
\dfrac{5\pi}{3}	300^\circ
\dfrac{11\pi}{6}	330^\circ
2\pi	360^\circ

Graph of y=\sin x

Steps

For this function b=1, so the period is 2\pi and the function is graphed over [0,2\pi].
Divide [0,2\pi] into twelve equal parts to obtain the x-values.
0,\dfrac{\pi}{6},\dfrac{\pi}{3},\dfrac{\pi}{2},\dfrac{2\pi}{3},\dfrac{5\pi}{6},\pi,\dfrac{7\pi}{6},\dfrac{4\pi}{3},\dfrac{3\pi}{2},\dfrac{5\pi}{3},\dfrac{11\pi}{6},2\pi
Construct the table of values (below).
Plot the points.
(0,0),\left(\dfrac{\pi}{6},0.5\right),\left(\dfrac{\pi}{3},0.87\right),\left(\dfrac{\pi}{2},1\right),\left(\dfrac{2\pi}{3},0.87\right),\left(\dfrac{5\pi}{6},0.5\right),(\pi,0),\left(\dfrac{7\pi}{6},-0.5\right),\left(\dfrac{4\pi}{3},-0.87\right),\left(\dfrac{3\pi}{2},-1\right),\left(\dfrac{5\pi}{3},-0.87\right),\left(\dfrac{11\pi}{6},-0.5\right),(2\pi,0)
Join the points with a sinusoidal curve of amplitude 1.
Extend the graph by repeating the cycle, to the left from 0 to -2\pi and to the right beyond 2\pi.

Table of values of y=\sin x.
x	0	\dfrac{\pi}{6}	\dfrac{\pi}{3}	\dfrac{\pi}{2}	\dfrac{2\pi}{3}	\dfrac{5\pi}{6}	\pi	\dfrac{7\pi}{6}	\dfrac{4\pi}{3}	\dfrac{3\pi}{2}	\dfrac{5\pi}{3}	\dfrac{11\pi}{6}	2\pi
\sin x	0	0.5	0.87	1	0.87	0.5	0	-0.5	-0.87	-1	-0.87	-0.5	0

Table of values of y=\sin x.
x	\sin x
0	0
\dfrac{\pi}{6}	0.5
\dfrac{\pi}{3}	0.87
\dfrac{\pi}{2}	1
\dfrac{2\pi}{3}	0.87
\dfrac{5\pi}{6}	0.5
\pi	0
\dfrac{7\pi}{6}	-0.5
\dfrac{4\pi}{3}	-0.87
\dfrac{3\pi}{2}	-1
\dfrac{5\pi}{3}	-0.87
\dfrac{11\pi}{6}	-0.5
2\pi	0

Graph of y=\sin x. The dots are the points of the table, which make one cycle [0,2\pi] (the cycle). The dashed lines are y=1 and y=-1. The curve repeats every 2\pi on both sides.

Property	Value
Domain	(-\infty,\infty)=\mathbb{R}
Range	[-1,1]
Period	2\pi
Amplitude	1
Nature	odd function

The graph in the interval [0,2\pi] is called a cycle. Since the period of the sine function is 2\pi, the sine graph can be extended on both sides of the x-axis through every interval of 2\pi.

Graph of y=\cos x

Steps

For this function b=1, so the period is 2\pi and the function is graphed over [0,2\pi].
Divide [0,2\pi] into twelve equal parts: the same x-values as for the sine graph.
0,\dfrac{\pi}{6},\dfrac{\pi}{3},\dfrac{\pi}{2},\dfrac{2\pi}{3},\dfrac{5\pi}{6},\pi,\dfrac{7\pi}{6},\dfrac{4\pi}{3},\dfrac{3\pi}{2},\dfrac{5\pi}{3},\dfrac{11\pi}{6},2\pi
Construct the table of values (below).
Plot the points.
(0,1),\left(\dfrac{\pi}{6},0.87\right),\left(\dfrac{\pi}{3},0.5\right),\left(\dfrac{\pi}{2},0\right),\left(\dfrac{2\pi}{3},-0.5\right),\left(\dfrac{5\pi}{6},-0.87\right),(\pi,-1),\left(\dfrac{7\pi}{6},-0.87\right),\left(\dfrac{4\pi}{3},-0.5\right),\left(\dfrac{3\pi}{2},0\right),\left(\dfrac{5\pi}{3},0.5\right),\left(\dfrac{11\pi}{6},0.87\right),(2\pi,1)
Join the points with a sinusoidal curve of amplitude 1.
Extend the graph by repeating the cycle, from 0 to -2\pi and beyond.

Table of values of y=\cos x.
x	0	\dfrac{\pi}{6}	\dfrac{\pi}{3}	\dfrac{\pi}{2}	\dfrac{2\pi}{3}	\dfrac{5\pi}{6}	\pi	\dfrac{7\pi}{6}	\dfrac{4\pi}{3}	\dfrac{3\pi}{2}	\dfrac{5\pi}{3}	\dfrac{11\pi}{6}	2\pi
\cos x	1	0.87	0.5	0	-0.5	-0.87	-1	-0.87	-0.5	0	0.5	0.87	1

Table of values of y=\cos x.
x	\cos x
0	1
\dfrac{\pi}{6}	0.87
\dfrac{\pi}{3}	0.5
\dfrac{\pi}{2}	0
\dfrac{2\pi}{3}	-0.5
\dfrac{5\pi}{6}	-0.87
\pi	-1
\dfrac{7\pi}{6}	-0.87
\dfrac{4\pi}{3}	-0.5
\dfrac{3\pi}{2}	0
\dfrac{5\pi}{3}	0.5
\dfrac{11\pi}{6}	0.87
2\pi	1

Graph of y=\cos x. The dots make one cycle [0,2\pi]. The cosine curve is the sine curve moved \dfrac{\pi}{2} to the left, and it is symmetric about the y-axis.

Property	Value
Domain	(-\infty,\infty)=\mathbb{R}
Range	[-1,1]
Period	2\pi
Amplitude	1
Nature	even function

The graph in the interval [0,2\pi] is a cycle. Since the period of the cosine function is 2\pi, the cosine graph can be extended on both sides of the x-axis through every interval of 2\pi.

Graph of y=\tan x

Steps

For this function b=1, so the period is \pi. The graph is first drawn over [-\pi,\pi], that is over two periods.
Take the x-values from 0 to \pm\pi in equal parts of \dfrac{\pi}{6}.
0,\pm\dfrac{\pi}{6},\pm\dfrac{\pi}{3},\pm\dfrac{\pi}{2},\pm\dfrac{2\pi}{3},\pm\dfrac{5\pi}{6},\pm\pi
Construct the table of values (below).
Plot the points.
(-\pi,0),\left(-\dfrac{5\pi}{6},0.58\right),\left(-\dfrac{2\pi}{3},1.73\right),\left(-\dfrac{\pi}{3},-1.73\right),\left(-\dfrac{\pi}{6},-0.58\right),(0,0),\left(\dfrac{\pi}{6},0.58\right),\left(\dfrac{\pi}{3},1.73\right),\left(\dfrac{2\pi}{3},-1.73\right),\left(\dfrac{5\pi}{6},-0.58\right),(\pi,0)
Join the points by a free hand curve. At x=\pm\dfrac{\pi}{2} the function is not defined: the curve rises (or falls) without bound towards the vertical lines x=\pm\dfrac{\pi}{2}, which are asymptotes.
Since tangent has period \pi, the graph in [-\pi,\pi] is repeated in the same way over every further interval of length \pi, for example from -2\pi to 2\pi.

Table of values of y=\tan x.
x	-\pi	-\dfrac{5\pi}{6}	-\dfrac{2\pi}{3}	-\dfrac{\pi}{2}	-\dfrac{\pi}{3}	-\dfrac{\pi}{6}	0	\dfrac{\pi}{6}	\dfrac{\pi}{3}	\dfrac{\pi}{2}	\dfrac{2\pi}{3}	\dfrac{5\pi}{6}	\pi
\tan x	0	0.58	1.73	undefined	-1.73	-0.58	0	0.58	1.73	undefined	-1.73	-0.58	0

Table of values of y=\tan x.
x	\tan x
-\pi	0
-\dfrac{5\pi}{6}	0.58
-\dfrac{2\pi}{3}	1.73
-\dfrac{\pi}{2}	undefined
-\dfrac{\pi}{3}	-1.73
-\dfrac{\pi}{6}	-0.58
0	0
\dfrac{\pi}{6}	0.58
\dfrac{\pi}{3}	1.73
\dfrac{\pi}{2}	undefined
\dfrac{2\pi}{3}	-1.73
\dfrac{5\pi}{6}	-0.58
\pi	0

Graph of y=\tan x. The dots are the table points on [-\pi,\pi]. The dashed vertical lines are the asymptotes x=\pm\dfrac{\pi}{2},\pm\dfrac{3\pi}{2},\dots; between two of them the curve takes every real value.

Property	Value
Domain	\{x: x\in\mathbb{R}\text{ and }x\neq(2n+1)\frac{\pi}{2},\ n\text{ an integer}\}
Range	(-\infty,\infty)=\mathbb{R}
Period	\pi
Amplitude	nil (there is no greatest or least value)
Nature	odd function

The graph in the interval [0,\pi] is called a cycle. Since the period of the tangent function is \pi, the tangent graph can be extended on both sides of the x-axis through every interval of \pi.

Graph of y=3\sin2x

Steps

\sin x has period 2\pi. For y=3\sin2x, b=2, so the period is \dfrac{2\pi}{2}=\pi and the function is graphed over [0,\pi].
Divide [0,\pi] into eight equal parts to obtain the x-values.
0,\dfrac{\pi}{6},\dfrac{\pi}{4},\dfrac{\pi}{3},\dfrac{\pi}{2},\dfrac{2\pi}{3},\dfrac{3\pi}{4},\dfrac{5\pi}{6},\pi
Construct the table of values (below).
Plot the points and join them with a sinusoidal curve of amplitude 3.
(0,0),\left(\dfrac{\pi}{6},2.60\right),\left(\dfrac{\pi}{4},3\right),\left(\dfrac{\pi}{3},2.60\right),\left(\dfrac{\pi}{2},0\right),\left(\dfrac{2\pi}{3},-2.60\right),\left(\dfrac{3\pi}{4},-3\right),\left(\dfrac{5\pi}{6},-2.60\right),(\pi,0)
Extend the graph by repeating the cycle, for example over the interval -2\pi to 2\pi.

Table of values of y=3\sin2x (the book rounds \frac{3\sqrt3}{2}=2.598 to 2.61).
x	\dfrac{\pi}{6}	\dfrac{\pi}{4}	\dfrac{\pi}{3}	\dfrac{\pi}{2}	\dfrac{2\pi}{3}	\dfrac{3\pi}{4}	\dfrac{5\pi}{6}	\pi
2x	\dfrac{\pi}{3}	\dfrac{\pi}{2}	\dfrac{2\pi}{3}	\pi	\dfrac{4\pi}{3}	\dfrac{3\pi}{2}	\dfrac{5\pi}{3}	2\pi
\sin2x	\dfrac{\sqrt3}{2}	1	\dfrac{\sqrt3}{2}	0	-\dfrac{\sqrt3}{2}	-1	-\dfrac{\sqrt3}{2}	0
3\sin2x	2.60	3	2.60	0	-2.60	-3	-2.60	0

Table of values of y=3\sin2x (the book rounds \frac{3\sqrt3}{2}=2.598 to 2.61).
x	2x	\sin2x	3\sin2x
0	0	0	0
\dfrac{\pi}{6}	\dfrac{\pi}{3}	\dfrac{\sqrt3}{2}	2.60
\dfrac{\pi}{4}	\dfrac{\pi}{2}	1	3
\dfrac{\pi}{3}	\dfrac{2\pi}{3}	\dfrac{\sqrt3}{2}	2.60
\dfrac{\pi}{2}	\pi	0	0
\dfrac{2\pi}{3}	\dfrac{4\pi}{3}	-\dfrac{\sqrt3}{2}	-2.60
\dfrac{3\pi}{4}	\dfrac{3\pi}{2}	-1	-3
\dfrac{5\pi}{6}	\dfrac{5\pi}{3}	-\dfrac{\sqrt3}{2}	-2.60
\pi	2\pi	0	0

Graph of y=3\sin2x. The dots make one cycle [0,\pi]. The amplitude is 3 and the period is \pi, so the wave repeats twice as often as \sin x.

Property	Value
Domain	(-\infty,\infty)=\mathbb{R}
Range	[-3,3]
Period	\pi
Amplitude	3
Nature	odd function

Graph of y=3\cos2x

Steps

\cos x has period 2\pi. For y=3\cos2x, b=2, so the period is \dfrac{2\pi}{2}=\pi and the function is graphed over [0,\pi].
Divide [0,\pi] into eight equal parts to obtain the x-values.
0,\dfrac{\pi}{6},\dfrac{\pi}{4},\dfrac{\pi}{3},\dfrac{\pi}{2},\dfrac{2\pi}{3},\dfrac{3\pi}{4},\dfrac{5\pi}{6},\pi
Construct the table of values (below).
Plot the points and join them with a sinusoidal curve of amplitude 3.
(0,3),\left(\dfrac{\pi}{6},1.5\right),\left(\dfrac{\pi}{4},0\right),\left(\dfrac{\pi}{3},-1.5\right),\left(\dfrac{\pi}{2},-3\right),\left(\dfrac{2\pi}{3},-1.5\right),\left(\dfrac{3\pi}{4},0\right),\left(\dfrac{5\pi}{6},1.5\right),(\pi,3)
The graph can be extended by repeating the cycle.

Table of values of y=3\cos2x.
x	0	\dfrac{\pi}{6}	\dfrac{\pi}{4}	\dfrac{\pi}{3}	\dfrac{\pi}{2}	\dfrac{2\pi}{3}	\dfrac{3\pi}{4}	\dfrac{5\pi}{6}	\pi
2x	0	\dfrac{\pi}{3}	\dfrac{\pi}{2}	\dfrac{2\pi}{3}	\pi	\dfrac{4\pi}{3}	\dfrac{3\pi}{2}	\dfrac{5\pi}{3}	2\pi
\cos2x	1	\dfrac12	0	-\dfrac12	-1	-\dfrac12	0	\dfrac12	1
3\cos2x	3	1.5	0	-1.5	-3	-1.5	0	1.5	3

Table of values of y=3\cos2x.
x	2x	\cos2x	3\cos2x
0	0	1	3
\dfrac{\pi}{6}	\dfrac{\pi}{3}	\dfrac12	1.5
\dfrac{\pi}{4}	\dfrac{\pi}{2}	0	0
\dfrac{\pi}{3}	\dfrac{2\pi}{3}	-\dfrac12	-1.5
\dfrac{\pi}{2}	\pi	-1	-3
\dfrac{2\pi}{3}	\dfrac{4\pi}{3}	-\dfrac12	-1.5
\dfrac{3\pi}{4}	\dfrac{3\pi}{2}	0	0
\dfrac{5\pi}{6}	\dfrac{5\pi}{3}	\dfrac12	1.5
\pi	2\pi	1	3

Graph of y=3\cos2x. The dots make one cycle [0,\pi]. The amplitude is 3 and the period is \pi.

Property	Value
Domain	(-\infty,\infty)=\mathbb{R}
Range	[-3,3]
Period	\pi
Amplitude	3
Nature	even function

Graph of y=\operatorname{cosec}x

Steps

The cosecant is the reciprocal of the sine, which is periodic of period 2\pi. So cosecant is also periodic of period 2\pi.
For this function b=1, so the period is 2\pi. As in the book, first draw the graph from 0 to \pm\pi.
Take the x-values 0,\pm\dfrac{\pi}{6},\pm\dfrac{\pi}{3},\pm\dfrac{\pi}{2},\pm\dfrac{2\pi}{3},\pm\dfrac{5\pi}{6},\pm\pi and construct the table (below).
Plot the points.
\left(-\dfrac{5\pi}{6},-2\right),\left(-\dfrac{2\pi}{3},-1.15\right),\left(-\dfrac{\pi}{2},-1\right),\left(-\dfrac{\pi}{3},-1.15\right),\left(-\dfrac{\pi}{6},-2\right),\left(\dfrac{\pi}{6},2\right),\left(\dfrac{\pi}{3},1.15\right),\left(\dfrac{\pi}{2},1\right),\left(\dfrac{2\pi}{3},1.15\right),\left(\dfrac{5\pi}{6},2\right)
Join these points by a free hand curve. Where \sin x=0, that is at x=0,\pm\pi, the cosecant is not defined and the curve goes off along the vertical lines (asymptotes).
Extend the graph by repeating the cycle, for example to the interval -2\pi to 2\pi.

Table of values of y=\operatorname{cosec}x.
x	-\pi	-\dfrac{5\pi}{6}	-\dfrac{2\pi}{3}	-\dfrac{\pi}{2}	-\dfrac{\pi}{3}	-\dfrac{\pi}{6}	0	\dfrac{\pi}{6}	\dfrac{\pi}{3}	\dfrac{\pi}{2}	\dfrac{2\pi}{3}	\dfrac{5\pi}{6}	\pi
\operatorname{cosec}x	undefined	-2	-1.15	-1	-1.15	-2	undefined	2	1.15	1	1.15	2	undefined

Table of values of y=\operatorname{cosec}x.
x	\operatorname{cosec}x
-\pi	undefined
-\dfrac{5\pi}{6}	-2
-\dfrac{2\pi}{3}	-1.15
-\dfrac{\pi}{2}	-1
-\dfrac{\pi}{3}	-1.15
-\dfrac{\pi}{6}	-2
0	undefined
\dfrac{\pi}{6}	2
\dfrac{\pi}{3}	1.15
\dfrac{\pi}{2}	1
\dfrac{2\pi}{3}	1.15
\dfrac{5\pi}{6}	2
\pi	undefined

Graph of y=\operatorname{cosec}x. The curve never enters the band -1<y<1. The asymptotes are the lines x=n\pi (the y-axis is one of them). The dots are the table points on [-\pi,\pi].

Property	Value
Domain	\{x: x\in\mathbb{R}\text{ and }x\neq n\pi,\ n\text{ an integer}\}
Range	(-\infty,-1]\cup[1,\infty), that is \|y\|\ge1
Period	2\pi
Amplitude	nil
Nature	odd function

The graph in the interval [0,2\pi] is called a cycle. Since the period of the cosecant function is 2\pi, the cosecant graph can be extended on both sides of the x-axis through every interval of 2\pi.

Graph of y=\sec x

Steps

The secant is the reciprocal of the cosine, which is periodic of period 2\pi. So secant is also periodic of period 2\pi.
For this function b=1, so the period is 2\pi. As in the book, first draw the graph from 0 to \pm\pi.
Take the x-values 0,\pm\dfrac{\pi}{6},\pm\dfrac{\pi}{3},\pm\dfrac{\pi}{2},\pm\dfrac{2\pi}{3},\pm\dfrac{5\pi}{6},\pm\pi and construct the table (below).
Plot the points.
(-\pi,-1),\left(-\dfrac{5\pi}{6},-1.15\right),\left(-\dfrac{2\pi}{3},-2\right),\left(-\dfrac{\pi}{3},2\right),\left(-\dfrac{\pi}{6},1.15\right),(0,1),\left(\dfrac{\pi}{6},1.15\right),\left(\dfrac{\pi}{3},2\right),\left(\dfrac{2\pi}{3},-2\right),\left(\dfrac{5\pi}{6},-1.15\right),(\pi,-1)
Join these points by a free hand curve. At x=\pm\dfrac{\pi}{2} the cosine is 0, so the secant is not defined and the curve goes off along the vertical lines (asymptotes).
Extend the graph by repeating the cycle, for example to the interval -2\pi to 2\pi.

Table of values of y=\sec x.
x	-\pi	-\dfrac{5\pi}{6}	-\dfrac{2\pi}{3}	-\dfrac{\pi}{2}	-\dfrac{\pi}{3}	-\dfrac{\pi}{6}	0	\dfrac{\pi}{6}	\dfrac{\pi}{3}	\dfrac{\pi}{2}	\dfrac{2\pi}{3}	\dfrac{5\pi}{6}	\pi
\sec x	-1	-1.15	-2	undefined	2	1.15	1	1.15	2	undefined	-2	-1.15	-1

Table of values of y=\sec x.
x	\sec x
-\pi	-1
-\dfrac{5\pi}{6}	-1.15
-\dfrac{2\pi}{3}	-2
-\dfrac{\pi}{2}	undefined
-\dfrac{\pi}{3}	2
-\dfrac{\pi}{6}	1.15
0	1
\dfrac{\pi}{6}	1.15
\dfrac{\pi}{3}	2
\dfrac{\pi}{2}	undefined
\dfrac{2\pi}{3}	-2
\dfrac{5\pi}{6}	-1.15
\pi	-1

Graph of y=\sec x. The curve never enters the band -1<y<1 and touches y=1 and y=-1. The asymptotes are the lines x=(2n+1)\dfrac{\pi}{2}. The dots are the table points on [-\pi,\pi].

Property	Value
Domain	\{x: x\in\mathbb{R}\text{ and }x\neq\frac{\pi}{2}+n\pi,\ n\text{ an integer}\}
Range	(-\infty,-1]\cup[1,\infty), that is \|y\|\ge1
Period	2\pi
Amplitude	nil
Nature	even function

Graph of y=\cot x

Steps

\cot x is periodic of period \pi. For this function b=1, so the period is \pi and the function is graphed over [0,\pi].
Take the x-values 0,\pm\dfrac{\pi}{6},\pm\dfrac{\pi}{3},\pm\dfrac{\pi}{2},\pm\dfrac{2\pi}{3},\pm\dfrac{5\pi}{6},\pm\pi and construct the table (below).
Plot the points.
\left(-\dfrac{5\pi}{6},1.73\right),\left(-\dfrac{2\pi}{3},0.58\right),\left(-\dfrac{\pi}{2},0\right),\left(-\dfrac{\pi}{3},-0.58\right),\left(-\dfrac{\pi}{6},-1.73\right),\left(\dfrac{\pi}{6},1.73\right),\left(\dfrac{\pi}{3},0.58\right),\left(\dfrac{\pi}{2},0\right),\left(\dfrac{2\pi}{3},-0.58\right),\left(\dfrac{5\pi}{6},-1.73\right)
Join these points by a free hand curve. At x=0,\pm\pi the sine is 0, so the cotangent is not defined and the curve goes off along the vertical lines (asymptotes).
Since cotangent has period \pi, first draw the graph on [-\pi,\pi]. In the same way draw the graph for the interval from -2\pi to 2\pi.

Table of values of y=\cot x (corrected, see the note below).
x	-\pi	-\dfrac{5\pi}{6}	-\dfrac{2\pi}{3}	-\dfrac{\pi}{2}	-\dfrac{\pi}{3}	-\dfrac{\pi}{6}	0	\dfrac{\pi}{6}	\dfrac{\pi}{3}	\dfrac{\pi}{2}	\dfrac{2\pi}{3}	\dfrac{5\pi}{6}	\pi
\cot x	undefined	1.73	0.58	0	-0.58	-1.73	undefined	1.73	0.58	0	-0.58	-1.73	undefined

Table of values of y=\cot x (corrected, see the note below).
x	\cot x
-\pi	undefined
-\dfrac{5\pi}{6}	1.73
-\dfrac{2\pi}{3}	0.58
-\dfrac{\pi}{2}	0
-\dfrac{\pi}{3}	-0.58
-\dfrac{\pi}{6}	-1.73
0	undefined
\dfrac{\pi}{6}	1.73
\dfrac{\pi}{3}	0.58
\dfrac{\pi}{2}	0
\dfrac{2\pi}{3}	-0.58
\dfrac{5\pi}{6}	-1.73
\pi	undefined

Graph of y=\cot x. The asymptotes are the lines x=n\pi (the y-axis is one of them). The curve falls from left to right in every piece and crosses the x-axis at x=(2n+1)\dfrac{\pi}{2}. The dots are the table points on [-\pi,\pi].

Property	Value
Domain	\{x: x\in\mathbb{R}\text{ and }x\neq n\pi,\ n\text{ an integer}\}
Range	(-\infty,\infty)=\mathbb{R}
Period	\pi
Amplitude	nil
Nature	odd function

The graph in the interval [0,\pi] is called a cycle. Since the period of the cotangent function is \pi, the cotangent graph can be extended on both sides of the x-axis through every interval of \pi.

Worked example: graph of y=-4\sin\dfrac x2

Find the amplitude and period of y=-4\sin\dfrac{x}{2}, make a table of five points and sketch one cycle.

Read the amplitude from a in y=a\sin bx: it is |a|. The minus sign does not change the amplitude.
\text{amplitude}=|-4|=4
Find the period from b=\dfrac12.
\text{period}=\dfrac{2\pi}{b}=\dfrac{2\pi}{1/2}=4\pi
Divide one cycle [0,4\pi] into four equal parts, each of length \pi. Evaluate \sin\dfrac x2 at the five points and multiply by -4.

x	\pi	2\pi	3\pi	4\pi
\dfrac x2	\dfrac\pi2	\pi	\dfrac{3\pi}2	2\pi
\sin\dfrac x2	1	0	-1	0
y=-4\sin\dfrac x2	-4	0	4	0

x	\dfrac x2	\sin\dfrac x2	y=-4\sin\dfrac x2
0	0	0	0
\pi	\dfrac\pi2	1	-4
2\pi	\pi	0	0
3\pi	\dfrac{3\pi}2	-1	4
4\pi	2\pi	0	0

One full period of y=-4\sin\dfrac x2 on [0,4\pi]. The minus sign turns the sine wave upside down: it falls to -4 first, then rises to 4.

Plot the points and join them with a smooth wave. Compared with y=4\sin\frac x2 the curve is reflected in the x-axis.

\text{amplitude }4,\ \text{period }4\pi,\ \text{range }[-4,4]

9.5Even and odd trigonometric functions

All functions, including the trigonometric functions, can be sorted as even, odd or neither.

Odd function: f is odd if and only if f(-x)=-f(x) for every x in the domain. Its graph is symmetric about the origin.
Even function: f is even if and only if f(-x)=f(x) for every x in the domain. Its graph is symmetric about the y-axis.

O means odd and E means even.
O or E	Identity	Meaning
O	\sin(-x)=-\sin x	sine is odd
E	\cos(-x)=\cos x	cosine is even
O	\tan(-x)=-\tan x	tangent is odd
O	\cot(-x)=-\cot x	cotangent is odd
E	\sec(-x)=\sec x	secant is even
O	\operatorname{cosec}(-x)=-\operatorname{cosec}x	cosecant is odd

Example 8

Check whether the function f(x)=x^3\sin x is odd or even.

Replace x by -x.
f(-x)=(-x)^3\sin(-x)
Use (-x)^3=-x^3 and \sin(-x)=-\sin x.
f(-x)=(-x^3)(-\sin x)
The two minus signs cancel.
f(-x)=x^3\sin x=f(x)
Since f(-x)=f(x), the function is even. (Both factors are odd, and odd \times odd is even.)
\therefore f\text{ is an even function}

\text{even}

Example 9

Check whether the following functions are odd or even: (i) y=3\sin x+4\cos x (ii) y=\dfrac{\tan x}{x-\sin x}.

(i)

Replace x by -x.
f(-x)=3\sin(-x)+4\cos(-x)
Use \sin(-x)=-\sin x and \cos(-x)=\cos x.
f(-x)=-3\sin x+4\cos x=-(3\sin x-4\cos x)
This is not f(x)=3\sin x+4\cos x, and it is not -f(x)=-3\sin x-4\cos x either. (Check with numbers: f\left(\frac{\pi}{2}\right)=3 but f\left(-\frac{\pi}{2}\right)=-3, so f is not even; f(0)=4 but -f(0)=-4, so f is not odd.)
f(-x)\neq f(x)\quad\text{and}\quad f(-x)\neq-f(x)

(ii)

Replace x by -x.
f(-x)=\dfrac{\tan(-x)}{(-x)-\sin(-x)}
Use \tan(-x)=-\tan x and \sin(-x)=-\sin x.
f(-x)=\dfrac{-\tan x}{-x+\sin x}
Take out -1 from the denominator; the two minus signs cancel.
f(-x)=\dfrac{-\tan x}{-(x-\sin x)}=\dfrac{\tan x}{x-\sin x}=f(x)
Since f(-x)=f(x), the function is even.
\therefore f\text{ is an even function}

\text{(i) neither even nor odd; (ii) even}

9.6Application

Many real situations repeat in a regular way, and a sine or cosine function can describe them. This section applies the graphs of the unit to such situations.

Example 10

A Ferris wheel has a diameter of 30\,\mathrm{m} with a centre 17\,\mathrm{m} above the ground. It makes one complete revolution every 60 seconds. (1) Graph one complete period of the Ferris wheel that models the height in relation to time. Assume a rider starts at the lowest point. (2) Find the equation of the graph using the cosine function. (3) What is the height of the rider at 45 seconds? (4) At what time or times is the rider at a height of 15\,\mathrm{m}?

(1) Graph one complete period

The Ferris wheel of Example 10: diameter 30 m, centre 17 m above the ground, so the lowest point is 2 m and the highest point is 32 m above the ground. The rider starts at the lowest point.

The diameter is 30\,\mathrm{m}, so the radius is 15\,\mathrm{m}. The centre is 17\,\mathrm{m} above the ground, so the height of the rider varies about the midline h=17.
\text{lowest}=17-15=2\,\mathrm{m},\qquad\text{highest}=17+15=32\,\mathrm{m}
One revolution takes 60\,\mathrm{s}, so the period is 60\,\mathrm{s}. The rider starts at the lowest point, so the graph starts at its minimum, rises to the midline after 15\,\mathrm{s}, reaches the maximum after 30\,\mathrm{s}, is back at the midline after 45\,\mathrm{s} and returns to the minimum after 60\,\mathrm{s}.
(0,2),\ (15,17),\ (30,32),\ (45,17),\ (60,2)

Height of the rider against time for one revolution. The dashed line is the midline h=17. The book's sketch is only approximate: the true start is 2\,\mathrm{m} and the true top is 32\,\mathrm{m}.

(2) The equation using the cosine function

Use the model y=A\cos(Bx-D)+C. With no phase shift and the variable t, it reads
h(t)=A\cos(Bt)+C
The amplitude is |A|=15. The rider starts at the lowest point, where an ordinary cosine would be at its highest, so the cosine is turned upside down: A=-15.
|A|=15\;\Longrightarrow\;A=-15
The period is \dfrac{2\pi}{B}=60.
60=\dfrac{2\pi}{B}\;\Longrightarrow\;B=\dfrac{\pi}{30}
The midline is C=17. Substitute A, B and C.
h(t)=-15\cos\left(\dfrac{\pi}{30}t\right)+17=-15\cos\left(\dfrac{\pi t}{30}\right)+17

(3) Height after 45 seconds

Put t=45 in the equation.
h(45)=-15\cos\left(\dfrac{45\pi}{30}\right)+17=-15\cos\left(\dfrac{3\pi}{2}\right)+17
\cos\dfrac{3\pi}{2}=0.
h(45)=-15(0)+17=17\,\mathrm{m}

(4) Times when the height is 15 m

Put h(t)=15.
15=-15\cos\left(\dfrac{\pi t}{30}\right)+17
Subtract 17.
-2=-15\cos\left(\dfrac{\pi t}{30}\right)
Divide by -15.
\cos\left(\dfrac{\pi t}{30}\right)=\dfrac{2}{15}
Take the inverse cosine. The calculator (in radians) gives the first angle \theta_1=1.437.
\dfrac{\pi t}{30}=\cos^{-1}\left(\dfrac{2}{15}\right)=1.437
The cosine is positive in the first and fourth quadrants, so the second angle in one turn is \theta_2=2\pi-1.437=4.846. In one revolution (0\le t\le60, that is 0\le\theta\le2\pi) there are only these two.
\theta_1=1.437,\qquad\theta_2=2\pi-1.437=4.846
Solve for t with t=\dfrac{30}{\pi}\theta.
t_1=\dfrac{30}{\pi}(1.437)=13.7\,\mathrm{s},\qquad t_2=\dfrac{30}{\pi}(4.846)=46.3\,\mathrm{s}

h(t)=-15\cos\left(\dfrac{\pi t}{30}\right)+17;\quad h(45)=17\,\mathrm{m};\quad t\approx13.7\,\mathrm{s}\text{ and }46.3\,\mathrm{s}

Transforming sine and cosine

We can transform the sine and cosine to fit a specific scenario using

y=a\,f\big[b(x-c)\big]+d

The four constants of a sinusoidal model (book's Key Facts).
Constant	Meaning	How to read it
a	vertical stretch, or amplitude	a=\dfrac{\text{Maximum}-\text{Minimum}}{2}
b	related to the horizontal stretch (\frac1b) and to the period	\text{period}=\dfrac1b\times2\pi or \text{period}=\dfrac1b\times360^\circ
c	horizontal translation, or phase shift	the graph moves c units to the right
d	vertical translation, which is the midline	d=\dfrac{\text{Maximum}+\text{Minimum}}{2}

Example 11

The most used current is alternating current and it reverses direction in a cyclic fashion. In Canada the standard frequency is 60\,\mathrm{Hz} (60 cycles per second). The maximum voltage is about 170\,\mathrm{V}. The voltage can be modelled by V(t)=a\sin[k(t-d)]+c. (a) What is the period of 60\,\mathrm{Hz} AC? (b) Determine the value of k. (c) What is the amplitude of the voltage function? (d) Model the voltage with a suitably transformed sine function.

(a)

There are 60 cycles in one second, so each cycle takes
\text{period}=\dfrac{1}{60}\,\mathrm{s}

(b)

k is the number of degrees turned per second, since one cycle is 360^\circ.
k=\dfrac{360^\circ}{\text{period}}=\dfrac{360^\circ}{\frac{1}{60}}=21\,600

(c)

The maximum is 170 and, by symmetry, the minimum is -170.
\text{Maximum}=170,\qquad\text{Minimum}=-170
Use amplitude =\dfrac{\text{Maximum}-\text{Minimum}}{2}.
\text{amplitude}=\dfrac{170-(-170)}{2}=170

(d)

The voltage oscillates about 0 with no shift: a=170, k=21\,600, c=0 and d=0.
V(t)=a\sin[k(t-d)]+c
Substitute.
V(t)=170\sin(21\,600\,t)

V(t)=170\sin(21\,600\,t)

The unit-opening problem (page 216)

A Ferris wheel is 60\,\mathrm{ft} in diameter. It makes one revolution every 100 seconds. Yasir climbs up 6\,\mathrm{ft} of stairs to get on the wheel at its lowest point. Model the height of a rider as a sinusoidal function and graph one revolution. (The book gives the answer h(t)=-30\cos3.6t+36 in the figure.)

The diameter is 60\,\mathrm{ft}, so the radius is 30\,\mathrm{ft}. Yasir climbs 6\,\mathrm{ft} of stairs to the lowest point, so the wheel's lowest point is 6\,\mathrm{ft} above the ground and the centre is 30+6=36\,\mathrm{ft} above the ground.
\text{lowest}=6\,\mathrm{ft},\quad\text{centre}=36\,\mathrm{ft},\quad\text{highest}=66\,\mathrm{ft}
The amplitude is the radius and the midline is the height of the centre. The rider starts at the lowest point, so the cosine is upside down: a=-30, d=36.
a=-30,\qquad d=36
One revolution takes 100\,\mathrm{s}, so the angle turned per second is \dfrac{360^\circ}{100}=3.6^\circ (that is \dfrac{\pi}{50} radian).
k=3.6^\circ\text{ per second}
Put the constants into the model.
h(t)=-30\cos(3.6\,t)+36

Height of the rider against time for one revolution: from 6\,\mathrm{ft} up to 66\,\mathrm{ft} and back in 100\,\mathrm{s}.

h(t)=-30\cos(3.6\,t)+36

Worked example: a Ferris wheel model with a cosine

A Ferris wheel has radius 8\,\mathrm{m} and its lowest point is 3\,\mathrm{m} above the ground. It makes one revolution every 24\,\mathrm{s} and a rider starts at the lowest point. (1) Find h(t). (2) Find the height at t=10\,\mathrm{s}. (3) When is the rider first at 15\,\mathrm{m}?

(1) The equation

Use h=a\cos[b(t-c)]+d. The midline is halfway between the lowest and highest points: the centre is 3+8=11\,\mathrm{m} up, so d=11.
d=3+8=11
The rider starts at a minimum, and \cos0=1 is a maximum, so the cosine must be turned upside down: a=-8 (the amplitude is the radius). There is no horizontal shift, c=0.
a=-8,\quad c=0
The period is 24\,\mathrm{s}, so b=\dfrac{2\pi}{24}=\dfrac\pi{12}.
h(t)=11-8\cos\dfrac{\pi t}{12}
Check: at t=0, h=11-8=3 (lowest point) and at t=12, h=11+8=19 (top).

Height of the rider, h(t)=11-8\cos\dfrac{\pi t}{12}, for one revolution. The dashed line is the midline h=11 and the dashed line h=15 meets the curve at t=8 and t=16.

(2) Height at 10 s

Substitute t=10. The angle is \dfrac{10\pi}{12}=\dfrac{5\pi}{6}=150^\circ and \cos150^\circ=-\dfrac{\sqrt3}2.
h(10)=11-8\left(-\dfrac{\sqrt3}{2}\right)=11+4\sqrt3\approx17.9\ \mathrm{m}

(3) When is the height 15 m?

Put h=15 and solve for the cosine.
11-8\cos\dfrac{\pi t}{12}=15\ \Longrightarrow\ \cos\dfrac{\pi t}{12}=-\dfrac12
The smallest positive angle with cosine -\dfrac12 is \dfrac{2\pi}3 (120^\circ).
\dfrac{\pi t}{12}=\dfrac{2\pi}{3}\ \Longrightarrow\ t=8\ \mathrm{s}
The graph shows a second time, t=16\,\mathrm{s} (angle \dfrac{4\pi}3), on the way down.

h(t)=11-8\cos\dfrac{\pi t}{12};\ h(10)\approx17.9\,\mathrm{m};\ t=8\,\mathrm{s}\ (\text{also }16\,\mathrm{s})

Formula sheet

Range of sine and cosine

-1\le\sin\theta\le1,\qquad-1\le\cos\theta\le1\qquad(|\sin\theta|\le1,\ |\cos\theta|\le1)

\theta: any real number

Unit: dimensionless

Range of secant and cosecant

|\sec\theta|\ge1,\qquad|\operatorname{cosec}\theta|\ge1

values in (-1,1) are never taken

Unit: dimensionless

Domain of the six functions

\tan,\sec:\ \theta\neq(2n+1)\dfrac{\pi}{2};\qquad\cot,\operatorname{cosec}:\ \theta\neq n\pi;\qquad\sin,\cos:\ \theta\in\mathbb{R}

n: any integer

Unit: angle in radians

Periodic function

f(x+p)=f(x)

p>0: the period (least such p)

Unit: same as x

Periods of the basic functions

\sin,\cos,\sec,\operatorname{cosec}:\ 2\pi;\qquad\tan,\cot:\ \pi

period of the reciprocal equals the period of the function

Unit: radian

Period of f(ax+b)

\text{period}=\dfrac{p}{|a|}

p: period of f; \sin(ax+b), \cos(ax+b): \dfrac{2\pi}{|a|}; \tan(ax+b), \cot(ax+b): \dfrac{\pi}{|a|}

Unit: radian

Period of a sum

\text{Period}=\dfrac{\text{L.C.M. of the numerators}}{\text{H.C.F. of the denominators}}

the periods are written as fractions, for example \dfrac{\pi}{3} and \dfrac{3\pi}{1} give \dfrac{3\pi}{1}=3\pi

Unit: radian

Period of powers

\cos^{n}x,\ \sin^{n}x:\ \pi\ (n\text{ even}),\ 2\pi\ (n\text{ odd});\qquad\tan^{n}x:\ \pi

n: positive integer

Unit: radian

Maximum and minimum

M=a+|b|,\qquad m=a-|b|

for a+b\sin\theta, a+b\cos\theta, a+b\sin(c\theta+d), a+b\cos(c\theta+d)

Unit: same as y

Reciprocal of a positive function

M'=\dfrac1m,\qquad m'=\dfrac1M

valid when M>0 and m>0

Unit: same as y

Amplitude and period of a\sin bx

\text{amplitude}=|a|,\qquad\text{period}=\dfrac{2\pi}{b}

b>0; the same for a\cos bx

Unit: radian

Even and odd functions

\text{even: }f(-x)=f(x);\qquad\text{odd: }f(-x)=-f(x)

even: symmetric about the y-axis; odd: symmetric about the origin

Unit: none

Parity of the trigonometric functions

\cos,\sec\ \text{even};\qquad\sin,\tan,\cot,\operatorname{cosec}\ \text{odd}

e.g. \sin(-x)=-\sin x, \cos(-x)=\cos x

Unit: none

Transformed sine and cosine

y=a\,f\big[b(x-c)\big]+d

a: amplitude =\dfrac{\max-\min}{2}; b: period =\dfrac1b\times2\pi; c: phase shift; d: midline =\dfrac{\max+\min}{2}

Unit: same as y

Key terms

Domain: The set of all possible values of x for which a function f(x) is defined.
Range: The set of all values that f(x) takes when x is any number from the domain.
Closed interval: [a,b]=\{x: a\le x\le b\}, the numbers from a to b including both ends.
Open interval: (a,b)=\{x: a<x<b\}, the numbers between a and b without the ends.
Periodic function: A function with f(x+p)=f(x) for some p>0, so that its values repeat after a fixed interval.
Period: The smallest positive number p for which f(x+p)=f(x) for all x.
Periodicity: The repeating behaviour of the trigonometric functions: the values at \theta and 2n\pi\pm\theta are the same.
Maximum and minimum value: The greatest and the least value of a function; for a+b\sin\theta they are a+|b| and a-|b|.
Cycle: The part of the graph of a periodic function over one period, for example [0,2\pi] for sine and cosine.
Amplitude: For y=a\sin bx or y=a\cos bx, the number |a|: half the distance between the maximum and the minimum.
Sinusoidal curve: A wave-shaped curve of the form of the sine or cosine graph.
Even function: A function with f(-x)=f(x); its graph is symmetric about the y-axis.
Odd function: A function with f(-x)=-f(x); its graph is symmetric about the origin.
Phase shift: The horizontal translation c in y=a\,f[b(x-c)]+d.
Midline: The horizontal line y=d=\dfrac{\max+\min}{2} about which the graph of y=a\,f[b(x-c)]+d oscillates.

Exercise 9.1p. 228

Maximum and minimum values of a+b\sin\theta and a+b\cos\theta (and their reciprocals), domain and range, and periods. Rewrite each function in the standard form, then use M=a+|b|, m=a-|b|, the rule that \sin and \cos stay in [-1,1], and period =\frac{2\pi}{|a|} for sine and cosine or \frac{\pi}{|a|} for tangent and cotangent.

Find the maximum and minimum values of the following trigonometric functions.

Solution

(i) y=2-2\cos\theta

Compare with y=a+b\cos\theta.
a=2,\quad b=-2,\quad |b|=2
The largest value of a+b\cos\theta is a+|b| (the book's rule: it is largest when b\cos\theta is largest).
M=a+|b|=2+2=4
The smallest value is a-|b|.
m=a-|b|=2-2=0
Where they occur: because b=-2 is negative, M occurs when \cos\theta=-1 (then y=2+2=4) and m occurs when \cos\theta=1 (then y=2-2=0).

M=4,\quad m=0

(ii) y=\dfrac23-\dfrac12\sin\theta

Compare with y=a+b\sin\theta.
a=\dfrac23,\quad b=-\dfrac12,\quad |b|=\dfrac12
Maximum value.
M=a+|b|=\dfrac23+\dfrac12=\dfrac{4+3}{6}=\dfrac76
Minimum value.
m=a-|b|=\dfrac23-\dfrac12=\dfrac{4-3}{6}=\dfrac16
Where they occur: M when \sin\theta=-1 and m when \sin\theta=1, because b is negative.

M=\dfrac76,\quad m=\dfrac16

(iii) y=\dfrac15-2\sin(3\theta-7)

Compare with y=a+b\sin(c\theta+d).
a=\dfrac15,\quad b=-2,\quad c=3,\quad d=-7,\quad |b|=2
The value of \sin(3\theta-7) always lies in [-1,1], whatever c and d are, so only a and |b| matter.
M=a+|b|=\dfrac15+2=\dfrac{11}{5}
Minimum value.
m=a-|b|=\dfrac15-2=-\dfrac95
Where they occur: M when \sin(3\theta-7)=-1 and m when \sin(3\theta-7)=1.

M=\dfrac{11}{5},\quad m=-\dfrac95

(iv) y=7+\dfrac35\cos(2\theta-1)

Compare with y=a+b\cos(c\theta+d).
a=7,\quad b=\dfrac35,\quad c=2,\quad d=-1,\quad |b|=\dfrac35
Maximum value.
M=a+|b|=7+\dfrac35=\dfrac{38}{5}
Minimum value.
m=a-|b|=7-\dfrac35=\dfrac{32}{5}
Where they occur: M when \cos(2\theta-1)=1 and m when \cos(2\theta-1)=-1.

M=\dfrac{38}{5},\quad m=\dfrac{32}{5}

Book answer. (i) M = 4, m = 0; (ii) M = 7/6, m = 1/6; (iii) M = 11/5, m = −9/5; (iv) M = 38/5, m = 32/5.

Find the maximum and minimum values of the following reciprocal trigonometric functions.

Solution

(i) y=\dfrac{1}{4+3\sin\theta}

Work with the denominator 4+3\sin\theta, which is a+b\sin\theta with a=4, b=3.
M_d=4+3=7,\quad m_d=4-3=1
Both extreme values of the denominator are positive, so the denominator is never zero. For positive numbers the reciprocal reverses the order: the largest denominator gives the smallest y and the smallest denominator gives the largest y (book's key fact: if M>0 and m>0 then M'=\frac1m, m'=\frac1M).
M=\dfrac{1}{m_d}=\dfrac11=1,\qquad m=\dfrac{1}{M_d}=\dfrac17

M=1,\quad m=\dfrac17

(ii) y=\dfrac{1}{\frac12-5\cos\theta}

Work with the denominator \frac12-5\cos\theta, which is a+b\cos\theta with a=\frac12, b=-5.
M_d=\dfrac12+5=\dfrac{11}{2},\quad m_d=\dfrac12-5=-\dfrac92
Take the reciprocal of each extreme value of the denominator, as in the book's Example 7(iii).
M'=\dfrac{1}{M_d}=\dfrac{2}{11},\qquad m'=\dfrac{1}{m_d}=-\dfrac29

The denominator changes sign, so y has two branches. The branch above the axis never falls below \frac{2}{11}, and the branch below the axis never rises above -\frac29.

M=\dfrac{2}{11},\quad m=-\dfrac29

(iii) y=\dfrac{1}{\frac13-4\sin(2\theta-5)}

The denominator is a+b\sin(c\theta+d) with a=\frac13, b=-4, c=2, d=-5.
M_d=\dfrac13+4=\dfrac{13}{3},\quad m_d=\dfrac13-4=-\dfrac{11}{3}
Take the reciprocals.
M'=\dfrac{1}{M_d}=\dfrac{3}{13},\qquad m'=\dfrac{1}{m_d}=-\dfrac{3}{11}

M=\dfrac{3}{13},\quad m=-\dfrac{3}{11}

(iv) y=\dfrac{1}{3+\frac25\sin(5\theta-7)}

The denominator is a+b\sin(c\theta+d) with a=3, b=\frac25, c=5, d=-7.
M_d=3+\dfrac25=\dfrac{17}{5},\quad m_d=3-\dfrac25=\dfrac{13}{5}
Both are positive, so the denominator never vanishes and the order reverses: the largest value of y comes from the smallest denominator.
M=\dfrac{1}{m_d}=\dfrac{5}{13},\qquad m=\dfrac{1}{M_d}=\dfrac{5}{17}

M=\dfrac{5}{13},\quad m=\dfrac{5}{17}

Book answer. (i) M = 1, m = 1/7; (ii) M = 2/11, m = −2/9; (iii) M = 3/13, m = −3/11; (iv) M = 5/13, m = 5/17.

Note. In (ii) and (iii) the denominator takes both positive and negative values, so it passes through zero and y is not bounded: the true range is (-\infty,-\frac29]\cup[\frac{2}{11},\infty) in (ii). The book's 'maximum' and 'minimum' here are the reciprocals of the extreme values of the denominator, which is what its Example 7(iii) does, and they agree with the answer key. Parts (i) and (iv) are genuine maximum and minimum values.

Find domain and range of the following:

Solution

(i) y=7\cos4x

Cosine is defined for every real number, so 4x can be any real number and x is unrestricted.
\text{Dom}=\mathbb{R}=(-\infty,\infty)
For the range put \theta=4x (as the book does in Example 1). Then y=7\cos\theta and -1\le\cos\theta\le1.
-1\le\cos4x\le1
Multiply every part by 7 (a positive number, so the inequalities keep their direction).
-7\le7\cos4x\le7
So y lies between -7 and 7 and takes every value in between.
\text{Range}=[-7,7]

The curve repeats every \frac{\pi}{2} and stays between -7 and 7.

\text{Dom}=\mathbb{R},\quad \text{Range}=[-7,7]

(ii) y=\cos\dfrac{x}{3}

\cos\frac{x}{3} exists for every real x.
\text{Dom}=\mathbb{R}
Put \theta=\frac{x}{3}. Then y=\cos\theta, and the range of cosine is [-1,1].
-1\le\cos\dfrac{x}{3}\le1\;\Rightarrow\;\text{Range}=[-1,1]

The period is 6\pi, so one wave is stretched across a wide interval, but the height is still 1.

\text{Dom}=\mathbb{R},\quad \text{Range}=[-1,1]

(iii) y=\sin\dfrac{2x}{3}

\sin\frac{2x}{3} exists for every real x.
\text{Dom}=\mathbb{R}
Put \theta=\frac{2x}{3}. Then y=\sin\theta, and -1\le\sin\theta\le1.
-1\le\sin\dfrac{2x}{3}\le1\;\Rightarrow\;\text{Range}=[-1,1]

The period is 3\pi and the graph stays between -1 and 1.

\text{Dom}=\mathbb{R},\quad \text{Range}=[-1,1]

(iv) y=7\cot\dfrac{\pi}{2}x

Cotangent is not defined where its angle is a multiple of \pi (there \sin=0). Here the angle is \frac{\pi x}{2}.
\dfrac{\pi x}{2}=n\pi\;\Rightarrow\;x=2n,\quad n\in\mathbb{Z}
So all even integers are removed from the domain.
\text{Dom}=\mathbb{R}-\{2n:\,n\in\mathbb{Z}\}
Cotangent takes every real value, and multiplying by 7 does not restrict this.
\text{Range}=\mathbb{R}=(-\infty,\infty)

Vertical asymptotes at the even integers x=0,\pm2,\dots. The curve covers every y.

\text{Dom}=\mathbb{R}-\{2n:\,n\in\mathbb{Z}\},\quad \text{Range}=\mathbb{R}

(v) y=4\tan\pi x

Tangent is not defined at odd multiples of \frac{\pi}{2} (there \cos=0). Here the angle is \pi x.
\pi x=(2n+1)\dfrac{\pi}{2}\;\Rightarrow\;x=\dfrac{2n+1}{2}=n+\dfrac12,\quad n\in\mathbb{Z}
Remove these values.
\text{Dom}=\mathbb{R}-\left\{\dfrac{2n+1}{2}:\,n\in\mathbb{Z}\right\}
Tangent takes every real value, so 4\tan\pi x does too.
\text{Range}=\mathbb{R}

The asymptotes are at x=\pm\frac12,\pm\frac32,\dots and the period is 1.

\text{Dom}=\mathbb{R}-\left\{\tfrac{2n+1}{2}\right\},\quad \text{Range}=\mathbb{R}

(vi) y=\operatorname{cosec}4x

\operatorname{cosec}4x=\dfrac{1}{\sin4x} is not defined where \sin4x=0, that is where 4x is a multiple of \pi.
4x=n\pi\;\Rightarrow\;x=\dfrac{n\pi}{4},\quad n\in\mathbb{Z}
Remove these values.
\text{Dom}=\mathbb{R}-\left\{\dfrac{n\pi}{4}:\,n\in\mathbb{Z}\right\}
Since |\sin4x|\le1, taking reciprocals gives |\operatorname{cosec}4x|\ge1 (book's Remark III). So y never lies in (-1,1).
|y|\ge1\;\Rightarrow\;\text{Range}=(-\infty,-1]\cup[1,\infty)

Every branch lies on or beyond the lines y=\pm1, and the asymptotes are at the multiples of \frac{\pi}{4}.

\text{Dom}=\mathbb{R}-\left\{\tfrac{n\pi}{4}\right\},\quad \text{Range}=(-\infty,-1]\cup[1,\infty)

Book answer. (i) R, [−7, 7]; (ii) R, [−1, 1]; (iii) R, [−1, 1]; (iv) R, range 0; (v) R, range 0; (vi) R, [−6, 6].

Note. The printed key is wrong for (iv), (v) and (vi). It gives 'range = 0' for (iv) and (v) and '[−6, 6]' for (vi), and it lists the domain as all of R for these three. Substituting back shows otherwise: 7\cot\frac{\pi}{2}x and 4\tan\pi x take every real value (their domains exclude the points above), and \operatorname{cosec}4x has range |y|\ge1, so [-6,6] cannot be its range. Parts (i) to (iii) agree with the key.

Find the periods of the following:

Solution

(i) y=6\sec(2x-3)

The period of \sec x is 2\pi, and multiplying by 6 does not change the period. For \sec(ax+b) the period is \dfrac{2\pi}{|a|}; here a=2.
\text{Period}=\dfrac{2\pi}{|2|}=\pi

\pi

(ii) y=\cos(5x+4)

The period of \cos(ax+b) is \dfrac{2\pi}{|a|}; here a=5. The shift +4 moves the graph but does not change the period.
\text{Period}=\dfrac{2\pi}{5}

\dfrac{2\pi}{5}

(iii) y=\cot4x+\sin\dfrac{5x}{2}

For a sum, find the period of each term. \cot ax has period \dfrac{\pi}{|a|}.
\text{Period of }\cot4x=\dfrac{\pi}{4}
\sin ax has period \dfrac{2\pi}{|a|}, with a=\frac52.
\text{Period of }\sin\dfrac{5x}{2}=\dfrac{2\pi}{5/2}=\dfrac{4\pi}{5}
The period of the sum is the least number that both periods divide into. As in the book's Example 3, write the periods as fractions and use the rule: LCM of the numerators divided by HCF of the denominators.
\text{Period}=\dfrac{\text{LCM of }\pi\text{ and }4\pi}{\text{HCF of }4\text{ and }5}=\dfrac{4\pi}{1}=4\pi
Check: 4\pi is 16 times \frac{\pi}{4} and 5 times \frac{4\pi}{5}, both whole numbers, so both terms repeat after 4\pi.

4\pi

(iv) y=7\sin(3x+3)

\sin(ax+b) has period \dfrac{2\pi}{|a|}; here a=3. The factor 7 and the shift 3 do not change it.
\text{Period}=\dfrac{2\pi}{3}

\dfrac{2\pi}{3}

(v) y=5\sin(2x+3)

Here a=2.
\text{Period}=\dfrac{2\pi}{|2|}=\pi

\pi

(vi) y=2\tan3x+7\cos5x

\tan ax has period \dfrac{\pi}{|a|}.
\text{Period of }2\tan3x=\dfrac{\pi}{3}
\cos ax has period \dfrac{2\pi}{|a|}.
\text{Period of }7\cos5x=\dfrac{2\pi}{5}
Combine with the rule LCM of numerators over HCF of denominators.
\text{Period}=\dfrac{\text{LCM of }\pi\text{ and }2\pi}{\text{HCF of }3\text{ and }5}=\dfrac{2\pi}{1}=2\pi
Check: 2\pi=6\cdot\frac{\pi}{3} and 2\pi=5\cdot\frac{2\pi}{5}.

2\pi

Book answer. (i) π; (ii) 2π/5; (iii) 4π; (iv) 2π/3; (v) π; (vi) 2π.

Exercise 9.2p. 244–245

Even and odd functions, graphs of trigonometric functions and their characteristics, graphical solution of equations, and modelling with y=a\sin[k(t-d)]+c or h(t)=A\cos(Bt)+C. Test parity with f(-x)=\pm f(x); draw graphs from a table over one period; for models find the amplitude \frac{\text{max}-\text{min}}{2}, the midline \frac{\text{max}+\text{min}}{2} and the period.

Check whether the following functions are odd or even?

Solution

(i) y=\sin x+x\cos x

Name the function f(x) and replace x by -x.
f(x)=\sin x+x\cos x,\qquad f(-x)=\sin(-x)+(-x)\cos(-x)
Use \sin(-x)=-\sin x and \cos(-x)=\cos x.
f(-x)=-\sin x-x\cos x
Take out the common factor -1.
f(-x)=-(\sin x+x\cos x)=-f(x)
f(-x)=-f(x), so the function is odd.

\text{odd function}

(ii) y=x^3\sin x\cos x

Replace x by -x.
f(-x)=(-x)^3\sin(-x)\cos(-x)
(-x)^3=-x^3, \sin(-x)=-\sin x, \cos(-x)=\cos x.
f(-x)=(-x^3)(-\sin x)(\cos x)
Two minus signs cancel.
f(-x)=x^3\sin x\cos x=f(x)
f(-x)=f(x), so the function is even.

\text{even function}

(iii) y=\dfrac{x^2\tan x}{x+\sin x}

Replace x by -x.
f(-x)=\dfrac{(-x)^2\tan(-x)}{(-x)+\sin(-x)}
(-x)^2=x^2, \tan(-x)=-\tan x, \sin(-x)=-\sin x.
f(-x)=\dfrac{x^2(-\tan x)}{-x-\sin x}
Take -1 out of the denominator; it cancels the minus in the numerator.
f(-x)=\dfrac{-x^2\tan x}{-(x+\sin x)}=\dfrac{x^2\tan x}{x+\sin x}=f(x)
f(-x)=f(x), so the function is even.

\text{even function}

(iv) y=x^3\sin x\cos^2x

Replace x by -x.
f(-x)=(-x)^3\sin(-x)\cos^2(-x)
(-x)^3=-x^3, \sin(-x)=-\sin x and \cos^2(-x)=(\cos x)^2=\cos^2x.
f(-x)=(-x^3)(-\sin x)\cos^2x=x^3\sin x\cos^2x=f(x)
f(-x)=f(x), so the function is even.

\text{even function}

(v) y=\dfrac{\sin^2x}{x+\tan x}

Replace x by -x.
f(-x)=\dfrac{\sin^2(-x)}{(-x)+\tan(-x)}
\sin^2(-x)=(-\sin x)^2=\sin^2x and \tan(-x)=-\tan x.
f(-x)=\dfrac{\sin^2x}{-x-\tan x}
Take -1 out of the denominator.
f(-x)=-\dfrac{\sin^2x}{x+\tan x}=-f(x)
f(-x)=-f(x), so the function is odd. (The numerator is even, the denominator is odd, and even divided by odd is odd.)

\text{odd function}

(vi) y=\dfrac{\tan x-\sin x}{\sin^3x}

Replace x by -x.
f(-x)=\dfrac{\tan(-x)-\sin(-x)}{\sin^3(-x)}
\tan(-x)=-\tan x, \sin(-x)=-\sin x, \sin^3(-x)=-\sin^3x.
f(-x)=\dfrac{-\tan x+\sin x}{-\sin^3x}
Take -1 out of the numerator; it cancels the minus in the denominator.
f(-x)=\dfrac{-(\tan x-\sin x)}{-\sin^3x}=\dfrac{\tan x-\sin x}{\sin^3x}=f(x)
f(-x)=f(x), so the function is even.

\text{even function}

(vii) y=\dfrac{\sec x}{x+\tan x}

Replace x by -x.
f(-x)=\dfrac{\sec(-x)}{(-x)+\tan(-x)}
\sec(-x)=\sec x (secant is even) and \tan(-x)=-\tan x.
f(-x)=\dfrac{\sec x}{-x-\tan x}
Take -1 out of the denominator.
f(-x)=-\dfrac{\sec x}{x+\tan x}=-f(x)
f(-x)=-f(x), so the function is odd.

\text{odd function}

(viii) y=x^2\sin x-\cot x

Replace x by -x.
f(-x)=(-x)^2\sin(-x)-\cot(-x)
(-x)^2=x^2, \sin(-x)=-\sin x, \cot(-x)=-\cot x.
f(-x)=x^2(-\sin x)-(-\cot x)=-x^2\sin x+\cot x
Take out -1.
f(-x)=-(x^2\sin x-\cot x)=-f(x)
f(-x)=-f(x), so the function is odd.

\text{odd function}

Book answer. (i) odd; (ii) even; (iii) even; (iv) even; (v) even; (vi) even; (vii) odd; (viii) odd.

Note. The key gives 'even' for (v), but substituting -x gives f(-x)=-f(x), so y=\dfrac{\sin^2x}{x+\tan x} is odd (numerator even, denominator odd). Every other part agrees with the key.

Draw the graph of each of the following functions:

Solution

(i) y=2\sin x

This has the form y=a\sin bx with a=2, b=1 (book's procedure for sine and cosine graphs). Step 1: the period is \dfrac{2\pi}{b}.
\text{Period}=\dfrac{2\pi}{1}=2\pi,\qquad\text{amplitude }|a|=2
Step 2: divide the interval [0,2\pi] into twelve equal parts, so the x-values go up in steps of \frac{\pi}{6}.
Step 3: make a table. Each value of \sin x is doubled.

x	0	\dfrac{\pi}{6}	\dfrac{\pi}{3}	\dfrac{\pi}{2}	\dfrac{2\pi}{3}	\dfrac{5\pi}{6}	\pi	\dfrac{7\pi}{6}	\dfrac{4\pi}{3}	\dfrac{3\pi}{2}	\dfrac{5\pi}{3}	\dfrac{11\pi}{6}	2\pi
\sin x	0	\tfrac12	\tfrac{\sqrt3}{2}	1	\tfrac{\sqrt3}{2}	\tfrac12	0	-\tfrac12	-\tfrac{\sqrt3}{2}	-1	-\tfrac{\sqrt3}{2}	-\tfrac12	0
2\sin x	0	1	\sqrt3	2	\sqrt3	1	0	-1	-\sqrt3	-2	-\sqrt3	-1	0

x	\sin x	2\sin x
0	0	0
\dfrac{\pi}{6}	\tfrac12	1
\dfrac{\pi}{3}	\tfrac{\sqrt3}{2}	\sqrt3
\dfrac{\pi}{2}	1	2
\dfrac{2\pi}{3}	\tfrac{\sqrt3}{2}	\sqrt3
\dfrac{5\pi}{6}	\tfrac12	1
\pi	0	0
\dfrac{7\pi}{6}	-\tfrac12	-1
\dfrac{4\pi}{3}	-\tfrac{\sqrt3}{2}	-\sqrt3
\dfrac{3\pi}{2}	-1	-2
\dfrac{5\pi}{3}	-\tfrac{\sqrt3}{2}	-\sqrt3
\dfrac{11\pi}{6}	-\tfrac12	-1
2\pi	0	0

Step 4: plot the points of the table. They are the maximum point \left(\frac{\pi}{2},2\right), the minimum point \left(\frac{3\pi}{2},-2\right) and the x-intercepts 0,\pi,2\pi, with the points between. Join them by a smooth sinusoidal curve of amplitude 2.
Step 5: since the period is 2\pi, repeat the cycle to the left and right.

One cycle on [0,2\pi] (dots from the table), repeated to the left. The curve touches y=2 and y=-2.

Characteristic	Value
Domain	(-\infty,\infty)=\mathbb{R}
Range	[-2,2]
Period	2\pi
Amplitude	2
Nature	odd function

(ii) y=2\cos3x

Here a=2 and b=3. The period is \dfrac{2\pi}{b}.
\text{Period}=\dfrac{2\pi}{3},\qquad\text{amplitude}=2
Divide the interval \left[0,\frac{2\pi}{3}\right] into eight equal parts, so x goes up in steps of \frac{\pi}{12}, and make a table.

x	0	\dfrac{\pi}{12}	\dfrac{\pi}{6}	\dfrac{\pi}{4}	\dfrac{\pi}{3}	\dfrac{5\pi}{12}	\dfrac{\pi}{2}	\dfrac{7\pi}{12}	\dfrac{2\pi}{3}
3x	0	\dfrac{\pi}{4}	\dfrac{\pi}{2}	\dfrac{3\pi}{4}	\pi	\dfrac{5\pi}{4}	\dfrac{3\pi}{2}	\dfrac{7\pi}{4}	2\pi
\cos3x	1	\tfrac{\sqrt2}{2}	0	-\tfrac{\sqrt2}{2}	-1	-\tfrac{\sqrt2}{2}	0	\tfrac{\sqrt2}{2}	1
2\cos3x	2	\sqrt2	0	-\sqrt2	-2	-\sqrt2	0	\sqrt2	2

x	3x	\cos3x	2\cos3x
0	0	1	2
\dfrac{\pi}{12}	\dfrac{\pi}{4}	\tfrac{\sqrt2}{2}	\sqrt2
\dfrac{\pi}{6}	\dfrac{\pi}{2}	0	0
\dfrac{\pi}{4}	\dfrac{3\pi}{4}	-\tfrac{\sqrt2}{2}	-\sqrt2
\dfrac{\pi}{3}	\pi	-1	-2
\dfrac{5\pi}{12}	\dfrac{5\pi}{4}	-\tfrac{\sqrt2}{2}	-\sqrt2
\dfrac{\pi}{2}	\dfrac{3\pi}{2}	0	0
\dfrac{7\pi}{12}	\dfrac{7\pi}{4}	\tfrac{\sqrt2}{2}	\sqrt2
\dfrac{2\pi}{3}	2\pi	1	2

Plot the points and join them with a sinusoidal curve of amplitude 2. It starts at the maximum (0,2), crosses the axis at \frac{\pi}{6}, reaches the minimum at \frac{\pi}{3}, crosses again at \frac{\pi}{2} and is back at the maximum at \frac{2\pi}{3}. Then repeat the cycle on both sides.

Cycle from the table on [0,\frac{2\pi}{3}], repeated on [-\frac{2\pi}{3},0].

Characteristic	Value
Domain	(-\infty,\infty)=\mathbb{R}
Range	[-2,2]
Period	\dfrac{2\pi}{3}
Amplitude	2
Nature	even function

(iii) y=2\tan2x

The period of \tan bx is \dfrac{\pi}{b}. Here b=2, so one cycle of the curve has length \frac{\pi}{2} and the factor 2 only stretches the graph upwards (tangent has no amplitude).
\text{Period}=\dfrac{\pi}{2}
Where is it undefined? \tan2x fails when 2x is an odd multiple of \frac{\pi}{2}, that is when x=\frac{\pi}{4}+\frac{n\pi}{2}. So the cycle from -\frac{\pi}{4} to \frac{\pi}{4} lies between two vertical asymptotes.
Divide this interval into equal parts and make a table. At the end-points the function is undefined and the curve goes off to \pm\infty.

x	-\dfrac{\pi}{4}	-\dfrac{\pi}{6}	-\dfrac{\pi}{12}	\dfrac{\pi}{12}	\dfrac{\pi}{6}	\dfrac{\pi}{4}
2x	-\dfrac{\pi}{2}	-\dfrac{\pi}{3}	-\dfrac{\pi}{6}	\dfrac{\pi}{6}	\dfrac{\pi}{3}	\dfrac{\pi}{2}
\tan2x	\text{undefined}	-\sqrt3	-\tfrac{1}{\sqrt3}	\tfrac{1}{\sqrt3}	\sqrt3	\text{undefined}
2\tan2x	\text{undefined}	-3.46	-1.15	1.15	3.46	\text{undefined}

x	2x	\tan2x	2\tan2x
-\dfrac{\pi}{4}	-\dfrac{\pi}{2}	\text{undefined}	\text{undefined}
-\dfrac{\pi}{6}	-\dfrac{\pi}{3}	-\sqrt3	-3.46
-\dfrac{\pi}{12}	-\dfrac{\pi}{6}	-\tfrac{1}{\sqrt3}	-1.15
0	0	0	0
\dfrac{\pi}{12}	\dfrac{\pi}{6}	\tfrac{1}{\sqrt3}	1.15
\dfrac{\pi}{6}	\dfrac{\pi}{3}	\sqrt3	3.46
\dfrac{\pi}{4}	\dfrac{\pi}{2}	\text{undefined}	\text{undefined}

Plot the points, draw the asymptotes x=\pm\frac{\pi}{4} as dashed lines and join the points by a free-hand curve that rises through the origin. Repeat the branch every \frac{\pi}{2}.

One branch between the asymptotes x=-\frac{\pi}{4} and x=\frac{\pi}{4}, repeated every \frac{\pi}{2}.

Characteristic	Value
Domain	\{x:\,x\in\mathbb{R},\ x\ne\dfrac{\pi}{4}+\dfrac{n\pi}{2},\ n\in\mathbb{Z}\}
Range	(-\infty,\infty)=\mathbb{R}
Period	\dfrac{\pi}{2}
Amplitude	nil (tangent has no maximum)
Nature	odd function

(iv) y=\cos\dfrac{x}{2}

Here a=1 and b=\frac12. The period is \dfrac{2\pi}{b}.
\text{Period}=\dfrac{2\pi}{1/2}=4\pi,\qquad\text{amplitude}=1
Divide [0,4\pi] into eight equal parts (steps of \frac{\pi}{2}) and make a table.

x	0	\dfrac{\pi}{2}	\pi	\dfrac{3\pi}{2}	2\pi	\dfrac{5\pi}{2}	3\pi	\dfrac{7\pi}{2}	4\pi
\tfrac{x}{2}	0	\dfrac{\pi}{4}	\dfrac{\pi}{2}	\dfrac{3\pi}{4}	\pi	\dfrac{5\pi}{4}	\dfrac{3\pi}{2}	\dfrac{7\pi}{4}	2\pi
\cos\tfrac{x}{2}	1	\tfrac{\sqrt2}{2}	0	-\tfrac{\sqrt2}{2}	-1	-\tfrac{\sqrt2}{2}	0	\tfrac{\sqrt2}{2}	1

x	\tfrac{x}{2}	\cos\tfrac{x}{2}
0	0	1
\dfrac{\pi}{2}	\dfrac{\pi}{4}	\tfrac{\sqrt2}{2}
\pi	\dfrac{\pi}{2}	0
\dfrac{3\pi}{2}	\dfrac{3\pi}{4}	-\tfrac{\sqrt2}{2}
2\pi	\pi	-1
\dfrac{5\pi}{2}	\dfrac{5\pi}{4}	-\tfrac{\sqrt2}{2}
3\pi	\dfrac{3\pi}{2}	0
\dfrac{7\pi}{2}	\dfrac{7\pi}{4}	\tfrac{\sqrt2}{2}
4\pi	2\pi	1

Plot the points and join them by a smooth curve. It starts at the maximum (0,1), crosses the axis at \pi, has its minimum at 2\pi, crosses at 3\pi and returns to 1 at 4\pi. Repeat to the left.

One cycle takes 4\pi: the cosine wave stretched to twice its usual length.

Characteristic	Value
Domain	(-\infty,\infty)=\mathbb{R}
Range	[-1,1]
Period	4\pi
Amplitude	1
Nature	even function

(v) y=2\sin3x

Here a=2 and b=3.
\text{Period}=\dfrac{2\pi}{3},\qquad\text{amplitude}=2
Divide \left[0,\frac{2\pi}{3}\right] into eight equal parts (steps of \frac{\pi}{12}) and make a table.

x	\dfrac{\pi}{12}	\dfrac{\pi}{6}	\dfrac{\pi}{4}	\dfrac{\pi}{3}	\dfrac{5\pi}{12}	\dfrac{\pi}{2}	\dfrac{7\pi}{12}	\dfrac{2\pi}{3}
3x	\dfrac{\pi}{4}	\dfrac{\pi}{2}	\dfrac{3\pi}{4}	\pi	\dfrac{5\pi}{4}	\dfrac{3\pi}{2}	\dfrac{7\pi}{4}	2\pi
\sin3x	\tfrac{\sqrt2}{2}	1	\tfrac{\sqrt2}{2}	0	-\tfrac{\sqrt2}{2}	-1	-\tfrac{\sqrt2}{2}	0
2\sin3x	\sqrt2	2	\sqrt2	0	-\sqrt2	-2	-\sqrt2	0

x	3x	\sin3x	2\sin3x
0	0	0	0
\dfrac{\pi}{12}	\dfrac{\pi}{4}	\tfrac{\sqrt2}{2}	\sqrt2
\dfrac{\pi}{6}	\dfrac{\pi}{2}	1	2
\dfrac{\pi}{4}	\dfrac{3\pi}{4}	\tfrac{\sqrt2}{2}	\sqrt2
\dfrac{\pi}{3}	\pi	0	0
\dfrac{5\pi}{12}	\dfrac{5\pi}{4}	-\tfrac{\sqrt2}{2}	-\sqrt2
\dfrac{\pi}{2}	\dfrac{3\pi}{2}	-1	-2
\dfrac{7\pi}{12}	\dfrac{7\pi}{4}	-\tfrac{\sqrt2}{2}	-\sqrt2
\dfrac{2\pi}{3}	2\pi	0	0

Plot the points and join them. The curve rises from the origin to the maximum \left(\frac{\pi}{6},2\right), crosses at \frac{\pi}{3}, has its minimum \left(\frac{\pi}{2},-2\right) and returns to the axis at \frac{2\pi}{3}. Repeat the cycle to the left.

Cycle from the table on [0,\frac{2\pi}{3}] and its extension to the left.

Characteristic	Value
Domain	(-\infty,\infty)=\mathbb{R}
Range	[-2,2]
Period	\dfrac{2\pi}{3}
Amplitude	2
Nature	odd function

(vi) y=3\cos x

Here a=3 and b=1.
\text{Period}=2\pi,\qquad\text{amplitude}=3
Divide [0,2\pi] into twelve equal parts (steps of \frac{\pi}{6}) and make a table.

x	0	\dfrac{\pi}{6}	\dfrac{\pi}{3}	\dfrac{\pi}{2}	\dfrac{2\pi}{3}	\dfrac{5\pi}{6}	\pi	\dfrac{7\pi}{6}	\dfrac{4\pi}{3}	\dfrac{3\pi}{2}	\dfrac{5\pi}{3}	\dfrac{11\pi}{6}	2\pi
\cos x	1	\tfrac{\sqrt3}{2}	\tfrac12	0	-\tfrac12	-\tfrac{\sqrt3}{2}	-1	-\tfrac{\sqrt3}{2}	-\tfrac12	0	\tfrac12	\tfrac{\sqrt3}{2}	1
3\cos x	3	\tfrac{3\sqrt3}{2}	\tfrac32	0	-\tfrac32	-\tfrac{3\sqrt3}{2}	-3	-\tfrac{3\sqrt3}{2}	-\tfrac32	0	\tfrac32	\tfrac{3\sqrt3}{2}	3

Plot the points and join them by a smooth curve. It starts at the maximum (0,3), crosses the axis at \frac{\pi}{2}, has its minimum (\pi,-3), crosses at \frac{3\pi}{2} and returns to 3 at 2\pi. Repeat to the left.

The cosine wave stretched upwards by the factor 3.

Characteristic	Value
Domain	(-\infty,\infty)=\mathbb{R}
Range	[-3,3]
Period	2\pi
Amplitude	3
Nature	even function

(vii) y=\cos^2x

Use the double-angle identity to see the shape and the period.
\cos^2x=\dfrac{1+\cos2x}{2}=\dfrac12+\dfrac12\cos2x
So the graph is a cosine curve with b=2, raised by \frac12 and with half-height \frac12. The period is \dfrac{2\pi}{2}=\pi. (The book's key fact: \cos^n x with n even has period \pi.)
\text{Period}=\pi,\qquad 0\le\cos^2x\le1
Divide [0,\pi] into six equal parts (steps of \frac{\pi}{6}) and make a table.

x	0	\dfrac{\pi}{6}	\dfrac{\pi}{3}	\dfrac{\pi}{2}	\dfrac{2\pi}{3}	\dfrac{5\pi}{6}	\pi
\cos x	1	\tfrac{\sqrt3}{2}	\tfrac12	0	-\tfrac12	-\tfrac{\sqrt3}{2}	-1
\cos^2x	1	\tfrac34	\tfrac14	0	\tfrac14	\tfrac34	1

Plot the points and join them. The curve starts at the maximum (0,1), falls to 0 at \frac{\pi}{2} (where it only touches the axis) and rises to 1 at \pi. It never goes below the x-axis. Repeat on both sides.

The curve stays between 0 and 1 and repeats every \pi.

Characteristic	Value
Domain	(-\infty,\infty)=\mathbb{R}
Range	[0,1]
Period	\pi
Amplitude	\tfrac12 (half of max minus min)
Nature	even function

(viii) y=\sin^2x

Use the double-angle identity.
\sin^2x=\dfrac{1-\cos2x}{2}=\dfrac12-\dfrac12\cos2x
The graph is a cosine curve turned upside down, with b=2, raised by \frac12. The period is \pi and 0\le\sin^2x\le1.
\text{Period}=\pi
Divide [0,\pi] into six equal parts and make a table.

x	0	\dfrac{\pi}{6}	\dfrac{\pi}{3}	\dfrac{\pi}{2}	\dfrac{2\pi}{3}	\dfrac{5\pi}{6}	\pi
\sin x	0	\tfrac12	\tfrac{\sqrt3}{2}	1	\tfrac{\sqrt3}{2}	\tfrac12	0
\sin^2x	0	\tfrac14	\tfrac34	1	\tfrac34	\tfrac14	0

Plot the points and join them. The curve starts at (0,0), rises to the maximum \left(\frac{\pi}{2},1\right) and comes back to 0 at \pi. Repeat on both sides.

The curve stays between 0 and 1 and repeats every \pi.

Characteristic	Value
Domain	(-\infty,\infty)=\mathbb{R}
Range	[0,1]
Period	\pi
Amplitude	\tfrac12 (half of max minus min)
Nature	even function

(ix) y=\tan^2x

The period of \tan x is \pi, so \tan^2x also repeats every \pi (the book's key fact: the periodicity of \tan^n x is that of \tan x, whether n is even or odd).
\text{Period}=\pi
\tan x is undefined at x=\frac{\pi}{2}+n\pi, so these are vertical asymptotes. Squaring makes every value non-negative, so y\ge0 and the graph never goes below the axis.
Make a table over one cycle [0,\pi]. The value at \frac{\pi}{2} is undefined.

x	0	\dfrac{\pi}{6}	\dfrac{\pi}{4}	\dfrac{\pi}{3}	\dfrac{\pi}{2}	\dfrac{2\pi}{3}	\dfrac{3\pi}{4}	\dfrac{5\pi}{6}	\pi
\tan x	0	\tfrac{1}{\sqrt3}	1	\sqrt3	\text{undefined}	-\sqrt3	-1	-\tfrac{1}{\sqrt3}	0
\tan^2x	0	\tfrac13	1	3	\text{undefined}	3	1	\tfrac13	0

Plot the points, draw the asymptotes x=\pm\frac{\pi}{2} as dashed lines and join the points. The curve touches the axis at 0 and shoots up on both sides of each asymptote. Repeat every \pi.

Each branch lies above the x-axis and has vertical asymptotes at the odd multiples of \frac{\pi}{2}.

Characteristic	Value
Domain	\{x:\,x\in\mathbb{R},\ x\ne(2n+1)\dfrac{\pi}{2}\}
Range	[0,\infty)
Period	\pi
Amplitude	nil
Nature	even function

(x) y=\sin\dfrac{x}{2}

Here a=1 and b=\frac12.
\text{Period}=\dfrac{2\pi}{1/2}=4\pi,\qquad\text{amplitude}=1
Divide [0,4\pi] into eight equal parts (steps of \frac{\pi}{2}) and make a table.

x	0	\dfrac{\pi}{2}	\pi	\dfrac{3\pi}{2}	2\pi	\dfrac{5\pi}{2}	3\pi	\dfrac{7\pi}{2}	4\pi
\tfrac{x}{2}	0	\dfrac{\pi}{4}	\dfrac{\pi}{2}	\dfrac{3\pi}{4}	\pi	\dfrac{5\pi}{4}	\dfrac{3\pi}{2}	\dfrac{7\pi}{4}	2\pi
\sin\tfrac{x}{2}	0	\tfrac{\sqrt2}{2}	1	\tfrac{\sqrt2}{2}	0	-\tfrac{\sqrt2}{2}	-1	-\tfrac{\sqrt2}{2}	0

Plot the points and join them. The curve rises from the origin to the maximum (\pi,1), comes back to the axis at 2\pi, has the minimum (3\pi,-1) and returns to 0 at 4\pi. Repeat to the left.

One cycle takes 4\pi.

Characteristic	Value
Domain	(-\infty,\infty)=\mathbb{R}
Range	[-1,1]
Period	4\pi
Amplitude	1
Nature	odd function

Book answer. Graphs with characteristics: (i) period 2π, amplitude 2; (ii) period 2π/3, amplitude 2; (iii) period π/2; (iv) period 4π; (v) period 2π/3, amplitude 2; (vi) period 2π, amplitude 3; (vii) and (viii) range [0, 1], period π, even; (ix) period π, y ≥ 0, even; (x) period 4π, amplitude 1.

Note. Where the key's characteristics differ from the working above: for (x) it prints 'even function', but \sin\frac{x}{2} is odd because \sin\left(-\frac{x}{2}\right)=-\sin\frac{x}{2}. For (vii) and (viii) it prints amplitude 1; using the book's own rule amplitude =\frac{\text{max}-\text{min}}{2} gives \frac12 (the height of the wave is 1, from 0 to 1). The domain printed for (iii) is garbled; the correct domain is every x except \frac{\pi}{4}+\frac{n\pi}{2}.

Draw the graphs of y=\sin x and y=\sin2x in [0,2\pi] on the same scale.

Solution

y=\sin x has period 2\pi and y=\sin2x has period \frac{2\pi}{2}=\pi, so in [0,2\pi] the second curve completes two full waves while the first completes one. Both have amplitude 1.
Divide [0,2\pi] into twelve equal parts (steps of \frac{\pi}{6}) and make one table for both functions.

x	\dfrac{\pi}{6}	\dfrac{\pi}{3}	\dfrac{\pi}{2}	\dfrac{2\pi}{3}	\dfrac{5\pi}{6}	\pi	\dfrac{7\pi}{6}	\dfrac{4\pi}{3}	\dfrac{3\pi}{2}	\dfrac{5\pi}{3}	\dfrac{11\pi}{6}	2\pi
\sin x	\tfrac12	\tfrac{\sqrt3}{2}	1	\tfrac{\sqrt3}{2}	\tfrac12	0	-\tfrac12	-\tfrac{\sqrt3}{2}	-1	-\tfrac{\sqrt3}{2}	-\tfrac12	0
2x	\dfrac{\pi}{3}	\dfrac{2\pi}{3}	\pi	\dfrac{4\pi}{3}	\dfrac{5\pi}{3}	2\pi	\dfrac{7\pi}{3}	\dfrac{8\pi}{3}	3\pi	\dfrac{10\pi}{3}	\dfrac{11\pi}{3}	4\pi
\sin2x	\tfrac{\sqrt3}{2}	\tfrac{\sqrt3}{2}	0	-\tfrac{\sqrt3}{2}	-\tfrac{\sqrt3}{2}	0	\tfrac{\sqrt3}{2}	\tfrac{\sqrt3}{2}	0	-\tfrac{\sqrt3}{2}	-\tfrac{\sqrt3}{2}	0

Plot both sets of points on the same axes and join each by a smooth curve. The curves meet at x=0, \pi and 2\pi (the common zeros) and also wherever \sin x=\sin2x. Note that \sin2x has maximum points at \frac{\pi}{4} and \frac{5\pi}{4}, and minimum points at \frac{3\pi}{4} and \frac{7\pi}{4}.

On [0,2\pi] the graph of y=\sin2x has twice as many waves as y=\sin x.

Book answer. Graphs of y = sin x and y = sin 2x on one set of axes over [0, 2π] (no numerical answer).

Draw the graphs of y=\cos x and y=\cos2x in [0,2\pi] on the same scale.

Solution

y=\cos x has period 2\pi and y=\cos2x has period \pi, so in [0,2\pi] the second curve completes two waves while the first completes one. Both have amplitude 1 and both start at the maximum 1.
Divide [0,2\pi] into twelve equal parts (steps of \frac{\pi}{6}) and make one table.

x	0	\dfrac{\pi}{6}	\dfrac{\pi}{3}	\dfrac{\pi}{2}	\dfrac{2\pi}{3}	\dfrac{5\pi}{6}	\pi	\dfrac{7\pi}{6}	\dfrac{4\pi}{3}	\dfrac{3\pi}{2}	\dfrac{5\pi}{3}	\dfrac{11\pi}{6}	2\pi
\cos x	1	\tfrac{\sqrt3}{2}	\tfrac12	0	-\tfrac12	-\tfrac{\sqrt3}{2}	-1	-\tfrac{\sqrt3}{2}	-\tfrac12	0	\tfrac12	\tfrac{\sqrt3}{2}	1
2x	0	\dfrac{\pi}{3}	\dfrac{2\pi}{3}	\pi	\dfrac{4\pi}{3}	\dfrac{5\pi}{3}	2\pi	\dfrac{7\pi}{3}	\dfrac{8\pi}{3}	3\pi	\dfrac{10\pi}{3}	\dfrac{11\pi}{3}	4\pi
\cos2x	1	\tfrac12	-\tfrac12	-1	-\tfrac12	\tfrac12	1	\tfrac12	-\tfrac12	-1	-\tfrac12	\tfrac12	1

Plot both sets of points on the same axes and join each by a smooth curve. Both start at (0,1) and both end at (2\pi,1). The curve y=\cos2x has minima at \frac{\pi}{2} and \frac{3\pi}{2} and a maximum again at \pi.

y=\cos2x completes two waves in the interval where y=\cos x completes one.

Book answer. Graphs of y = cos x and y = cos 2x on one set of axes over [0, 2π] (no numerical answer).

Solve graphically:

Solution

(i) \sin x=\cos x

To solve graphically, draw the graphs of the two sides, y=\sin x and y=\cos x, on the same axes. The x-coordinates of the points where they cross are the solutions.
Draw both over [0,2\pi] using the tables of values (steps of \frac{\pi}{6} from the book's procedure). The curves cross twice: once where both are positive, between \frac{\pi}{6} and \frac{\pi}{3}, and once where both are negative, between \frac{7\pi}{6} and \frac{4\pi}{3}.
Read off the crossings and check by substitution: at x=\frac{\pi}{4} both are \frac{\sqrt2}{2}\approx0.71; at x=\frac{5\pi}{4} both are -\frac{\sqrt2}{2}\approx-0.71.
x=\dfrac{\pi}{4}\approx0.79,\qquad x=\dfrac{5\pi}{4}\approx3.93
Since both functions have period 2\pi, the same crossings repeat: x=\frac{\pi}{4}+n\pi, n\in\mathbb{Z}.

The two curves intersect at x=\frac{\pi}{4} and x=\frac{5\pi}{4} in [0,2\pi].

x=\dfrac{\pi}{4},\ \dfrac{5\pi}{4}\ \text{in }[0,2\pi]

(ii) \cos x=x

Draw y=\cos x and the straight line y=x (through the origin with gradient 1) on the same axes. The x-coordinate of the intersection is the solution. (Here x is in radians.)
The cosine curve starts at y=1 when x=0 and decreases, while the line y=x starts at 0 and increases, so they must cross once, for x between 0 and \frac{\pi}{2}. For x>\frac{\pi}{2} the cosine is at most 1 but the line is already above 1.57, and for x<0 the line is negative while \cos x is positive, so there is no other crossing.
Locate the crossing by trial from the table.
\begin{array}{c|ccc}x&0.7&0.74&0.8\\\hline\cos x&0.765&0.738&0.697\end{array}
At x=0.7 the cosine is above the line and at x=0.8 it is below, so the crossing is near 0.74, where \cos x\approx0.74.
x\approx0.74

The line y=x meets y=\cos x once, at about x=0.74.

x\approx0.74

(iii) \sin x=x

Draw y=\sin x and the line y=x on the same axes.
Both pass through the origin, so x=0 is a solution. Near the origin the line y=x is the tangent to y=\sin x, and for x>0 the curve stays below it: for example at x=1 the sine is 0.84 but the line is 1. For x>1 the sine never exceeds 1 while the line keeps rising, so they cannot meet again. By symmetry about the origin (both functions are odd), there is no crossing for negative x either.
x=0

The line y=x touches the sine curve only at the origin.

x=0

(iv) \tan x=x

Draw y=\tan x and the line y=x on the same axes.
Both pass through the origin, so x=0 is a solution. For 0<x<\frac{\pi}{2} the tangent curve lies above the line: for example \tan1=1.56>1, and it goes to \infty at \frac{\pi}{2}, so there is no second crossing in this branch. By symmetry there is none for -\frac{\pi}{2}<x<0 either. The curve and the line meet only at the origin on the branch through the origin.
x=0
The curve reappears after each asymptote, so the line meets each later branch once more (the first such crossing is near x\approx4.49 and, by symmetry, near -4.49); the book's graph, drawn on [0,\pi], shows only the solution x=0.

Between the asymptotes the tangent curve is above the line for x>0 and below it for x<0; they meet at the origin.

x=0

Book answer. Graphs only: (i) sin x and cos x meet; (ii) y = x meets cos x once (x ≈ 0.74); (iii) y = x meets sin x at the origin; (iv) y = x meets tan x at the origin.

Alternating current cyclically reverses direction. The maximum voltage is about 180 volts when the standard frequency is 56 Hz (56 cycles per second). The voltage can be modeled by V(t)=a\sin(k(t-d))+c. Determine each of the following:

Solution

a. Period of 56 Hz AC

The period is the time for one cycle, the reciprocal of the frequency.
T=\dfrac{1}{f}=\dfrac{1}{56}\ \text{second}
So each cycle lasts \frac{1}{56} of a second.

T=\dfrac{1}{56}\ \text{s}

b. The value of k

In V(t)=a\sin(k(t-d))+c the number k turns the time into an angle: one period must give 360^\circ.
k=\dfrac{360^\circ}{\text{period}}
Put in the period \frac{1}{56} s.
k=\dfrac{360^\circ}{1/56}=360\times56=20\,160

k=20\,160

c. The amplitude of the voltage function

The maximum voltage is 180 V and, since the current reverses direction, the minimum is -180 V.
\text{Max}=180,\quad\text{Min}=-180
Amplitude is half the difference between the maximum and minimum (book's key fact: a=\frac{\text{Maximum}-\text{Minimum}}{2}).
a=\dfrac{180-(-180)}{2}=180

a=180\ \text{V}

d. Model the voltage with an appropriate transformed Sine function.

The midline is halfway between the maximum and the minimum, so the vertical shift is c=\frac{180+(-180)}{2}=0.
The wave starts at zero at t=0, as the sine does, so there is no horizontal shift: d=0.
Substitute a=180, k=20\,160, c=0, d=0 in V(t)=a\sin(k(t-d))+c.
V(t)=180\sin(20\,160\,t)\quad(\text{angle in degrees})

Two cycles of the voltage: each cycle takes \frac{1}{56} s and the voltage swings between -180 V and 180 V.

V(t)=180\sin(20\,160\,t)

Book answer. a. each cycle is 1/56 second; b. k = 20,160; c. 180; d. V(t) = 180 Sin 20,160t.

Note. The key agrees with all four parts (its 'Sin 20°,160t' is \sin(20\,160\,t) with k in degrees per second). If the angle is measured in radians instead, k=2\pi\times56=112\pi\approx351.9 and V(t)=180\sin(112\pi t), the same function.

A Ferris Wheel has a diameter of 17 m. You board at the bottom of the Ferris Wheel from a platform 2 m off the ground. If it takes 25 seconds to reach the top, determine an equation for the height of a rider with respect to time.

Solution

Set-up. The wheel has diameter 17 m, so its radius is 8.5 m. You board at the bottom, which is on a platform 2 m above the ground, so the lowest point of the rider is 2 m.
Highest point and midline. The top is a diameter higher than the bottom, and the centre is halfway.
\text{Highest}=2+17=19\ \text{m},\qquad\text{midline }C=\dfrac{19+2}{2}=10.5\ \text{m}
Amplitude. It is half the difference between the highest and lowest heights.
|A|=\dfrac{19-2}{2}=8.5
Period. Going from the bottom to the top is half a turn. It takes 25 s, so a complete revolution takes twice as long.
\text{Period}=2\times25=50\ \text{s}
**Find B.** Use \text{period}=\dfrac{2\pi}{B}.
B=\dfrac{2\pi}{50}=\dfrac{\pi}{25}
Choose the form. Use h(t)=A\cos(Bt)+C as in the book's Example 10. A cosine starts at its maximum, but the rider starts at the bottom (the minimum), so A is negative: A=-8.5.
h(t)=-8.5\cos\left(\dfrac{\pi}{25}t\right)+10.5
Check. At t=0 the cosine is 1, so h=-8.5+10.5=2 (the platform). At t=25, \cos\pi=-1, so h=8.5+10.5=19 (the top). Both are correct.

One revolution of the wheel: the rider starts at 2 m, reaches 19 m after 25 s and is back at 2 m after 50 s.

h(t)=-8.5\cos\left(\dfrac{\pi}{25}t\right)+10.5

Book answer. h(t) = −8.5 Cos(πt/25) + 10.5

The pendulum of a clock swings with a periodic motion that can be represented by a trigonometric function. At rest, the pendulum is 16 cm above the base. The highest point of the swing is 20 cm above the base, and it take 2 seconds for the pendulum to swing back and forth to its starting point once. Assume the pendulum is released from its highest point. Write a cosine equation that models the height of the pendulum as a function of time.

Solution

Read the data. At rest the pendulum hangs at its lowest point, 16 cm above the base. The highest point of the swing is 20 cm. One complete swing (back and forth to the starting point) takes 2 s, so the period is 2 s.
Midline and amplitude. The midline is halfway between the lowest and highest heights, and the amplitude is half their difference.
C=\dfrac{20+16}{2}=18\ \text{cm},\qquad A=\dfrac{20-16}{2}=2\ \text{cm}
**Find B.** Use \text{period}=\dfrac{2\pi}{B}.
2=\dfrac{2\pi}{B}\;\Rightarrow\;B=\pi
Choose the form. The pendulum is released from its highest point, where a cosine curve with positive coefficient also starts, so A=+2 and there is no phase shift.
h(t)=2\cos(\pi t)+18
Check. h(0)=2+18=20 (highest). h(1)=2\cos\pi+18=16 (lowest, half a period later). h(2)=20 (back at the start).

The pendulum starts at its highest point 20 cm, is lowest (16 cm) after 1 s and is back at the top after 2 s.

h(t)=2\cos(\pi t)+18

Book answer. y = 2 Cos[2πx] + 18

Note. The key's y=2\cos(2\pi x)+18 has period \frac{2\pi}{2\pi}=1 s, which contradicts the 2 s in the question. With a period of 2 s the correct value is B=\frac{2\pi}{2}=\pi, giving h(t)=2\cos(\pi t)+18. The amplitude 2 and the midline 18 in the key are right.

Review Exercise 9p. 246–249

Revision of the whole unit: trigonometric ratios and identities from the earlier units, even and odd functions, periods, maximum and minimum values, domain and range, and modelling with sine and cosine curves.

1 (i)

If \cos\theta=-\dfrac{\sqrt3}{2} and the terminal arm of the angle is in the III quadrant. Then \sin\theta=\ldots

A\dfrac12
B-\dfrac12Answer
C\sqrt3
D-\dfrac{2}{\sqrt3}

Why

Use \sin^2\theta+\cos^2\theta=1 to find the size of \sin\theta.
\sin^2\theta=1-\left(-\dfrac{\sqrt3}{2}\right)^2=1-\dfrac34=\dfrac14\;\Rightarrow\;|\sin\theta|=\dfrac12
In the III quadrant both sine and cosine are negative, so \sin\theta=-\frac12.
\sin\theta=-\dfrac12

Option A has the right size but the wrong sign (it would be the answer in the II quadrant). Options C and D are larger than 1 in size, which no sine can be.

Book answer. b

1 (ii)

The exact value of the trigonometric function \tan(-15\pi)=\ldots

A0Answer
B-1
C1
DUndefined

Why

Tangent is odd, so \tan(-x)=-\tan x.
\tan(-15\pi)=-\tan15\pi
Tangent has period \pi, so \tan(n\pi)=\tan0=0 for every integer n.
\tan15\pi=0\;\Rightarrow\;\tan(-15\pi)=0

It is not undefined: tangent is undefined only at odd multiples of \frac{\pi}{2}, and 15\pi is an even multiple of \frac{\pi}{2}.

Book answer. a

1 (iii)

If 2\sin\theta+\dfrac12\operatorname{cosec}\theta and \theta=45^\circ, then the value of the given trigonometric identity is:

A\dfrac{1}{\sqrt2}
B\dfrac13
C\dfrac{3}{\sqrt2}Answer
D\dfrac{\sqrt2}{3}

Why

At 45^\circ: \sin45^\circ=\dfrac{1}{\sqrt2} and \operatorname{cosec}45^\circ=\sqrt2.
Substitute.
2\left(\dfrac{1}{\sqrt2}\right)+\dfrac12(\sqrt2)=\sqrt2+\dfrac{\sqrt2}{2}
Add, and write the answer in the form of the options (\sqrt2=\frac{2}{\sqrt2}).
\dfrac{3\sqrt2}{2}=\dfrac{3}{\sqrt2}

Book answer. c

1 (iv)

If \sin(270^\circ+\theta)=x and the terminal side of an angle \theta is in IV quadrant, then x=\ldots

A\cos\theta
B-\cos\thetaAnswer
C\sin\theta
D-\sin\theta

Why

Expand with the formula for \sin(A+B).
\sin(270^\circ+\theta)=\sin270^\circ\cos\theta+\cos270^\circ\sin\theta
Put in \sin270^\circ=-1 and \cos270^\circ=0.
=(-1)\cos\theta+0=-\cos\theta

This is an identity true for every \theta. The angle 270^\circ+\theta lies in the III quadrant when \theta is in the IV quadrant, where sine is negative; and -\cos\theta is negative because \cos\theta>0 in the IV quadrant. So the value is -\cos\theta, not \cos\theta.

Book answer. a

Note. The answer key gives (a) \cos\theta. That cannot be right: \sin(270^\circ+\theta)=-\cos\theta for every \theta, and for \theta in the IV quadrant (where \cos\theta>0) the value is negative, so it cannot equal \cos\theta. The correct option is (b).

1 (v)

The trigonometric identity \dfrac{\sin\alpha+\sin2\alpha}{1+\cos\alpha+\cos2\alpha}=\ldots

A\sin\alpha
B\cos\alpha
C\tan\alphaAnswer
D\cot\alpha

Why

Write \sin2\alpha=2\sin\alpha\cos\alpha and \cos2\alpha=2\cos^2\alpha-1.
\dfrac{\sin\alpha+2\sin\alpha\cos\alpha}{1+\cos\alpha+2\cos^2\alpha-1}
Factorise: numerator \sin\alpha(1+2\cos\alpha); denominator \cos\alpha+2\cos^2\alpha=\cos\alpha(1+2\cos\alpha).
=\dfrac{\sin\alpha(1+2\cos\alpha)}{\cos\alpha(1+2\cos\alpha)}
Cancel the common factor 1+2\cos\alpha.
=\dfrac{\sin\alpha}{\cos\alpha}=\tan\alpha

Book answer. c

1 (vi)

Express 2\sin3x\sin7x as a sum or difference:

A\cos4x-\cos10xAnswer
B\cos10x-\cos4x
C\cos4x+\cos10x
D\cos10x+\cos4x

Why

Use the product formula 2\sin A\sin B=\cos(A-B)-\cos(A+B) with A=3x, B=7x.
2\sin3x\sin7x=\cos(3x-7x)-\cos(3x+7x)
Cosine is even, so \cos(-4x)=\cos4x.
=\cos4x-\cos10x

Option B has the two terms in the wrong order, which changes the sign of the whole expression.

Book answer. a

1 (vii)

Express \sin5x+\sin7x as a product:

A2\sin6x\cos xAnswer
B2\sin x\cos6x
C2\cos7x\sin5x
D2\cos5x\sin7x

Why

Use \sin C+\sin D=2\sin\dfrac{C+D}{2}\cos\dfrac{C-D}{2} with C=7x, D=5x.
\sin7x+\sin5x=2\sin\dfrac{7x+5x}{2}\cos\dfrac{7x-5x}{2}
Simplify.
=2\sin6x\cos x

Check with x=\frac{\pi}{12} if in doubt: both sides equal 2\sin\frac{\pi}{2}\cos\frac{\pi}{12}.

Book answer. a

1 (viii)

If \tan A=\dfrac17 and \tan B=\dfrac13, then \cos2A is equal to:

A\sin B
B\sin4BAnswer
C\sin3B
D\cos2B

Why

Use \cos2A=\dfrac{1-\tan^2A}{1+\tan^2A}.
\cos2A=\dfrac{1-\frac{1}{49}}{1+\frac{1}{49}}=\dfrac{48}{50}=\dfrac{24}{25}
Now find \sin2B and \cos2B from \tan B=\frac13 using \sin2B=\dfrac{2\tan B}{1+\tan^2B} and \cos2B=\dfrac{1-\tan^2B}{1+\tan^2B}.
\sin2B=\dfrac{2/3}{10/9}=\dfrac35,\qquad\cos2B=\dfrac{8/9}{10/9}=\dfrac45
Double again: \sin4B=2\sin2B\cos2B.
\sin4B=2\cdot\dfrac35\cdot\dfrac45=\dfrac{24}{25}
So \cos2A=\sin4B.

The others are not \frac{24}{25}: \sin B=\frac{1}{\sqrt{10}}\approx0.32 and \cos2B=\frac45.

Book answer. b

1 (ix)

Whether the function f(x)=\dfrac{\sin^3x}{x^2+\tan x} is:

Aeven
Bodd
Cneither even nor oddAnswer
Dboth even and odd

Why

Replace x by -x.
f(-x)=\dfrac{\sin^3(-x)}{(-x)^2+\tan(-x)}=\dfrac{-\sin^3x}{x^2-\tan x}
Compare with f(x)=\dfrac{\sin^3x}{x^2+\tan x}. The denominator changes from x^2+\tan x to x^2-\tan x (it is a mixture of an even and an odd term), so f(-x) is neither f(x) nor -f(x).
f(-x)\ne f(x)\ \text{and}\ f(-x)\ne-f(x)

A quick check: at x=1, f(1)=\frac{0.596}{1+1.557}=0.233 while f(-1)=\frac{-0.596}{1-1.557}=1.07, which is neither 0.233 nor -0.233.

Book answer. c

1 (x)

The period of \cos\dfrac{x}{5} is:

A10\piAnswer
B\dfrac{2\pi}{5}
C2\pi
D4\pi

Why

The period of \cos bx is \dfrac{2\pi}{|b|}. Here b=\frac15.
\text{Period}=\dfrac{2\pi}{1/5}=10\pi

Option B is what you get by wrongly using b=5.

Book answer. a

1 (xi)

2\cos5x\cdot\sin3x=\ldots

A\sin8x-\sin2xAnswer
B\sin8x+\sin2x
C\cos8x+\cos2x
D\sin4x-\sin x

Why

Use the product formula 2\cos A\sin B=\sin(A+B)-\sin(A-B) with A=5x, B=3x.
2\cos5x\sin3x=\sin(5x+3x)-\sin(5x-3x)
Simplify.
=\sin8x-\sin2x

Book answer. a

1 (xii)

The trigonometric functions which are even and having period =2\pi are:

A\sin x & \cos x
B\sec x & \cos xAnswer
C\sin x & \operatorname{cosec}x
D\tan x & \cot x

Why

The even trigonometric functions are cosine and its reciprocal secant, because \cos(-x)=\cos x and \sec(-x)=\sec x. Both have period 2\pi.

Option A has \sin x which is odd; option C has two odd functions; option D has two odd functions with period \pi.

Book answer. b

1 (xiii)

If f is a periodic function and its period is \pi, then f(\theta) could be equal to:

A2\cos x
B2\cos3x
C3\cos2xAnswer
D\cos4x

Why

The period of \cos bx is \dfrac{2\pi}{|b|}. Find it for each option.
\text{A: }2\pi,\quad\text{B: }\dfrac{2\pi}{3},\quad\text{C: }\dfrac{2\pi}{2}=\pi,\quad\text{D: }\dfrac{2\pi}{4}=\dfrac{\pi}{2}
Only 3\cos2x has period \pi (the factor 3 does not change the period).

Book answer. c

1 (xiv)

If function f(x)=\sin8x is a periodic function and its period equals:

A\pi
B\dfrac{\pi}{4}Answer
C2\pi
D\dfrac{\pi}{2}

Why

The period of \sin bx is \dfrac{2\pi}{|b|} with b=8.
\text{Period}=\dfrac{2\pi}{8}=\dfrac{\pi}{4}

Book answer. b

1 (xv)

If the range of the function f(\theta)=a\sin(2\theta)+b, where a>0, is [3,5], then 3a+2b=\ldots

A11Answer
B14
C9
D5

Why

Since -1\le\sin2\theta\le1 and a>0, the function runs from b-a to b+a.
\text{Range}=[b-a,\ b+a]=[3,5]
Set up two equations and add them.
b+a=5,\quad b-a=3\;\Rightarrow\;2b=8\;\Rightarrow\;b=4
Then find a.
a=5-4=1
Substitute.
3a+2b=3(1)+2(4)=11

Book answer. a

1 (xvi)

The minimum value of the trigonometric function f(\theta)=17\sin(4\theta) is:

A4
B-4
C-17Answer
D17

Why

Here a=0 and b=17 in the form a+b\sin(c\theta+d), so m=a-|b|.
m=0-17=-17

The 4 inside the sine changes only the period. 17 is the maximum value, not the minimum.

Book answer. c

1 (xvii)

If the given figure represent the curve y=3\sin x, then |a|+|b|=\ldots

The figure of the question: the maximum is at \left(\frac{\pi}{2},a\right) and the minimum at \left(\frac{3\pi}{2},b\right).

A1
B2
C3
D6Answer

Why

The point \left(\frac{\pi}{2},a\right) is the top of the wave and \left(\frac{3\pi}{2},b\right) is the bottom. Substitute the x-values in y=3\sin x.
a=3\sin\dfrac{\pi}{2}=3,\qquad b=3\sin\dfrac{3\pi}{2}=-3
Add the absolute values.
|a|+|b|=3+3=6

Option C (3) is the amplitude alone; the answer needs both the maximum and the minimum height.

Book answer. d

1 (xviii)

The maximum value of 7\cos x+24\sin x is:

A25Answer
B-25
C7
D24

Why

Write the expression as a single cosine. Put 7=R\cos\alpha and 24=R\sin\alpha; squaring and adding gives R.
R=\sqrt{7^2+24^2}=\sqrt{49+576}=\sqrt{625}=25
Then 7\cos x+24\sin x=25\cos(x-\alpha), with \tan\alpha=\frac{24}{7}. A cosine has maximum value 1, so the maximum is 25\times1=25.
\text{Max}=25

The maximum is not 7+24=31, because the two terms reach their own maxima at different values of x. And -25 is the minimum.

Book answer. a

If \cos\theta-\sin\theta=\sqrt2\sin\theta, then show that \cos\theta+\sin\theta=\sqrt2\cos\theta.

Solution

Given.
\cos\theta-\sin\theta=\sqrt2\sin\theta
Move -\sin\theta to the right side, so that all the sine terms are together.
\cos\theta=\sqrt2\sin\theta+\sin\theta=(\sqrt2+1)\sin\theta
Divide both sides by \cos\theta and by (\sqrt2+1) to get \tan\theta.
\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{1}{\sqrt2+1}
Rationalise by multiplying top and bottom by \sqrt2-1 (note (\sqrt2+1)(\sqrt2-1)=2-1=1).
\tan\theta=\dfrac{\sqrt2-1}{(\sqrt2+1)(\sqrt2-1)}=\sqrt2-1\quad\Rightarrow\quad\sin\theta=(\sqrt2-1)\cos\theta
To show. Start with the left side of \cos\theta+\sin\theta=\sqrt2\cos\theta and replace \sin\theta.
\cos\theta+\sin\theta=\cos\theta+(\sqrt2-1)\cos\theta
Collect the \cos\theta terms.
=\left[1+(\sqrt2-1)\right]\cos\theta=\sqrt2\cos\theta
This is the right side, so \cos\theta+\sin\theta=\sqrt2\cos\theta. Hence proved.

Verify the following Trigonometric identities:

Solution

(a) \dfrac{\tan x-\cot x}{\sin x\cos x}=\sec^2x-\operatorname{cosec}^2x

Start with the left side. Write \tan x and \cot x in terms of \sin x and \cos x.
\text{LHS}=\dfrac{\dfrac{\sin x}{\cos x}-\dfrac{\cos x}{\sin x}}{\sin x\cos x}
Combine the numerator over the common denominator \sin x\cos x.
=\dfrac{\dfrac{\sin^2x-\cos^2x}{\sin x\cos x}}{\sin x\cos x}=\dfrac{\sin^2x-\cos^2x}{\sin^2x\cos^2x}
Split into two fractions.
=\dfrac{\sin^2x}{\sin^2x\cos^2x}-\dfrac{\cos^2x}{\sin^2x\cos^2x}=\dfrac{1}{\cos^2x}-\dfrac{1}{\sin^2x}
Use \sec x=\frac{1}{\cos x} and \operatorname{cosec}x=\frac{1}{\sin x}.
=\sec^2x-\operatorname{cosec}^2x=\text{RHS}

\text{LHS}=\text{RHS}

(b) \dfrac{\sec^4x-\tan^4x}{\sec^2x+\tan^2x}=\sec^2x-\tan^2x

Start with the left side. The numerator is a difference of squares, \sec^4x-\tan^4x=(\sec^2x)^2-(\tan^2x)^2.
\text{LHS}=\dfrac{(\sec^2x-\tan^2x)(\sec^2x+\tan^2x)}{\sec^2x+\tan^2x}
Cancel the common factor \sec^2x+\tan^2x (it is never zero, as \sec^2x\ge1).
=\sec^2x-\tan^2x=\text{RHS}

Also \sec^2x-\tan^2x=1, so both sides equal 1.

\text{LHS}=\text{RHS}

Start with the left side and take the common denominator (1-\cos t)(1+\cos t).
\text{LHS}=\dfrac{\sin t(1+\cos t)-\sin t\cos t(1-\cos t)}{(1-\cos t)(1+\cos t)}
Take \sin t out of the numerator and multiply out the brackets.
=\dfrac{\sin t\left[1+\cos t-\cos t+\cos^2t\right]}{1-\cos^2t}=\dfrac{\sin t\,(1+\cos^2t)}{1-\cos^2t}
Use 1-\cos^2t=\sin^2t in the denominator.
=\dfrac{\sin t\,(1+\cos^2t)}{\sin^2t}=\dfrac{1+\cos^2t}{\sin t}
Write \frac{1}{\sin t}=\operatorname{cosec}t.
=\operatorname{cosec}t\,(1+\cos^2t)=\text{RHS}

\text{LHS}=\text{RHS}

Prove that:

Solution

(i) \dfrac{\tan(\alpha+\beta)-\tan\beta}{1+\tan(\alpha+\beta)\tan\beta}=\tan\alpha

Recall the formula for the tangent of a difference.
\tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}
The left side has exactly this form with A=\alpha+\beta and B=\beta.
\text{LHS}=\tan\left[(\alpha+\beta)-\beta\right]
Simplify the angle.
=\tan\alpha=\text{RHS}

\text{LHS}=\text{RHS}

(ii) \dfrac{1+\sin2\theta-\cos2\theta}{1+\sin2\theta+\cos2\theta}=\tan\theta

Use the double-angle identities 1-\cos2\theta=2\sin^2\theta, 1+\cos2\theta=2\cos^2\theta and \sin2\theta=2\sin\theta\cos\theta. Group the numerator as (1-\cos2\theta)+\sin2\theta and the denominator as (1+\cos2\theta)+\sin2\theta.
\text{LHS}=\dfrac{2\sin^2\theta+2\sin\theta\cos\theta}{2\cos^2\theta+2\sin\theta\cos\theta}
Take out the common factors 2\sin\theta and 2\cos\theta.
=\dfrac{2\sin\theta(\sin\theta+\cos\theta)}{2\cos\theta(\cos\theta+\sin\theta)}
Cancel 2(\sin\theta+\cos\theta).
=\dfrac{\sin\theta}{\cos\theta}=\tan\theta=\text{RHS}

\text{LHS}=\text{RHS}

A Ferris wheel is 40 meters in diameter and boarded from a platform that is 4 meter above the ground. The six o'clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 16 minutes. The function h(t) gives a person's height in meters above the ground t minutes after the wheel begins to turn.

Solution

a. Find the period, amplitude and vertical shift of h(t).

The period is the time for one full revolution.
\text{Period}=16\ \text{minutes}
The amplitude is half the difference between the highest and lowest heights (equal to the radius).
\text{Amplitude}=\dfrac{44-4}{2}=20\ \text{m}
The vertical shift is the midline, the height of the centre of the wheel.
\text{Vertical shift}=\dfrac{44+4}{2}=24\ \text{m}

\text{Period }16\text{ min},\ \text{amplitude }20\text{ m},\ \text{vertical shift }24\text{ m}

b. Find a formula for the height function h(t).

Use h(t)=A\cos(Bt)+C (book's Example 10). The vertical shift gives C=24 and the amplitude gives |A|=20.
C=24,\quad|A|=20
The rider starts at the lowest point (the minimum), where a cosine with positive coefficient would be at its maximum. So A is negative.
A=-20
Find B from the period \dfrac{2\pi}{B}=16.
B=\dfrac{2\pi}{16}=\dfrac{\pi}{8}
Substitute.
h(t)=-20\cos\left(\dfrac{\pi t}{8}\right)+24
Check: h(0)=-20+24=4 (platform), h(8)=20+24=44 (top), h(16)=4 (back at the start).

One revolution (16 minutes): the height rises from 4 m to 44 m and back.

h(t)=-20\cos\left(\dfrac{\pi t}{8}\right)+24

c. How high off the ground is a person after 5 minutes?

Put t=5 in the formula.
h(5)=-20\cos\left(\dfrac{5\pi}{8}\right)+24
\frac{5\pi}{8}=112.5^\circ is in the second quadrant, so the cosine is negative: \cos\frac{5\pi}{8}=-0.3827 (calculator in radian mode).
h(5)=-20(-0.3827)+24=7.65+24
Add.
h(5)\approx31.7\ \text{m}\approx32\ \text{m}

h(5)\approx31.7\ \text{m}\approx32\text{ m}

Book answer. a. Period = π/6; amplitude = −20; vertical shift = 24. b. h(t) = −20 Cos(πt/8) + 24. c. The height is 32 m after 5 minutes.

Note. The key prints the period as \frac{\pi}{6}, which is a misprint: the question says one revolution takes 16 minutes, so the period is 16 minutes (and then B=\frac{\pi}{8}, as the key's own formula for h(t) uses). The key writes the amplitude as -20; the amplitude is 20, and the - sign is the coefficient A in the formula. Parts b and c agree (32 m is 31.65 m rounded to the nearest metre).

The h (in meters) above the ground of a rider on a Ferris wheel, t (in seconds) after the rider begins is given: h(t)=10\sin(3(t-30))+12. Determine each of the following:

Solution

a. The maximum and minimum heights of the rider above the ground.

Compare with h(t)=a\sin[k(t-d)]+c: here a=10, k=3, d=30, c=12.
The sine factor lies between -1 and 1, so M=c+|a| and m=c-|a|.
\text{Max}=12+10=22\ \text{m},\qquad\text{Min}=12-10=2\ \text{m}

\text{Max}=22\text{ m},\ \text{Min}=2\text{ m}

b. The height of the rider above the ground 30 seconds after start.

Put t=30 in h(t)=10\sin(3(t-30))+12.
h(30)=10\sin(3\cdot0)+12
\sin0=0.
h(30)=0+12=12\ \text{m}

h(30)=12\ \text{m}

c. The time required for the Ferris wheel to complete one complete revolution.

The period is the time for one revolution. As in the book's Example 11, the angle is measured in degrees, so one cycle is 360^\circ and k=\dfrac{360^\circ}{\text{period}}.
Solve for the period with k=3.
\text{Period}=\dfrac{360}{k}=\dfrac{360}{3}=120\ \text{seconds}

Two revolutions: the height repeats every 120 s and stays between 2 m and 22 m.

120\ \text{s}

Book answer. a. Maximum height = 22 m, minimum height = 2 m. b. The height is 12 m after 30 seconds. c. One complete revolution takes place in 120 seconds.

Note. The key's 120 s for part c needs the angle 3(t-30) to be in degrees (\frac{360}{3}=120), the same convention as the book's Example 11. If the angle were in radians the period would be \frac{2\pi}{3}\approx2.1 s, far too short for a Ferris wheel, so degrees are intended. Parts a and b do not depend on the unit.

The top of the flagpole sways back and forth in high winds. The top sways 8 cm to the right (+8 cm) and 8 cm to the left (-8 cm) of its resting position. It moves back and forth 260 every minute. The pole was momentarily at its rest position at t=0, before it started moving to the right. Find:

Solution

a. The equation of the sinusoidal function that describes the distance the top of the pole is from resting position in terms of time elapsed.

Let y be the displacement (cm) from the resting position, positive to the right, at time t seconds. Since the pole is at rest position at t=0 and moves to the right, y starts at 0 and increases: this is a sine curve, y=a\sin kt (no phase shift, no vertical shift).
Amplitude. It sways 8 cm each side.
a=8
Period. The pole moves back and forth 260 times each minute, that is \dfrac{260}{60}=\dfrac{13}{3} cycles each second, so one cycle takes
\text{Period}=\dfrac{60}{260}=\dfrac{3}{13}\ \text{s}
**Find k.** As in the book's Example 11 (angle in degrees), k=\dfrac{360^\circ}{\text{period}}.
k=\dfrac{360}{3/13}=360\times\dfrac{13}{3}=1560
Write the equation.
y=8\sin(1560\,t)\quad(\text{angle in degrees})
In radians the same function is y=8\sin\left(\dfrac{26\pi}{3}t\right), because k=2\pi\times\dfrac{13}{3}=\dfrac{26\pi}{3}.

Two complete swings: the top moves to +8 cm, back through the rest position to -8 cm, and returns, once every \frac{3}{13} s.

y=8\sin(1560\,t)

b. The domain and range corresponding to the situation described.

Domain. Time elapsed cannot be negative, and the swaying goes on, so every time from the start onwards is allowed.
\text{Domain}=\{t:\ t\ge0,\ t\in\mathbb{R}\}
Range. The displacement is between -8 cm and 8 cm.
\text{Range}=\{y:\ -8\le y\le8,\ y\in\mathbb{R}\}

t\ge0,\quad -8\le y\le8

Book answer. a. y = 10 Sin 1440 t°, y = 10 Sin 1440 (t − 1/16)°; b. Domain = {t | t ≥ 0, t ∈ R}, Range = {y | −10 ≤ y ≤ 10, y ∈ R}

Note. The key does not match the printed question. The question gives an amplitude of 8 cm and 260 movements a minute, which give y=8\sin(1560\,t) and range [-8,8]. The key's y=10\sin(1440\,t) would be for 10 cm and \frac{1440}{360}\times60=240 movements a minute. Its domain t\ge0 agrees. The method is the same, so with the key's numbers one gets its answer.

Find the domain, range, and period of the following trigonometric functions:

Solution

(i) 2\sin\dfrac{x}{3}

Sine is defined for every real number.
\text{Domain}=\mathbb{R}
Range: -1\le\sin\frac{x}{3}\le1; multiply by 2.
-2\le2\sin\dfrac{x}{3}\le2\;\Rightarrow\;\text{Range}=[-2,2]
Period \dfrac{2\pi}{|b|} with b=\frac13.
\text{Period}=\dfrac{2\pi}{1/3}=6\pi

\text{Dom}=\mathbb{R},\ \text{Range}=[-2,2],\ \text{Period}=6\pi

(ii) 5\sin3x

Defined for all real x.
\text{Domain}=\mathbb{R}
Multiply -1\le\sin3x\le1 by 5.
\text{Range}=[-5,5]
Period with b=3.
\text{Period}=\dfrac{2\pi}{3}

\text{Dom}=\mathbb{R},\ \text{Range}=[-5,5],\ \text{Period}=\dfrac{2\pi}{3}

(iii) \dfrac12\sin\dfrac{2x}{3}

Defined for all real x.
\text{Domain}=\mathbb{R}
Multiply -1\le\sin\frac{2x}{3}\le1 by \frac12.
\text{Range}=\left[-\dfrac12,\dfrac12\right]
Period with b=\frac23.
\text{Period}=\dfrac{2\pi}{2/3}=3\pi

\text{Dom}=\mathbb{R},\ \text{Range}=\left[-\tfrac12,\tfrac12\right],\ \text{Period}=3\pi

(iv) \dfrac53\cos\dfrac{4x}{3}

Defined for all real x.
\text{Domain}=\mathbb{R}
Multiply -1\le\cos\frac{4x}{3}\le1 by \frac53.
\text{Range}=\left[-\dfrac53,\dfrac53\right]
Period with b=\frac43.
\text{Period}=\dfrac{2\pi}{4/3}=\dfrac{3\pi}{2}

\text{Dom}=\mathbb{R},\ \text{Range}=\left[-\tfrac53,\tfrac53\right],\ \text{Period}=\dfrac{3\pi}{2}

(v) 3\sin\pi x

Defined for all real x.
\text{Domain}=\mathbb{R}
Multiply -1\le\sin\pi x\le1 by 3.
\text{Range}=[-3,3]
Period with b=\pi.
\text{Period}=\dfrac{2\pi}{\pi}=2

\text{Dom}=\mathbb{R},\ \text{Range}=[-3,3],\ \text{Period}=2

(vi) 7\sin5x

Defined for all real x.
\text{Domain}=\mathbb{R}
Multiply -1\le\sin5x\le1 by 7.
\text{Range}=[-7,7]
Period with b=5.
\text{Period}=\dfrac{2\pi}{5}

\text{Dom}=\mathbb{R},\ \text{Range}=[-7,7],\ \text{Period}=\dfrac{2\pi}{5}

(vii) \dfrac32\cot\dfrac{2\pi}{3}x

Cotangent fails where its angle is a multiple of \pi. The angle is \frac{2\pi}{3}x.
\dfrac{2\pi}{3}x=n\pi\;\Rightarrow\;x=\dfrac{3n}{2},\quad n\in\mathbb{Z}
Remove these values.
\text{Domain}=\mathbb{R}-\left\{\dfrac{3n}{2}:\,n\in\mathbb{Z}\right\}
Cotangent takes every real value, so the multiple \frac32\cot(\cdot) does as well.
\text{Range}=\mathbb{R}
The period of \cot bx is \dfrac{\pi}{|b|} with b=\frac{2\pi}{3}.
\text{Period}=\dfrac{\pi}{2\pi/3}=\dfrac32

\text{Dom}=\mathbb{R}-\left\{\tfrac{3n}{2}\right\},\ \text{Range}=\mathbb{R},\ \text{Period}=\dfrac32

(viii) 9\cos(3x-2)

Defined for all real x.
\text{Domain}=\mathbb{R}
Multiply -1\le\cos(3x-2)\le1 by 9.
\text{Range}=[-9,9]
Period with b=3 (the -2 is only a shift).
\text{Period}=\dfrac{2\pi}{3}

\text{Dom}=\mathbb{R},\ \text{Range}=[-9,9],\ \text{Period}=\dfrac{2\pi}{3}

(ix) 8-\cos4x

Defined for all real x.
\text{Domain}=\mathbb{R}
Compare with a+b\cos\theta: a=8, b=-1, so M=8+1=9 and m=8-1=7.
\text{Range}=[7,9]
Period with b=4.
\text{Period}=\dfrac{2\pi}{4}=\dfrac{\pi}{2}

\text{Dom}=\mathbb{R},\ \text{Range}=[7,9],\ \text{Period}=\dfrac{\pi}{2}

(x) 7+5\sin\left(2x-\dfrac{\pi}{6}\right)

Defined for all real x.
\text{Domain}=\mathbb{R}
Compare with a+b\sin(c\theta+d): a=7, b=5, so M=12 and m=2.
\text{Range}=[2,12]
Period with c=2 (the shift -\frac{\pi}{6} does not matter).
\text{Period}=\dfrac{2\pi}{2}=\pi

\text{Dom}=\mathbb{R},\ \text{Range}=[2,12],\ \text{Period}=\pi

(xi) 6+4\cos\left(2x+\dfrac{\pi}{3}\right)

Defined for all real x.
\text{Domain}=\mathbb{R}
Here a=6, b=4, so M=10 and m=2.
\text{Range}=[2,10]
Period with c=2.
\text{Period}=\dfrac{2\pi}{2}=\pi

\text{Dom}=\mathbb{R},\ \text{Range}=[2,10],\ \text{Period}=\pi

Book answer. (i) R, [−2, 2], 6π; (ii) R, [−5, 5], 2π/3; (iii) R, [−1/2, 1/2], 3π; (iv) R, [−5/3, 5/3], 3π/2; (v) R, [−3, 3], 2; (vi) R, [−7, 7], 2π/5; (vii) R − {3n/2}, R, 3/2; (viii) R, [−9, 9], 2π/3; (ix) R, [7, 9], π/2; (x) R, [2, 12], π; (xi) R, [2, 10], π.

Note. In (vii) the printed expression \frac32\cot\frac{2\pi}{3}x is read as \frac32\cot\left(\frac{2\pi}{3}x\right), which is what the key's domain \mathbb{R}-\frac{3n}{2} and period \frac32 correspond to.

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, MCQ 11 mark

Period of \tan\dfrac{5x}{7} is:

A\dfrac{5\pi}{7}
B5\pi
C\dfrac{7\pi}{5}Answer
D7\pi

Why

The period of \tan x is \pi, and the period of \tan ax is \dfrac{\pi}{|a|}. Here a=\dfrac57.

\text{Period}=\dfrac{\pi}{\frac57}=\pi\cdot\dfrac75=\dfrac{7\pi}{5}

Option A, \dfrac{5\pi}{7}, multiplies by a instead of dividing, and D, 7\pi, ignores the 5.

Note. This is part (iii) of Example 2 of Section 9.2.

2026 Annual I, Set B, MCQ 21 mark

A function is periodic if:

Af(x)=f(-x)
Bf(x)=1
Cf(x)=0
Df(x+T)=f(x)Answer

Why

A function is periodic if there is a number T>0 with f(x+T)=f(x) for every x in the domain; the smallest such T is the period.

Option A, f(x)=f(-x), is the condition for an even function. B and C are constant functions, which are periodic only in a trivial way and are not the definition.

2026 Annual I, Set B, MCQ 31 mark

Range of a+b\sin 2\theta is:

Aa\pm bAnswer
Ba+b
Ca-b
D2(a\pm b)

Why

Since -1\le\sin2\theta\le1, the term b\sin2\theta runs from -|b| to |b|, so a+b\sin2\theta has minimum a-|b| and maximum a+|b|.

\text{Range}=[a-|b|,\ a+|b|]\quad\text{that is, } a\pm|b|

The factor 2 in 2\theta changes the period only, not the range. Option D multiplies the whole range by 2, and B and C give only one end.

Note. The option is printed a\pm b; it is read as a\pm|b|.

2025 Annual I, Set B, MCQ 161 mark

If graph of a function is symmetric about origin (e.g. \sin x), what type of function is it?

AEven
BOddAnswer
CNeither Even nor Odd
DPeriodic but not Symmetric

Why

A graph symmetric about the origin satisfies f(-x)=-f(x) and is an odd function; \sin x is the standard example, \sin(-x)=-\sin x.

An even function (A) has a graph symmetric about the y-axis.

2025 Annual I, Set B, MCQ 171 mark

The range of a trigonometric function 4\sin 4x is:

A[-1,1]
B[-4,4]Answer
C[-16,16]
D(-\infty,+\infty)

Why

The range of a\sin bx is [-|a|,|a|]. Here a=4, so -4\le4\sin4x\le4.

The factor 4 inside, in \sin4x, changes the period (to \dfrac\pi2) but not the range. Option C wrongly multiplies the two 4's to get 16.

2026 Annual I, Set B, Q.2 (i) OR4 marks

Sketch the graph of y=2\cos x for -\pi\le x\le\pi.

Solution

The cosine has period 2\pi and amplitude 2 here, so the graph on [-\pi,\pi] is one complete cycle. It is symmetric about the y-axis (cosine is even).
Make a table of values with x in steps of \dfrac\pi6, using y=2\cos x.

x	-\pi	-\dfrac{5\pi}{6}	-\dfrac{2\pi}{3}	-\dfrac{\pi}{2}	-\dfrac{\pi}{3}	-\dfrac{\pi}{6}	0	\dfrac{\pi}{6}	\dfrac{\pi}{3}	\dfrac{\pi}{2}	\dfrac{2\pi}{3}	\dfrac{5\pi}{6}	\pi
y=2\cos x	-2	-1.73	-1	0	1	1.73	2	1.73	1	0	-1	-1.73	-2

Plot the points (-\pi,-2), \left(-\dfrac\pi2,0\right), (0,2), \left(\dfrac\pi2,0\right), (\pi,-2) and the others, and join them with a smooth sinusoidal curve between y=-2 and y=2.

Graph of y=2\cos x for -\pi\le x\le\pi: amplitude 2, maximum (0,2), minima (\pm\pi,-2), zeros at x=\pm\dfrac\pi2.

Note. Exercise 9.2, question 2, draws graphs of the same kind, for example y=3\cos x in part (vi).

2025 Annual I, Set B, Q.2 (iii) OR4 marks

Find the maximum and minimum values of the function f(\theta)=\dfrac{1}{3+5\cos(2\theta+\pi)}.

Solution

The denominator 3+5\cos(2\theta+\pi) is a+b\cos(c\theta+d) with a=3, b=5.
M_d=3+5=8,\qquad m_d=3-5=-2
Take the reciprocals of the extreme values of the denominator.
M=\dfrac1{M_d}=\dfrac18,\qquad m=\dfrac1{m_d}=-\dfrac12

\text{Maximum}=\dfrac18,\qquad\text{Minimum}=-\dfrac12

Note. The true range of f is \left(-\infty,-\dfrac12\right]\cup\left[\dfrac18,\infty\right).

2025 Annual I, Set B, Q.2 (v)4 marks

Without drawing graph find range, amplitude, period and frequency of the function y=-4\cos(7x-\pi).

Solution

Compare y=-4\cos(7x-\pi) with y=a\cos(bx+c): a=-4, b=7.
Range. -1\le\cos(7x-\pi)\le1; multiplying by -4 turns the interval round, but the set of values is the same.
\text{Range}=[-4,4]
Amplitude =|a|.
\text{Amplitude}=4
Period =\dfrac{2\pi}{|b|}.
\text{Period}=\dfrac{2\pi}7
Frequency is the number of cycles in one unit of x, the reciprocal of the period.
\text{Frequency}=\dfrac1{\text{Period}}=\dfrac{7}{2\pi}

\text{Range}=[-4,4],\ \text{Amplitude}=4,\ \text{Period}=\dfrac{2\pi}{7},\ \text{Frequency}=\dfrac{7}{2\pi}

Note. The book uses frequency only in Section 9.6 (cycles per second); here it is the number of cycles per unit of x.

2026 Annual I, Set B, Q.2 (viii) OR2+2 = 4 marks

Find the maximum and minimum values of: y=\dfrac{1}{4+5\cos 3x}.

Solution

Work with the denominator 4+5\cos3x, which is a+b\cos(cx+d) with a=4, b=5, c=3.
Its greatest and least values are a+|b| and a-|b|.
M_d=4+5=9,\qquad m_d=4-5=-1
Take the reciprocals of these extreme values, as in the book's Example 7(iii).
M=\dfrac1{M_d}=\dfrac19,\qquad m=\dfrac1{m_d}=\dfrac1{-1}=-1

\text{Maximum}=\dfrac19,\qquad\text{Minimum}=-1

Note. The denominator is 0 when \cos3x=-\dfrac45, so the true range of y is (-\infty,-1]\cup\left[\dfrac19,\infty\right).

2026 Annual I, Set B, Q.2 (x)4 marks

Determine the properties of the graph of: y=\cos x i.e. Domain, Range, Period and Amplitude.

Solution

Domain. Cosine is defined for every real angle.
\text{Domain}=(-\infty,\infty)=\mathbb R
Range. -1\le\cos x\le1 for every x, and both values are reached.
\text{Range}=[-1,1]
Period. \cos(x+2\pi)=\cos x, and no smaller positive number works.
\text{Period}=2\pi
Amplitude. Half the difference between the greatest and least values, \dfrac{1-(-1)}{2}.
\text{Amplitude}=1

Property	y=\cos x
Domain	(-\infty,\infty)=\mathbb R
Range	[-1,1]
Period	2\pi
Amplitude	1
Nature	even function, symmetric about the y-axis

Graph of y=\cos x. One cycle is [0,2\pi], and the curve repeats every 2\pi.

2025 Annual I, Set B, Q.2 (x) OR (a)

A solar panel's sunlight reception is modeled as P(\theta)=100\cos(3\theta-90^\circ), with 0^\circ\le\theta\le 180^\circ. Find angle for the maximum sunlight and the max sunlight received.

Solution

P(\theta)=100\cos(3\theta-90^\circ) is largest when \cos(3\theta-90^\circ)=1, so the maximum sunlight is 100.
The cosine equals 1 at angles 0^\circ,360^\circ,720^\circ,\dots Since 0^\circ\le\theta\le180^\circ, the angle 3\theta-90^\circ lies between -90^\circ and 450^\circ, so only 0^\circ and 360^\circ occur.
3\theta-90^\circ=0^\circ\ \Rightarrow\ \theta=30^\circ;\qquad3\theta-90^\circ=360^\circ\ \Rightarrow\ \theta=150^\circ

\text{Maximum }P=100\ \text{at }\theta=30^\circ\text{ and }150^\circ

Note. Since \cos(3\theta-90^\circ)=\sin3\theta, the same results follow from \sin3\theta=1.

2025 Annual I, Set B, Q.2 (x) OR (b)

A solar panel's sunlight reception is modeled as P(\theta)=100\cos(3\theta-90^\circ), with 0^\circ\le\theta\le 180^\circ. Find angle for the minimum sunlight and the min sunlight received.

Solution

P(\theta) is smallest when \cos(3\theta-90^\circ)=-1, so the minimum is -100.
The cosine equals -1 at \pm180^\circ,540^\circ,\dots In the range -90^\circ\le3\theta-90^\circ\le450^\circ only 180^\circ occurs.
3\theta-90^\circ=180^\circ\ \Rightarrow\ 3\theta=270^\circ\ \Rightarrow\ \theta=90^\circ

\text{Minimum }P=-100\ \text{at }\theta=90^\circ

Note. A negative value of P cannot be physical sunlight received; the model gives -100 as the least value of the function.

2026 Annual I, Set B, Q.4 OR8 marks

In a robotics project, a mechanical arm moves in such a way that its vertical position h (in meters) is modeled by h(t)=15\cos\left(\dfrac t2\right), where t is time in seconds. Determine the first time t when the vertical position of the arm reaches 7.5 meters. Also determine the value of t for maximum height.

Solution

The height is h(t)=15\cos\dfrac t2 metres, with t in seconds. The amplitude is 15 and the period is \dfrac{2\pi}{1/2}=4\pi seconds.
**First time at 7.5\,\mathrm m.** Put h=7.5.
15\cos\dfrac t2=7.5\ \Rightarrow\ \cos\dfrac t2=\dfrac12
The cosine equals \dfrac12 first when its angle is \dfrac\pi3 (the cosine falls from 1 at t=0).
\dfrac t2=\dfrac\pi3\ \Rightarrow\ t=\dfrac{2\pi}{3}\approx2.09\ \text{s}
Maximum height. The greatest value of 15\cos\dfrac t2 is 15, when \cos\dfrac t2=1, that is \dfrac t2=0,2\pi,4\pi,\dots
t=0,\ 4\pi,\ 8\pi,\dots\ \text{seconds}
So the arm is at its maximum height 15\,\mathrm m at t=0 and again every 4\pi\approx12.57 s.

The height h(t)=15\cos\dfrac t2. It is 7.5 m at t=\dfrac{2\pi}3 and reaches 15 m again at t=4\pi.

t=\dfrac{2\pi}{3}\ \text{s}\approx2.09\ \text{s};\quad\text{maximum height at }t=0,\ 4\pi,\ 8\pi,\dots

Note. 'The value of t for maximum height' is read as the times at which h=15; the first is t=0 (the start) and the next is t=4\pi s.

2025 Annual I, Set B, Q.5 OR (a)

For y=3\cos 2x; -\dfrac{\pi}{2}\le x\le\dfrac{\pi}{2}, make table of values for given interval.

Solution

The period of y=3\cos2x is \dfrac{2\pi}{2}=\pi, so \left[-\dfrac\pi2,\dfrac\pi2\right] is exactly one cycle. Take the x-values 0,\pm\dfrac\pi6,\pm\dfrac\pi4,\pm\dfrac\pi3,\pm\dfrac\pi2 and find 2x, \cos2x and 3\cos2x.

x	-\dfrac{\pi}{2}	-\dfrac{\pi}{3}	-\dfrac{\pi}{4}	-\dfrac{\pi}{6}	0	\dfrac{\pi}{6}	\dfrac{\pi}{4}	\dfrac{\pi}{3}	\dfrac{\pi}{2}
2x	-\pi	-\dfrac{2\pi}{3}	-\dfrac{\pi}{2}	-\dfrac{\pi}{3}	0	\dfrac{\pi}{3}	\dfrac{\pi}{2}	\dfrac{2\pi}{3}	\pi
\cos2x	-1	-\dfrac12	0	\dfrac12	1	\dfrac12	0	-\dfrac12	-1
y=3\cos2x	-3	-1.5	0	1.5	3	1.5	0	-1.5	-3

Note. Section 9.4 of the book works y=3\cos2x on [0,\pi] with the same values.

2025 Annual I, Set B, Q.5 OR (b)

For y=3\cos 2x; -\dfrac{\pi}{2}\le x\le\dfrac{\pi}{2}, draw the graph of the function for given interval.

Solution

Plot the points of the table: \left(-\dfrac\pi2,-3\right), \left(-\dfrac\pi3,-1.5\right), \left(-\dfrac\pi4,0\right), \left(-\dfrac\pi6,1.5\right), (0,3), \left(\dfrac\pi6,1.5\right), \left(\dfrac\pi4,0\right), \left(\dfrac\pi3,-1.5\right), \left(\dfrac\pi2,-3\right).
Join them with a smooth sinusoidal curve of amplitude 3 and period \pi. The graph is symmetric about the y-axis (cosine is even), with maximum (0,3) and minima \left(\pm\dfrac\pi2,-3\right).

Graph of y=3\cos2x for -\dfrac\pi2\le x\le\dfrac\pi2: one complete cycle, amplitude 3, period \pi.

2026 Annual I, Set D, MCQ 161 mark

Period of \cos\dfrac{2x}{5} is:

A\dfrac52\pi
B2\pi
C\dfrac25\pi
D5\piAnswer

Why

The period of \cos x is 2\pi, so the period of \cos ax is \dfrac{2\pi}{|a|}. Here a=\dfrac25.

\text{Period}=\dfrac{2\pi}{\frac25}=2\pi\cdot\dfrac52=5\pi

Option A, \dfrac52\pi, is \dfrac{\pi}{2/5}, which is the period of a tangent, not a cosine; B is the period of \cos x itself.

Note. This is part (ii) of Example 2 of Section 9.2.

2025 Annual I, Set D, MCQ 161 mark

Which of the following is an odd function?

Af(x)=\cos x
Bf(x)=\sin x+\cos x
Cf(x)=\tan xAnswer
Df(x)=\sec x

Why

An odd function satisfies f(-x)=-f(x). \tan(-x)=-\tan x, so \tan x is odd. \cos x and \sec x are even, and \sin x+\cos x is neither: f(-x)=-\sin x+\cos x, which is neither f(x) nor -f(x) (Example 9(i) of the book).

2026 Annual I, Set D, MCQ 171 mark

Which of the given functions is even?

Af(x)=x^3
Bf(x)=\sin x
Cf(x)=x
Df(x)=\cos xAnswer

Why

A function is even if f(-x)=f(x). Only \cos(-x)=\cos x satisfies this. The others are odd: (-x)^3=-x^3, \sin(-x)=-\sin x, and f(x)=x gives f(-x)=-x.

2025 Annual I, Set D, MCQ 171 mark

The range of f(x)=3\cos x+4 is:

A[-1,1]
B[1,7]Answer
C[-3,3]
D[4,7]

Why

-1\le\cos x\le1, so -3\le3\cos x\le3 and adding 4: 1\le3\cos x+4\le7. In the form a+b\cos x the range is [a-|b|,a+|b|]=[4-3,4+3]=[1,7].

Option D, [4,7], forgets that 3\cos x is also negative; C is the range of 3\cos x without the +4.

2026 Annual I, Set D, MCQ 181 mark

Maximum value of y=3+2\sin\theta is:

A5Answer
B4
C3
D2

Why

Since \sin\theta\le1, the maximum of a+b\sin\theta is a+|b|=3+2=5 (when \sin\theta=1).

Option B, 4, is the value when \sin\theta=\dfrac12, and C is the value at \theta=0 (the midline).

2025 Annual I, Set D, MCQ 181 mark

Which function's graph is symmetric about y-axis?

Ay=\sin x
By=\tan x
Cy=\cos xAnswer
Dy=\csc x

Why

A graph symmetric about the y-axis is the graph of an even function. \cos(-x)=\cos x, so y=\cos x is even. The functions \sin x, \tan x and \csc x are odd, with graphs symmetric about the origin.

2026 Annual I, Set D, Q.2 (iii) OR4 marks

Find the maximum and minimum values of: y=\dfrac{1}{2+3\sin(2x+1)}.

Solution

The denominator 2+3\sin(2x+1) is a+b\sin(cx+d) with a=2, b=3.
M_d=2+3=5,\qquad m_d=2-3=-1
Take the reciprocals of the extreme values of the denominator.
M=\dfrac1{M_d}=\dfrac15,\qquad m=\dfrac1{m_d}=-1

\text{Maximum}=\dfrac15,\qquad\text{Minimum}=-1

Note. The true range of y is (-\infty,-1]\cup\left[\dfrac15,\infty\right).

2025 Annual I, Set D, Q.2 (iii) OR4 marks

Find the maximum and minimum values of the function f(\theta)=\dfrac{1}{3+4\sin(5\theta+\pi)}.

Solution

The denominator 3+4\sin(5\theta+\pi) is a+b\sin(c\theta+d) with a=3, b=4.
M_d=3+4=7,\qquad m_d=3-4=-1
Take the reciprocals of the extreme values of the denominator.
M=\dfrac1{M_d}=\dfrac17,\qquad m=\dfrac1{m_d}=-1

\text{Maximum}=\dfrac17,\qquad\text{Minimum}=-1

Note. The true range of f is (-\infty,-1]\cup\left[\dfrac17,\infty\right).

2026 Annual I, Set D, Q.2 (v)4 marks

Without drawing graph determine the amplitude, period and range of: y=2\sin(3x+1).

Solution

Compare y=2\sin(3x+1) with y=a\sin(bx+c): a=2, b=3.
Amplitude =|a|.
\text{Amplitude}=2
Period of \sin(bx+c) is \dfrac{2\pi}{|b|}. The shift +1 does not change it.
\text{Period}=\dfrac{2\pi}{3}
Range. -1\le\sin(3x+1)\le1, so -2\le y\le2.
\text{Range}=[-2,2]

\text{Amplitude}=2,\quad\text{Period}=\dfrac{2\pi}{3},\quad\text{Range}=[-2,2]

2025 Annual I, Set D, Q.2 (v)4 marks

Without drawing graph find range, amplitude, period and frequency of the function y=-3\sin\left(5x+\dfrac{\pi}{3}\right).

Solution

Compare y=-3\sin\left(5x+\dfrac\pi3\right) with y=a\sin(bx+c): a=-3, b=5.
Range.
-3\le y\le3,\quad\text{Range}=[-3,3]
Amplitude =|a|.
\text{Amplitude}=3
Period =\dfrac{2\pi}{|b|}.
\text{Period}=\dfrac{2\pi}5
Frequency is the reciprocal of the period.
\text{Frequency}=\dfrac5{2\pi}

\text{Range}=[-3,3],\ \text{Amplitude}=3,\ \text{Period}=\dfrac{2\pi}{5},\ \text{Frequency}=\dfrac{5}{2\pi}

2026 Annual I, Set D, Q.2 (ix) OR4 marks

Sketch the graph of y=3\sin x for -\pi\le x\le\pi.

Solution

y=3\sin x has period 2\pi and amplitude 3, so [-\pi,\pi] is one complete cycle. The graph is symmetric about the origin (sine is odd).
Table of values, x in steps of \dfrac\pi6, y=3\sin x.

x	-\pi	-\dfrac{5\pi}{6}	-\dfrac{2\pi}{3}	-\dfrac{\pi}{2}	-\dfrac{\pi}{3}	-\dfrac{\pi}{6}	0	\dfrac{\pi}{6}	\dfrac{\pi}{3}	\dfrac{\pi}{2}	\dfrac{2\pi}{3}	\dfrac{5\pi}{6}	\pi
y=3\sin x	0	-1.5	-2.60	-3	-2.60	-1.5	0	1.5	2.60	3	2.60	1.5	0

Plot the points and join them with a smooth curve: it passes through (0,0), rises to the maximum \left(\dfrac\pi2,3\right), returns to zero at \pi, and falls to the minimum \left(-\dfrac\pi2,-3\right) on the left.

Graph of y=3\sin x for -\pi\le x\le\pi: amplitude 3, zeros at -\pi,0,\pi.

Note. Compare Exercise 9.2, question 2(i), y=2\sin x.

2026 Annual I, Set D, Q.5 OR (a)3 marks

The height of seat on a Ferris wheel is modeled by the equation h(t)=10+8\sin\dfrac{\pi}{6}t. Find the period of motion.

Solution

For h(t)=a+b\sin(ct) the period is \dfrac{2\pi}{|c|}. Here c=\dfrac\pi6.
\text{Period}=\dfrac{2\pi}{\pi/6}=2\pi\cdot\dfrac6\pi=12\ \text{(seconds)}
One complete turn of the seat takes 12 time units: h(t+12)=h(t).

2026 Annual I, Set D, Q.5 OR (b)2 marks

For the Ferris wheel height h(t)=10+8\sin\dfrac{\pi}{6}t, determine the maximum and minimum height.

Solution

Compare h(t)=10+8\sin\dfrac\pi6t with a+b\sin(ct): a=10, b=8.
Maximum =a+|b|.
h_{\max}=10+8=18
Minimum =a-|b|.
h_{\min}=10-8=2

\text{Maximum}=18,\quad\text{Minimum}=2

2026 Annual I, Set D, Q.5 OR (c)3 marks

For the Ferris wheel height h(t)=10+8\sin\dfrac{\pi}{6}t, at what time is the height maximum?

Solution

The height is maximum when \sin\dfrac\pi6t=1.
The sine first equals 1 at the angle \dfrac\pi2.
\dfrac\pi6t=\dfrac\pi2\ \Rightarrow\ t=3
The motion repeats every 12 units, so the height is maximum again at t=3+12n, n=0,1,2,\dots

The height h(t)=10+8\sin\dfrac\pi6t: midline 10, maximum 18 at t=3,15,\dots, minimum 2 at t=9,21,\dots, period 12.

t=3\ \text{(first time), and then every period of }12

2025 Annual I, Set D, Q.5 OR (a)

For y=2\sin(2x); -\dfrac{\pi}{2}\le x\le\dfrac{\pi}{2}, make table of values for given interval.

Solution

The period of y=2\sin2x is \dfrac{2\pi}2=\pi, so \left[-\dfrac\pi2,\dfrac\pi2\right] is one cycle. Take x=0,\pm\dfrac\pi6,\pm\dfrac\pi4,\pm\dfrac\pi3,\pm\dfrac\pi2.

x	-\dfrac{\pi}{2}	-\dfrac{\pi}{3}	-\dfrac{\pi}{4}	-\dfrac{\pi}{6}	\dfrac{\pi}{6}	\dfrac{\pi}{4}	\dfrac{\pi}{3}	\dfrac{\pi}{2}
2x	-\pi	-\dfrac{2\pi}{3}	-\dfrac{\pi}{2}	-\dfrac{\pi}{3}	\dfrac{\pi}{3}	\dfrac{\pi}{2}	\dfrac{2\pi}{3}	\pi
\sin2x	0	-\dfrac{\sqrt3}{2}	-1	-\dfrac{\sqrt3}{2}	\dfrac{\sqrt3}{2}	1	\dfrac{\sqrt3}{2}	0
y=2\sin2x	0	-1.73	-2	-1.73	1.73	2	1.73	0

2025 Annual I, Set D, Q.5 OR (b)

For y=2\sin(2x); -\dfrac{\pi}{2}\le x\le\dfrac{\pi}{2}, draw the graph of the function for given interval.

Solution

Plot the points \left(-\dfrac\pi2,0\right), \left(-\dfrac\pi3,-1.73\right), \left(-\dfrac\pi4,-2\right), \left(-\dfrac\pi6,-1.73\right), (0,0), \left(\dfrac\pi6,1.73\right), \left(\dfrac\pi4,2\right), \left(\dfrac\pi3,1.73\right), \left(\dfrac\pi2,0\right).
Join them with a smooth sinusoidal curve of amplitude 2 and period \pi. The graph is symmetric about the origin (sine is odd): minimum \left(-\dfrac\pi4,-2\right), maximum \left(\dfrac\pi4,2\right).

Graph of y=2\sin2x for -\dfrac\pi2\le x\le\dfrac\pi2: one complete cycle, amplitude 2, period \pi.

Exam focus

Find the maximum and minimum values of a reciprocal function such as y=\dfrac1{2+3\sin(2x+1)}. (problem)
4-mark Q.2 (iii) or (viii) OR in all four sets: 2026 Set B Q.2 (viii) OR, 2026 Set D Q.2 (iii) OR, 2025 Set B Q.2 (iii) OR and Set D Q.2 (iii) OR. Find a\pm|b| for the denominator, then take reciprocals.
Go to answer
Without drawing a graph, find the amplitude, period, range and frequency of y=a\sin(bx+c) or a\cos(bx+c). (problem)
2026 Set D Q.2 (v), 2025 Set B Q.2 (v) and 2025 Set D Q.2 (v), 4 marks each. Amplitude |a|, period \dfrac{2\pi}{|b|}, range [-|a|,|a|], frequency the reciprocal of the period.
Go to answer
Make a table of values and draw the graph of y=a\sin bx or y=a\cos bx over a given interval. (long)
2025 Set B and Set D Q.5 OR (8 marks: table and graph of 3\cos2x and 2\sin2x on \left[-\dfrac\pi2,\dfrac\pi2\right]), and 4-mark sketches of y=2\cos x (2026 Set B Q.2 (i) OR) and y=3\sin x (2026 Set D Q.2 (ix) OR).
Go to answer
Solve an application with a sine or cosine model: a Ferris wheel, a robotic arm or a solar panel. (long)
2026 Set B Q.4 OR (robotic arm, 8 marks), 2026 Set D Q.5 OR (Ferris wheel, 3+2+3) and 2025 Set B Q.2 (x) OR (solar panel). Read period, amplitude, maximum and minimum, then solve \cos(\cdot)= a value.
Go to answer
State the domain, range, period and amplitude of y=\cos x (or of y=\sin x). (short)
2026 Set B Q.2 (x), 4 marks. These properties are tabulated for every graph in Section 9.4.
Go to answer
Find the period of f(ax+b) for sine, cosine and tangent, for example \tan\dfrac{5x}7 or \cos\dfrac{2x}5. (mcq)
2026 Set B MCQ 1 and Set D MCQ 16 (both are Example 2 of the book), and 2026 Set B MCQ 2 (definition of a periodic function). Exercise 9.1, question 4, practises more.
Go to answer
Decide whether a function is even or odd. (mcq)
2026 Set D MCQ 17, 2025 Set B MCQ 16, 2025 Set D MCQ 16 and MCQ 18. Replace x by -x; Exercise 9.2, question 1, has the longer products and quotients.
Go to answer
Find the range or the maximum of a+b\sin\theta and a\sin bx (MCQ). (mcq)
2026 Set B MCQ 3, 2026 Set D MCQ 18, 2025 Set B MCQ 17 and 2025 Set D MCQ 17. Use a\pm|b|; the number inside the sine changes the period only.
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9.1Domain and range of trigonometric functionsOften examined

Derivation Domain and range of y=sinθ

Derivation Domain and range of y=cosθ

Derivation Domain and range of y=tanθ

Derivation Domain and range of y=cotθ

Derivation Domain and range of y=secθ

Derivation Domain and range of y=cosecθ

Remarks

(i) y=4sin3x

(ii) y=2cosx−11​

(iii) y=1+2sinθ1​

(iv) y=2−sin3x1​

9.2Periodicity of trigonometric functionsOften examined

Periodicity of the sine function (9.2.1)

Derivation Period of \sin\theta

Periodicity of the tangent function (9.2.2)

Derivation Period of \tan\theta

(i)

(ii)

(iii)

9.3Maximum and minimum values of trigonometric functionsOften examined

Derivation Maximum and minimum of a+b\sin\theta and a+b\cos\theta

Derivation Maximum and minimum of a+b\sin(c\theta+d) and a+b\cos(c\theta+d)

(i) y=1+2\sin\theta

(ii) y=3+2\cos(3\theta-2)

(iii) y=\dfrac{1}{1+3\sin(2\theta-15)}

9.4Graphs of trigonometric functionsOften examined

Procedure for sketching the graphs of sine and cosine functions

Graph of y=\sin x

Graph of y=\cos x

Graph of y=\tan x

Graph of y=3\sin2x

Graph of y=3\cos2x

Graph of y=\operatorname{cosec}x

Graph of y=\sec x

Graph of y=\cot x

9.5Even and odd trigonometric functions

(i)

(ii)

9.6Application

(1) Graph one complete period

(2) The equation using the cosine function

(3) Height after 45 seconds

(4) Times when the height is 15 m

Transforming sine and cosine

(a)

(b)

(c)

(d)

(1) The equation

(2) Height at 10 s

(3) When is the height 15 m?

Formula sheet

Key terms

Exercise 9.1p. 228

Exercise 9.2p. 244–245

Review Exercise 9p. 246–249

FBISE 2025 and 2026 questions

Exam focus

Derivation Domain and range of $y = sin θ$

Derivation Domain and range of $y = cos θ$

Derivation Domain and range of $y = tan θ$

Derivation Domain and range of $y = cot θ$

Derivation Domain and range of $y = sec θ$

Derivation Domain and range of $y = cosec θ$

(i) $y = 4 sin 3 x$

(ii) $y = \frac{1}{2 cos x - 1}$

(iii) $y = \frac{1}{1 + 2 sin θ}$

(iv) $y = \frac{1}{2 - sin 3 x}$