TRIGONOMETRIC FUNCTIONS

A mathematician knows how to solve a problem, he can not solve it. - MILNE

3.1 Introduction

The word ’trigonometry’ is derived from the Greek words ’trigon’ and ‘metron’ and it means ‘measuring the sides of a triangle’. The subject was originally developed to solve geometric problems involving triangles. It was studied by sea captains for navigation, surveyor to map out the new lands, by engineers and others. Currently, trigonometry is used in many areas such as the science of seismology, designing electric circuits, describing the state of an atom, predicting the heights of tides in the ocean, analysing a musical tone and in many other areas.

In earlier classes, we have studied the trigonometric ratios of acute angles as the ratio of the sides of a right angled triangle. We have also studied the trigonometric identities and application of trigonometric ratios in solving the problems related to heights and distances. In this Chapter, we will generalise the concept of trigonometric ratios to trigonometric functions and study their properties.

3.2 Angles

Angle is a measure of rotation of a given ray about its initial point. The original ray is

called the initial side and the final position of the ray after rotation is called the terminal side of the angle. The point of rotation is called the vertex. If the direction of rotation is anticlockwise, the angle is said to be positive and if the direction of rotation is clockwise, then the angle is* negative* (Fig 3.1).

The measure of an angle is the amount of rotation performed to get the terminal side from the initial side. There are several units for measuring angles. The definition of an angle

Fig 3.2 suggests a unit, viz. one complete revolution from the position of the initial side as indicated in Fig 3.2.

This is often convenient for large angles. For example, we can say that a rapidly spinning wheel is making an angle of say 15 revolution per second. We shall describe two other units of measurement of an angle which are most commonly used, viz. degree measure and radian measure.

3.2.1 Degree measure

If a rotation from the initial side to terminal side is $(\frac{1}{360})^{\text{th }}$ of a revolution, the angle is said to have a measure of one degree, written as $1^{\circ}$. A degree is divided into 60 minutes, and a minute is divided into 60 seconds. One sixtieth of a degree is called a minute, written as $1^{\prime}$, and one sixtieth of a minute is called a second, written as $1^{\prime \prime}$. Thus, $\quad 1^{\circ}=60^{\prime}, \quad 1^{\prime}=60^{\prime \prime}$

Some of the angles whose measures are $360^{\circ}, 180^{\circ}, 270^{\circ}, 420^{\circ},-30^{\circ},-420^{\circ}$ are shown in Fig 3.3.

3.2.2 Radian measure

There is another unit for measurement of an angle, called the radian measure. Angle subtended at the centre by an arc of length 1 unit in a unit circle (circle of radius 1 unit) is said to have a measure of 1 radian. In the Fig 3.4(i) to (iv), $OA$ is the initial side and $OB$ is the terminal side. The figures show the angles whose measures are 1 radian, -1 radian, $1 \frac{1}{2}$ radian and $-1 \frac{1}{2}$ radian.

We know that the circumference of a circle of radius 1 unit is $2 \pi$. Thus, one complete revolution of the initial side subtends an angle of $2 \pi$ radian.

More generally, in a circle of radius $r$, an arc of length $r$ will subtend an angle of 1 radian. It is well-known that equal arcs of a circle subtend equal angle at the centre. Since in a circle of radius $r$, an arc of length $r$ subtends an angle whose measure is 1 radian, an arc of length $l$ will subtend an angle whose measure is $\frac{l}{r}$ radian. Thus, if in a circle of radius $r$, an arc of length $l$ subtends an angle $\theta$ radian at the centre, we have $\theta=\frac{l}{r}$ or $l=r \theta$.

3.2.3 Relation between radian and real numbers

Consider the unit circle with centre $O$. Let $A$ be any point on the circle. Consider OA as initial side of an angle. Then the length of an arc of the circle will give the radian measure of the angle which the arc will subtend at the centre of the circle. Consider the line PAQ which is tangent to the circle at A. Let the point A represent the real number zero, AP represents positive real number and AQ represents negative real numbers (Fig 3.5). If we rope the line $AP$ in the anticlockwise direction along the circle, and $AQ$ in the clockwise direction, then every real number will correspond to a radian measure and conversely. Thus, radian measures and real numbers can be considered as one and the same.

3.2.4 Relation between degree and radian Since a circle subtends at the centre

an angle whose radian measure is $2 \pi$ and its degree measure is $360^{\circ}$, it follows that$ 2 \pi \text{ radian }=360^{\circ} \quad \text{ or } \quad \pi \text{ radian }=180^{\circ} $

The above relation enables us to express a radian measure in terms of degree measure and a degree measure in terms of radian measure. Using approximate value of $\pi$ as $\frac{22}{7}$, we have

$ 1 \text{ radian }=\frac{180^{\circ}}{\pi}=57^{\circ} 16^{\prime} \text{ approximately. } $

Also $\quad 1^{\circ}=\frac{\pi}{180}$ radian $=0.01746$ radian approximately.

The relation between degree measures and radian measure of some common angles are given in the following table:

Degree	$30^{\circ}$	$45^{\circ}$	$60^{\circ}$	$90^{\circ}$	$180^{\circ}$	$270^{\circ}$	$360^{\circ}$
Radian	$\frac{\pi}{6}$	$\frac{\pi}{4}$	$\frac{\pi}{3}$	$\frac{\pi}{2}$	$\pi$	$\frac{3 \pi}{2}$	$2 \pi$

Notational Convention

Since angles are measured either in degrees or in radians, we adopt the convention that whenever we write angle $\theta^{\circ}$, we mean the angle whose degree measure is $\theta$ and whenever we write angle $\beta$, we mean the angle whose radian measure is $\beta$.

Note that when an angle is expressed in radians, the word ‘radian’ is frequently omitted. Thus, $\pi=180^{\circ}$ and $\frac{\pi}{4}=45^{\circ}$ are written with the understanding that $\pi$ and $\frac{\pi}{4}$ are radian measures. Thus, we can say that

$ \begin{aligned} & \text{ Radian measure }=\frac{\pi}{180} \times \text{ Degree measure } \\ & \text{ Degree measure }=\frac{180}{\pi} \times \text{ Radian measure } \end{aligned} $

Example 1 Convert $40^{\circ} 20^{\prime}$ into radian measure.

Solution We know that $180^{\circ}=\pi$ radian.

Hence $\quad 40^{\circ} 20^{\prime}=40 \frac{1}{3}$ degree $=\frac{\pi}{180} \times \frac{121}{3}$ radian $=\frac{121 \pi}{540}$ radian.

Therefore

$ 40^{\circ} 20^{\prime}=\frac{121 \pi}{540} \text{ radian. } $

**Example 2 ** Convert 6 radians into degree measure.

Solution We know that $\pi$ radian $=180^{\circ}$.

Hence

$ \begin{aligned} 6 \text{ radians } & =\frac{180}{\pi} \times 6 \text{ degree }=\frac{1080 \times 7}{22} \text{ degree } \\ & =343 \frac{7}{11} \text{ degree }=343^{\circ}+\frac{7 \times 60}{11} \text{ minute } \quad[\text{ as } 1^{\circ}=60^{\prime}] \\ & =343^{\circ}+38^{\prime}+\frac{2}{11} \text{ minute } \quad[\text{ as } 1^{\prime}=60^{\prime \prime}] \\ & =343^{\circ}+38^{\prime}+10.9^{\prime \prime} \quad=343^{\circ} 38^{\prime} 11^{\prime \prime} \text{ approximately. } \end{aligned} $

Hence $\quad 6$ radians $=343^{\circ} 38^{\prime} 11^{\prime \prime}$ approximately.

Example 3 Find the radius of the circle in which a central angle of $60^{\circ}$ intercepts an arc of length $37.4 cm$ (use $\pi=\frac{22}{7}$ ).

Solution Here $l=37.4 cm$ and $\theta=60^{\circ}=\frac{60 \pi}{180}$ radian $=\frac{\pi}{3}$

Hence, $\quad$ by $r=\frac{l}{\theta}$, we have

$ r=\frac{37.4 \times 3}{\pi}=\frac{37.4 \times 3 \times 7}{22}=35.7 cm $

Example 4 The minute hand of a watch is $1.5 cm$ long. How far does its tip move in 40 minutes? (Use $\pi=3.14$ ).

Solution In 60 minutes, the minute hand of a watch completes one revolution. Therefore, in 40 minutes, the minute hand turns through $\frac{2}{3}$ of a revolution. Therefore, $\theta=\frac{2}{3} \times 360^{\circ}$ or $\frac{4 \pi}{3}$ radian. Hence, the required distance travelled is given by

$ l=r \theta=1.5 \times \frac{4 \pi}{3} cm=2 \pi cm=2 \times 3.14 cm=6.28 cm . $

Example 5 If the arcs of the same lengths in two circles subtend angles $65^{\circ}$ and $110^{\circ}$ at the centre, find the ratio of their radii.

Solution Let $r_1$ and $r_2$ be the radii of the two circles. Given that

$ \theta_1=65^{\circ}=\frac{\pi}{180} \times 65=\frac{13 \pi}{36} \text{ radian } $

and

$ \theta_2=110^{\circ}=\frac{\pi}{180} \times 110=\frac{22 \pi}{36} \text{ radian } $

Let $l$ be the length of each of the arc. Then $l=r_1 \theta_1=r_2 \theta_2$, which gives

$ \frac{13 \pi}{36} \times r_1=\frac{22 \pi}{36} \times r_2 \text{, i.e., } \frac{r_1}{r_2}=\frac{22}{13} $

Hence $\quad r_1: r_2=22: 13$.

EXERCISE 3.1

1. Find the radian measures corresponding to the following degree measures:

(i) $25^{\circ}$

(ii) $-47^{\circ} 30^{\prime}$

(iii) $240^{\circ}$

(iv) $520^{\circ}$

Show Answer

Answer :

(i) $25^{\circ}$

We know that $180^{\circ}=\pi$ radian

$\therefore 25^{\circ}=\frac{\pi}{180} \times 25$ radian $=\frac{5 \pi}{36}$ radian

(ii) -$47^{\circ} 30^{\prime}$

$- 47^{\circ} 30^{\prime}={ }^{-47 \frac{1}{2}}$ degree $[1^{\circ}=60^{\prime}]$

$=\frac{-95}{2}$ degree

Since $180^{\circ}=\pi$ radian

$\frac{-95}{2}$ deg ree $=\frac{\pi}{180} \times(\frac{-95}{2})$ radian $=(\frac{-19}{36 \times 2}) \pi$ radian $=\frac{-19}{72} \pi$ radian

$\therefore-47^{\circ} 30^{\prime}=\frac{-19}{72} \pi$ radian

(iii) $240^{\circ}$

We know that $180^{\circ}=\pi$ radian

$\therefore 240^{\circ}=\frac{\pi}{180} \times 240$ radian $=\frac{4}{3} \pi$ radian

(iv) $520^{\circ}$

We know that $180^{\circ}=\pi$ radian

$\therefore 520^{\circ}=\frac{\pi}{180} \times 520$ radian $=\frac{26 \pi}{9}$ radian

2. Find the degree measures corresponding to the following radian measures (Use $\pi=\frac{22}{7}$ ).

(i) $\frac{11}{16}$

(ii) -4

(iii) $\frac{5 \pi}{3}$

(iv) $\frac{7 \pi}{6}$

Show Answer

Answer :

(i) $\frac{11}{16}$

We know that $\pi$ radian $=180^{\circ}$

$\therefore \frac{11}{16}$ radain $=\frac{180}{\pi} \times \frac{11}{16}$ deg ree $=\frac{45 \times 11}{\pi \times 4}$ deg ree

$=\frac{45 \times 11 \times 7}{22 \times 4}$ deg ree $=\frac{315}{8}$ deg ree

$=39 \frac{3}{8}$ deg ree

$=39^{\circ}+\frac{3 \times 60}{8}$ min utes $\quad[1^{\circ}=60^{\prime}]$

$=39^{\circ}+22^{\prime}+\frac{1}{2}$ min utes

$=39^{\circ} 22^{\prime} 30^{\prime \prime} \quad[1^{\prime}=60^{\prime \prime}]$

(ii) -4

We know that $\pi$ radian $=180^{\circ}$

$ \begin{aligned} -4 \text{ radian } & =\frac{180}{\pi} \times(-4) \text{ deg ree }=\frac{180 \times 7(-4)}{22} \text{ deg ree } \\ & =\frac{-2520}{11} \text{ deg ree }=-229 \frac{1}{11} \text{ deg ree } \\ & =-229^{\circ}+\frac{1 \times 60}{11} \text{ min utes } \quad[1^{\circ}=60^{\prime}] \\ & =-229^{\circ}+5^{\prime}+\frac{5}{11} \text{ min utes } \\ & =-229^{\circ} 5^{\prime} 27^{\prime \prime} \quad[1^{\prime}=60^{\prime \prime}] \end{aligned} $

(iii) $\frac{5 \pi}{3}$

We know that $\pi$ radian $=180^{\circ}$ $\therefore \frac{5 \pi}{3}$ radian $=\frac{180}{\pi} \times \frac{5 \pi}{3}$ deg ree $=300^{\circ}$

(iv) $\frac{7 \pi}{6}$

We know that $\pi$ radian $=180^{\circ}$

$\therefore \frac{7 \pi}{6}$ radian $=\frac{180}{\pi} \times \frac{7 \pi}{6}=210^{\circ}$

3. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?

Show Answer

Answer :

Number of revolutions made by the wheel in 1 minute $=360$

$\therefore$ Number of revolutions made by the wheel in 1 second $=\frac{360}{60}=6$

In one complete revolution, the wheel turns an angle of $2 \pi$ radian.

Hence, in 6 complete revolutions, it will turn an angle of $6 \times 2 \pi$ radian, i.e.,

$12 \pi$ radian

Thus, in one second, the wheel turns an angle of $12 \pi$ radian.

4. Find the degree measure of the angle subtended at the centre of a circle of radius $100 cm$ by an arc of length $22 cm$ (Use $\pi=\frac{22}{7}$ ).

Show Answer

Answer :

We know that in a circle of radius $r$ unit, if an arc of length / unit subtends an angle $\theta$ radian at the centre, then

$\theta=\frac{1}{r}$

Therefore, forr $=100 cm, I=22 cm$, we have

$\theta=\frac{22}{100}$ radian $=\frac{180}{\pi} \times \frac{22}{100}$ deg ree $=\frac{180 \times 7 \times 22}{22 \times 100}$ deg ree

$=\frac{126}{10}$ deg ree $=12 \frac{3}{5}$ deg ree $=12^{\circ} 36^{\prime} \quad[1^{\circ}=60^{\prime}]$

Thus, the required angle is $12^{\circ} 36^{u_2}$.

5. In a circle of diameter $40 cm$, the length of a chord is $20 cm$. Find the length of minor arc of the chord.

Show Answer

Answer :

Diameter of the circle $=40 cm$

$\therefore$ Radius $(r)$ of the circle $=\frac{40}{2} cm=20 cm$

Let $A B$ be a chord (length $=20 cm$ ) of the circle.

In $\triangle OAB, OA=OB=$ Radius of circle $=20 cm$

Also, $A B=20 cm$

Thus, $\triangle OAB$ is an equilateral triangle.

$\therefore \theta=60^{\circ}=\frac{\pi}{3}$ radian

We know that in a circle of radius $r$ unit, if an arc of length / unit subtends an angle $\theta$ radian at the centre, then

$\theta=\frac{l}{r}$

$\frac{\pi}{3}=\frac{\overline{AB}}{20} \rightarrow \overline{AB}=\frac{20 \pi}{3} cm$

6. If in two circles, arcs of the same length subtend angles $60^{\circ}$ and $75^{\circ}$ at the centre, find the ratio of their radii.

Show Answer

Answer :

Let the radii of the two circles be ${ }^{r_1}$ and ${ }^{r_2}$. Let an arc of length / subtend an angle of $60^{\circ}$ at the centre of the circle of radius $r_1$, while let an arc of length / subtend an angle of $75^{\circ}$ at the centre of the circle of radius $r_2$.

Now, $60^{\circ}=\frac{\pi}{3}$ radian and $75^{\circ}=\frac{5 \pi}{12}$ radian

We know that in a circle of radius $r$ unit, if an arc of length / unit subtends an angle $\theta$ radian at the centre, then

$\theta=\frac{l}{r}$ or $l=r \theta$

$\therefore l=\frac{r_1 \pi}{3}$ and $l=\frac{r_2 5 \pi}{12}$

$\Rightarrow \frac{r_1 \pi}{3}=\frac{r_2 5 \pi}{12}$

$\Rightarrow r_1=\frac{r_2 5}{4}$

$\Rightarrow \frac{r_1}{r_2}=\frac{5}{4}$

Thus, the ratio of the radii is 5:4.

7. Find the angle in radian through which a pendulum swings if its length is $75 cm$ and the tip describes an arc of length

(i) $10 cm$

(ii) $15 cm$

(iii) $21 cm$

Show Answer

Answer :

We know that in a circle of radius $r$ unit, if an arc of length / unit subtends an angle $\theta$ radian at the centre, then $\theta=\frac{l}{r}$

It is given that $r=75 cm$

(i) Here, $I=10 cm$

$\theta=\frac{10}{75}$ radian $=\frac{2}{15}$ radian

(ii) Here, $I=15 cm$

$\theta=\frac{15}{75}$ radian $=\frac{1}{5}$ radian

(iii) Here, $I=21 cm$

$\theta=\frac{21}{75}$ radian $=\frac{7}{25}$ radian

3.3 Trigonometric Functions

In earlier classes, we have studied trigonometric ratios for acute angles as the ratio of sides of a right angled triangle. We will now extend the definition of trigonometric ratios to any angle in terms of radian measure and study them as trigonometric functions.

Consider a unit circle with centre at origin of the coordinate axes. Let $P(a, b)$ be any point on the circle with angle $AOP=x$ radian, i.e., length of arc $AP=x$ (Fig 3.6).

We define $\cos x=a$ and $\sin x=b$ Since $\triangle OMP$ is a right triangle, we have $OM^{2}+MP^{2}=OP^{2}$ or $a^{2}+b^{2}=1$ Thus, for every point on the unit circle, we have

$ a^{2}+b^{2}=1 \text{ or } \cos ^{2} x+\sin ^{2} x=1 $

Since one complete revolution subtends an angle of $2 \pi$ radian at the centre of the circle, $\angle AOB=\frac{\pi}{2}$,

$\angle AOC=\pi$ and $\angle AOD=\frac{3 \pi}{2}$. All angles which are integral multiples of $\frac{\pi}{2}$ are called quadrantal angles. The coordinates of the points A, B, C and D are, respectively, $(1,0),(0,1),(-1,0)$ and $(0,-1)$. Therefore, for quadrantal angles, we have

$ \begin{aligned} & \cos 0^{\circ}=1 \quad \sin 0^{\circ}=0, \\ & \cos \frac{\pi}{2}=0 \quad \sin \frac{\pi}{2}=1 \\ & \cos \pi=-1 \quad \sin \pi=0 \\ & \cos \frac{3 \pi}{2}=0 \quad \sin \frac{3 \pi}{2}=-1 \\ & \cos 2 \pi=1 \quad \sin 2 \pi=0 \end{aligned} $

Now, if we take one complete revolution from the point $P$, we again come back to same point $P$. Thus, we also observe that if $x$ increases (or decreases) by any integral multiple of $2 \pi$, the values of sine and cosine functions do not change. Thus,

$ \sin (2 n \pi+x)=\sin x, n \in \mathbf{Z}, \cos (2 n \pi+x)=\cos x, n \in \mathbf{Z} $

Further, $\sin x=0$, if $x=0, \pm \pi, \pm 2 \pi, \pm 3 \pi$, …, i.e., when $x$ is an integral multiple of $\pi$ and $\cos x=0$, if $x= \pm \frac{\pi}{2}, \pm \frac{3 \pi}{2}, \pm \frac{5 \pi}{2}, \ldots$ i.e., $\cos x$ vanishes when $x$ is an odd multiple of $\frac{\pi}{2}$. Thus

$ \begin{aligned} & \sin x=0 \text{ implies } x=n \pi, \text{ where } n \text{ is any integer } \\ & \cos x=0 \text{ implies } x=(2 n+1) \frac{\pi}{2} \text{, where } n \text{ is any integer. } \end{aligned} $

We now define other trigonometric functions in terms of sine and cosine functions:

$\cosec x=\frac{1}{\sin x}, x \neq n \pi$, where $n$ is any integer.

$\sec x=\frac{1}{\cos x}, x \neq(2 n+1) \frac{\pi}{2}$, where $n$ is any integer.

$\tan x=\frac{\sin x}{\cos x}, x \neq(2 n+1) \frac{\pi}{2}$, where $n$ is any integer.

$\cot x=\frac{\cos x}{\sin x}, x \neq n \pi$, where $n$ is any integer.

We have shown that for all real $x, \sin ^{2} x+\cos ^{2} x=1$

It follows that

$ \begin{aligned} & 1+\tan ^{2} x=\sec ^{2} x \\ & 1+\cot ^{2} x=cosec^{2} x \end{aligned} $

In earlier classes, we have discussed the values of trigonometric ratios for $0^{\circ}$, $30^{\circ}, 45^{\circ}, 60^{\circ}$ and $90^{\circ}$. The values of trigonometric functions for these angles are same as that of trigonometric ratios studied in earlier classes. Thus, we have the following table:

	$0^{\circ}$	$\frac{\pi}{6}$	$\frac{\pi}{4}$	$\frac{\pi}{3}$	$\frac{\pi}{2}$	$\pi$	$\frac{3 \pi}{2}$	$2 \pi$
$\sin$	0	$\frac{1}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{\sqrt{3}}{2}$	1	0	-1	0
$\cos$	1	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{1}{2}$	0	-1	0	1
$\tan$	0	$\frac{1}{\sqrt{3}}$	1	$\sqrt{3}$	defined	0	not defined	0

The values of $cosec x, \sec x$ and $\cot x$ are the reciprocal of the values of $\sin x$, $\cos x$ and $\tan x$, respectively.

3.3.1 Sign of trigonometric functions

Let $P(a, b)$ be a point on the unit circle with centre at the origin such that $\angle AOP=x$. If $\angle AOQ=-x$, then the coordinates of the point $Q$ will be $(a,-b)$ (Fig 3.7). Therefore

$ \cos (-x)=\cos x $

and $\quad$ $ \sin (-x)=-\sin x $

Since for every point $P(a, b)$ on the unit circle, $-1 \leq a \leq 1$ and

$-1 \leq b \leq 1$, we have $-1 \leq \cos x \leq 1$ and $-1 \leq \sin x \leq 1$ for all $x$. We have learnt in previous classes that in the first quadrant $(0<x<\frac{\pi}{2}) a$ and $b$ are both positive, in the second quadrant $(\frac{\pi}{2}<x<\pi) a$ is negative and $b$ is positive, in the third quadrant $(\pi<x<\frac{3 \pi}{2}) a$ and $b$ are both negative and in the fourth quadrant $(\frac{3 \pi}{2}<x<2 \pi) a$ is positive and $b$ is negative. Therefore, $\sin x$ is positive for $0<x<\pi$, and negative for $\pi<x<2 \pi$. Similarly, $\cos x$ is positive for $0<x<\frac{\pi}{2}$, negative for $\frac{\pi}{2}<x<\frac{3 \pi}{2}$ and also positive for $\frac{3 \pi}{2}<x<2 \pi$. Likewise, we can find the signs of other trigonometric functions in different quadrants. In fact, we have the following table.

	I	II	III	IV
$\sin x$	+	+	-	-
$\cos x$	+	-	-	+
$\tan x$	+	-	+	-
$cosec x$	+	+	-	-
$\sec x$	+	-	-	+
$\cot x$	+	-	+	-

3.3.2 Domain and range of trigonometric functions

From the definition of sine and cosine functions, we observe that they are defined for all real numbers. Further, we observe that for each real number $x$,

$ -1 \leq \sin x \leq 1 \text{ and }-1 \leq \cos x \leq 1 $

Thus, domain of $y=\sin x$ and $y=\cos x$ is the set of all real numbers and range is the interval $[-1,1]$, i.e., $-1 \leq y \leq 1$.

Since $\cosec x=\frac{1}{\sin x}$, the domain of $y=cosec x$ is the set $\{x: x \in \mathbf{R}$ and $x \neq n \pi, n \in \mathbf{Z}\}$ and range is the set $\{y: y \in \mathbf{R}, y \geq 1$ or $y \leq-1\}$. Similarly, the domain of $y=\sec x$ is the set $\{x: x \in \mathbf{R}.$ and $.x \neq(2 n+1) \frac{\pi}{2}, n \in \mathbf{Z}\}$ and range is the set $\{y: y \in \mathbf{R}, y \leq-1$ or $y \geq 1\}$. The domain of $y=\tan x$ is the set $\{x: x \in \mathbf{R}$ and $.x \neq(2 n+1) \frac{\pi}{2}, n \in \mathbf{Z}\}$ and range is the set of all real numbers. The domain of $y=\cot x$ is the set $\{x: x \in \mathbf{R}$ and $x \neq n \pi, n \in \mathbf{Z}\}$ and the range is the set of all real numbers.

We further observe that in the first quadrant, as $x$ increases from 0 to $\frac{\pi}{2}, \sin x$ increases from 0 to 1 , as $x$ increases from $\frac{\pi}{2}$ to $\pi, \sin x$ decreases from 1 to 0 . In the third quadrant, as $x$ increases from $\pi$ to $\frac{3 \pi}{2}, \sin x$ decreases from 0 to -1 and finally, in the fourth quadrant, $\sin x$ increases from -1 to 0 as $x$ increases from $\frac{3 \pi}{2}$ to $2 \pi$. Similarly, we can discuss the behaviour of other trigonometric functions. In fact, we have the following table:

	I quadrant	II quadrant	III quadrant	IV quadrant
$\sin$	increases from 0 to 1	decreases from 1 to 0	decreases from 0 to -1	increases from -1 to 0
$\cos$	decreases from 1 to 0	decreases from 0 to -1	increases from -1 to 0	increases from 0 to 1
tan	increases from 0 to $\infty$	increases from $-\infty$ to 0	increases from 0 to $\infty$	increases from $-\infty$ to 0
$\cot$	decreases from $\infty$ to 0	decreases from 0 to- $-\infty$	decreases from $\infty$ to 0	decreases from 0 to $-\infty$
sec	increases from 1 to $\infty$	increases from $-\infty$ to -1	decreases from -1 to- $-\infty$	decreases from $\infty$ to 1
$cosec$	decreases from $\infty$ to 1	increases from 1 to $\infty$	increases from $-\infty$ to -1	decreases from-1 to- $\infty$

Remark In the above table, the statement $\tan x$ increases from 0 to $\infty$ (infinity) for $0<x<\frac{\pi}{2}$ simply means that $\tan x$ increases as $x$ increases for $0<x<\frac{\pi}{2}$ and assumes arbitraily large positive values as $x$ approaches to $\frac{\pi}{2}$. Similarly, to say that $cosec x$ decreases from -1 to $-\infty$ (minus infinity) in the fourth quadrant means that $cosec x$ decreases for $x \in(\frac{3 \pi}{2}, 2 \pi)$ and assumes arbitrarily large negative values as $x$ approaches to $2 \pi$. The symbols $\infty$ and $-\infty$ simply specify certain types of behaviour of functions and variables.

We have already seen that values of $\sin x$ and $\cos x$ repeats after an interval of $2 \pi$. Hence, values of $cosec x$ and $\sec x$ will also repeat after an interval of $2 \pi$. We

shall see in the next section that $\tan (\pi+x)=\tan x$. Hence, values of $\tan x$ will repeat after an interval of $\pi$. Since $\cot x$ is reciprocal of $\tan x$, its values will also repeat after an interval of $\pi$. Using this knowledge and behaviour of trigonometic functions, we can sketch the graph of these functions. The graph of these functions are given above:

Example 6 If $\cos x=-\frac{3}{5}, x$ lies in the third quadrant, find the values of other five trigonometric functions.

Solution Since $\cos x=-\frac{3}{5}$, we have $\sec x=-\frac{5}{3}$

Now $\quad \sin ^{2} x+\cos ^{2} x=1$, i.e., $\sin ^{2} x=1-\cos ^{2} x$

or

$ \sin ^{2} x=1-\frac{9}{25}=\frac{16}{25} $

Hence $\quad \sin x= \pm \frac{4}{5}$

Since $x$ lies in third quadrant, $\sin x$ is negative. Therefore

$ \sin x=-\frac{4}{5} $

which also gives

$ cosec x=-\frac{5}{4} $

Further, we have

$ \tan x=\frac{\sin x}{\cos x}=\frac{4}{3} \text{ and } \cot x=\frac{\cos x}{\sin x}=\frac{3}{4} $

Example 7 If $\cot x=-\frac{5}{12}, x$ lies in second quadrant, find the values of other five trigonometric functions.

Solution Since cot $x=-\frac{5}{12}$, we have $\tan x=-\frac{12}{5}$

Now $\quad \sec ^{2} x=1+\tan ^{2} x=1+\frac{144}{25}=\frac{169}{25}$

Hence

$ \sec x= \pm \frac{13}{5} $

Since $x$ lies in second quadrant, sec $x$ will be negative. Therefore

$ \sec x=-\frac{13}{5} $

which also gives

$ \cos x=-\frac{5}{13} $

Further, we have

$ \sin x=\tan x \cos x=(-\frac{12}{5}) \times(-\frac{5}{13})=\frac{12}{13} $

and $\quad \cosec x=\frac{1}{\sin x}=\frac{13}{12}$.

Example 8 Find the value of $\sin \frac{31 \pi}{3}$.

Solution We know that values of $\sin x$ repeats after an interval of $2 \pi$. Therefore

$ \sin \frac{31 \pi}{3}=\sin (10 \pi+\frac{\pi}{3})=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2} \text{. } $

Example 9 Find the value of $\cos (-1710^{\circ})$.

Solution We know that values of $\cos x$ repeats after an interval of $2 \pi$ or $360^{\circ}$. Therefore, $\cos (-1710^{\circ})=\cos (-1710^{\circ}+5 \times 360^{\circ})$

$ =\cos (-1710^{\circ}+1800^{\circ})=\cos 90^{\circ}=0 . $

EXERCISE 3.2

Find the values of other five trigonometric functions in Exercises 1 to 5.

1. $\cos x=-\frac{1}{2}, x$ lies in third quadrant.

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Answer :

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

$\therefore \sec x=\frac{1}{\cos x}=\frac{1}{(-\frac{1}{2})}=-2$

$\sin ^{2} x+\cos ^{2} x=1$

$\Rightarrow \sin ^{2} x=1-\cos ^{2} x$

$\Rightarrow \sin ^{2} x=1-(-\frac{1}{2})^{2}$

$\Rightarrow \sin ^{2} x=1-\frac{1}{4}=\frac{3}{4}$

$\Rightarrow \sin x= \pm \frac{\sqrt{3}}{2}$

Since $x$ lies in the $3^{\text{rd }}$ quadrant, the value of $\sin x$ will be negative.

$\therefore \sin x=-\frac{\sqrt{3}}{2}$

$cosec x=\frac{1}{\sin x}=\frac{1}{(-\frac{\sqrt{3}}{2})}=-\frac{2}{\sqrt{3}}$

$\tan x=\frac{\sin x}{\cos x}=\frac{(-\frac{\sqrt{3}}{2})}{(-\frac{1}{2})}=\sqrt{3}$

$\cot x=\frac{1}{\tan x}=\frac{1}{\sqrt{3}}$

2. $\sin x=\frac{3}{5}, x$ lies in second quadrant.

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Answer :

$\sin x=\frac{3}{5}$

$cosec x=\frac{1}{\sin x}=\frac{1}{(\frac{3}{5})}=\frac{5}{3}$

$\sin ^{2} x+\cos ^{2} x=1$

$\Rightarrow \cos ^{2} x=1-\sin ^{2} x$

$\Rightarrow \cos ^{2} x=1-(\frac{3}{5})^{2}$

$\Rightarrow \cos ^{2} x=1-\frac{9}{25}$

$\Rightarrow \cos ^{2} x=\frac{16}{25}$

$\Rightarrow \cos x= \pm \frac{4}{5}$

Since $x$ lies in the $2^{\text{nd }}$ quadrant, the value of $\cos x$ will be negative

$\therefore \cos x=-\frac{4}{5}$

$\sec x=\frac{1}{\cos x}=\frac{1}{(-\frac{4}{5})}=-\frac{5}{4}$

$\tan x=\frac{\sin x}{\cos x}=\frac{(\frac{3}{5})}{(-\frac{4}{5})}=-\frac{3}{4}$

$\cot x=\frac{1}{\tan x}=-\frac{4}{3}$

3. $\cot x=\frac{3}{4}, x$ lies in third quadrant.

Show Answer

Answer :

$\cot x=\frac{3}{4}$

$\tan x=\frac{1}{\cot x}=\frac{1}{(\frac{3}{4})}=\frac{4}{3}$

$1+\tan ^{2} x=\sec ^{2} x$

$\Rightarrow 1+(\frac{4}{3})^{2}=\sec ^{2} x$

$\Rightarrow 1+\frac{16}{9}=\sec ^{2} x$

$\Rightarrow \frac{25}{9}=\sec ^{2} x$

$\Rightarrow \sec x= \pm \frac{5}{3}$

Since $x$ lies in the $3^{\text{rd }}$ quadrant, the value of $\sec x$ will be negative.

$\therefore \sec x=-\frac{5}{3}$

$\cos x=\frac{1}{\sec x}=\frac{1}{(-\frac{5}{3})}=-\frac{3}{5}$

$\tan x=\frac{\sin x}{\cos x}$

$\Rightarrow \frac{4}{3}=\frac{\sin x}{(\frac{-3}{5})}$

$\Rightarrow \sin x=(\frac{4}{3}) \times(\frac{-3}{5})=-\frac{4}{5}$

$cosec x=\frac{1}{\sin x}=-\frac{5}{4}$

4. $\sec x=\frac{13}{5}, x$ lies in fourth quadrant.

Show Answer

Answer :

$\sec x=\frac{13}{5}$

$\cos x=\frac{1}{\sec x}=\frac{1}{(\frac{13}{5})}=\frac{5}{13}$

$\sin ^{2} x+\cos ^{2} x=1$

$\Rightarrow \sin ^{2} x=1-\cos ^{2} x$

$\Rightarrow \sin ^{2} x=1-(\frac{5}{13})^{2}$

$\Rightarrow \sin ^{2} x=1-\frac{25}{169}=\frac{144}{169}$

$\Rightarrow \sin x= \pm \frac{12}{13}$

Since $x$ lies in the $4^{\text{th }}$ quadrant, the value of $\sin x$ will be negative.

$\therefore \sin x=-\frac{12}{13}$

$cosec x=\frac{1}{\sin x}=\frac{1}{(-\frac{12}{13})}=-\frac{13}{12}$

$\tan x=\frac{\sin x}{\cos x}=\frac{(\frac{-12}{13})}{(\frac{5}{13})}=-\frac{12}{5}$

$\cot x=\frac{1}{\tan x}=\frac{1}{(-\frac{12}{5})}=-\frac{5}{12}$

5. $\tan x=-\frac{5}{12}, x$ lies in second quadrant.

Show Answer

Answer :

$\tan x=-\frac{5}{12}$ $\cot x=\frac{1}{\tan x}=\frac{1}{(-\frac{5}{12})}=-\frac{12}{5}$

$1+\tan ^{2} x=\sec ^{2} x$

$\Rightarrow 1+(-\frac{5}{12})^{2}=\sec ^{2} x$

$\Rightarrow 1+\frac{25}{144}=\sec ^{2} x$

$\Rightarrow \frac{169}{144}=\sec ^{2} x$

$\Rightarrow \sec x= \pm \frac{13}{12}$

Since $x$ lies in the $2^{\text{nd }}$ quadrant, the value of $\sec x$ will be negative.

$\therefore \sec x=-\frac{13}{12}$

$\cos x=\frac{1}{\sec x}=\frac{1}{(-\frac{13}{12})}=-\frac{12}{13}$

$\tan x=\frac{\sin x}{\cos x}$

$\Rightarrow-\frac{5}{12}=\frac{\sin x}{(-\frac{12}{13})}$

$\Rightarrow \sin x=(-\frac{5}{12}) \times(-\frac{12}{13})=\frac{5}{13}$

$cosec x=\frac{1}{\sin x}=\frac{1}{(\frac{5}{13})}=\frac{13}{5}$

Find the values of the trigonometric functions in Exercises 6 to 10.

6. $\sin 765^{\circ}$

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Answer :

It is known that the values of $\sin x$ repeat after an interval of $2 \pi$ or $360^{\circ}$.

$\therefore \sin 765^{\circ}=\sin (2 \times 360^{\circ}+45^{\circ})=\sin 45^{\circ}=\frac{1}{\sqrt{2}}$

7. $\cosec(-1410^{\circ})$

Show Answer

Answer :

It is known that the values of $cosec x$ repeat after an interval of $2 \pi$ or $360^{\circ}$.

$\therefore cosec(-1410^{\circ})=cosec(-1410^{\circ}+4 \times 360^{\circ})$

$=cosec(-1410^{\circ}+1440^{\circ})$

$=cosec 30^{\circ}=2$

8. $\tan \frac{19 \pi}{3}$

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Answer :

It isknown that the values of $\tan x$ repeat after an interval of $\pi$ or $180^{\circ}$.

$\therefore \tan \frac{19 \pi}{3}=\tan 6 \frac{1}{3} \pi=\tan (6 \pi+\frac{\pi}{3})=\tan \frac{\pi}{3}=\tan 60^{\circ}=\sqrt{3}$

9. $\sin (-\frac{11 \pi}{3})$

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Answer :

It is known that the values of $\sin x$ repeat after an interval of $2 \pi$ or $360^{\circ}$.

$\therefore \sin (-\frac{11 \pi}{3})=\sin (-\frac{11 \pi}{3}+2 \times 2 \pi)=\sin (\frac{\pi}{3})=\frac{\sqrt{3}}{2}$

10. $\cot (-\frac{15 \pi}{4})$

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Answer :

It is known that the values of $\cot x$ repeat after an interval of $\pi$ or $180^{\circ}$.

$\therefore \cot (-\frac{15 \pi}{4})=\cot (-\frac{15 \pi}{4}+4 \pi)=\cot \frac{\pi}{4}=1$

3.4 Trigonometric Functions of Sum and Difference of Two Angles

In this Section, we shall derive expressions for trigonometric functions of the sum and difference of two numbers (angles) and related expressions. The basic results in this connection are called trigonometric identities. We have seen that

1. $\sin (-x)=-\sin x$
2. $\cos (-x)=\cos x$

We shall now prove some more results:

3. $\cos (x+y)=\cos x \cos y-\sin x \sin y$

Consider the unit circle with centre at the origin. Let $x$ be the angle $P_4 OP_1$ and $y$ be the angle $P_1 OP_2$. Then $(x+y)$ is the angle $P_4 OP_2$. Also let $(-y)$ be the angle $P_4 OP_3$. Therefore, $P_1, P_2, P_3$ and $P_4$ will have the coordinates $P_1(\cos x, \sin x)$, $P_2[\cos (x+y), \sin (x+y)], P_3[\cos (-y), \sin (-y)]$ and $P_4(1,0)$ (Fig 3.14).

Consider the triangles $P_1 OP_3$ and $P_2 OP_4$. They are congruent (Why?). Therefore, $P_1 P_3$ and $P_2 P_4$ are equal. By using distance formula, we get

$ \begin{aligned} P_1 P_3^{2} & =[\cos x-\cos (-y)]^{2}+[\sin x-\sin (-y]^{2}. \\ & =(\cos x-\cos y)^{2}+(\sin x+\sin y)^{2} \\ & =\cos ^{2} x+\cos ^{2} y-2 \cos x \cos y+\sin ^{2} x+\sin ^{2} y+2 \sin x \sin y \\ & =2-2(\cos x \cos y-\sin x \sin y) \quad(\text{ Why?) } \end{aligned} $

Also, $\quad P_2 P_4{ }^{2}=[1-\cos (x+y)]^{2}+[0-\sin (x+y)]^{2}$

$ \begin{aligned} & =1-2 \cos (x+y)+\cos ^{2}(x+y)+\sin ^{2}(x+y) \\ & =2-2 \cos (x+y) \end{aligned} $

Since $P_1 P_3=P_2 P_4$, we have $P_1 P_3^{2}=P_2 P_4{ }^{2}$.

Therefore, $2-2(\cos x \cos y-\sin x \sin y)=2-2 \cos (x+y)$.

Hence $\cos (x+y)=\cos x \cos y-\sin x \sin y$

4. $\cos (x-y)=\cos x \cos y+\sin x \sin y$

Replacing $y$ by $-y$ in identity 3 , we get

$ \begin{aligned} & \cos (x+(-y))=\cos x \cos (-y)-\sin x \sin (-y) \\ & \text{ or } \quad \cos (x-y)=\cos x \cos y+\sin x \sin y \end{aligned} $

5. $\cos (\frac{\pi}{2}-x)=\sin x$

If we replace $x$ by $\frac{\pi}{2}$ and $y$ by $x$ in Identity (4), we get

$ \cos (\frac{\pi}{2}-x)=\cos \frac{\pi}{2} \cos x+\sin \frac{\pi}{2} \sin x=\sin x $

6. $\sin (\frac{\pi}{2}-x)=\cos x$

Using the Identity 5, we have

$ \sin (\frac{\pi}{2}-x)=\cos [\frac{\pi}{2}-(\frac{\pi}{2}-x)]=\cos x $

7. $\quad \sin (x+y)=\sin x \cos y+\cos x \sin y$

We know that

$ \begin{aligned} \sin (x+y) & =\cos (\frac{\pi}{2}-(x+y))=\cos ((\frac{\pi}{2}-x)-y) \\ & =\cos (\frac{\pi}{2}-x) \cos y+\sin (\frac{\pi}{2}-x) \sin y \\ & =\sin x \cos y+\cos x \sin y \end{aligned} $

8. $\begin{aligned} \quad \sin (x-y) & =\sin x \cos y-\cos x \sin y \end{aligned} $ If we replace $y$ by $-y$, in the Identity 7 , we get the result.

9. By taking suitable values of $x$ and $y$ in the identities $3,4,7$ and 8 , we get the following results:

$ \begin{aligned} \cos (\frac{\pi}{2}+x)=-\sin x & \sin (\frac{\pi}{2}+x)=\cos x \\ \cos (\pi-x)=-\cos x & \sin (\pi-x)=\sin x \end{aligned} $

$ \begin{aligned} \cos (\pi+x)=-\cos x & \sin (\pi+x)=-\sin x \\ \cos (2 \pi-x)=\cos x & \sin (2 \pi-x)=-\sin x \end{aligned} $

Similar results for $\tan x, \cot x, \sec x$ and $cosec x$ can be obtained from the results of $\sin$ $x$ and $\cos x$.

10. If none of the angles $x, y$ and $(x+y)$ is an odd multiple of $\frac{\pi}{2}$, then

$ \tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y} $

Since none of the $x, y$ and $(x+y)$ is an odd multiple of $\frac{\pi}{2}$, it follows that $\cos x$, $\cos y$ and $\cos (x+y)$ are non-zero. Now

$ \tan (x+y)=\frac{\sin (x+y)}{\cos (x+y)}=\frac{\sin x \cos y+\cos x \sin y}{\cos x \cos y-\sin x \sin y} \text{. } $

Dividing numerator and denominator by $\cos x \cos y$, we have

$ \begin{aligned} \tan (x+y) & =\frac{\frac{\sin x \cos y}{\cos x \cos y}+\frac{\cos x \sin y}{\cos x \cos y}}{\frac{\cos x \cos y}{\cos x \cos y}-\frac{\sin x \sin y}{\cos x \cos y}} \\ & =\frac{\tan x+\tan y}{1-\tan x \tan y} \end{aligned} $

11. $\tan (x-y)=\frac{\tan x-\tan y}{1+\tan x \tan y}$

If we replace $y$ by $-y$ in Identity 10 , we get

$ \begin{aligned} \tan (x-y) & =\tan [x+(-y)] \\ & =\frac{\tan x+\tan (-y)}{1-\tan x \tan (-y)}=\frac{\tan x-\tan y}{1+\tan x \tan y} \end{aligned} $

12. If none of the angles $x, y$ and $(x+y)$ is a multiple of $\pi$, then

$ \cot (x+y)=\frac{\cot x \cot y-1}{\cot y+\cot x} $

Since, none of the $x, y$ and $(x+y)$ is multiple of $\pi$, we find that $\sin x$ sin $y$ and $\sin (x+y)$ are non-zero. Now,

$ \cot (x+y)=\frac{\cos (x+y)}{\sin (x+y)}=\frac{\cos x \cos y-\sin x \sin y}{\sin x \cos y+\cos x \sin y} $

Dividing numerator and denominator by $\sin x \sin y$, we have

$ \cot (x+y)=\frac{\cot x \cot y-1}{\cot y+\cot x} $

13. $\cot (\boldsymbol{{}x}-\boldsymbol{{}y})=\frac{\cot \boldsymbol{{}x} \cot \boldsymbol{{}y}+\mathbf{1}}{\cot \boldsymbol{{}y}-\cot \boldsymbol{{}x}}$ if none of angles $x, y$ and $x-y$ is a multiple of $\pi$

If we replace $y$ by $-y$ in identity 12 , we get the result

14. $\cos 2 x=\cos ^{2} x-\sin ^{2} x=2 \cos ^{2} x-1=1-2 \sin ^{2} x=\frac{1-\tan ^{2} x}{1+\tan ^{2} x}$

We know that

$ \cos (x+y)=\cos x \cos y-\sin x \sin y $

Replacing $y$ by $x$, we get

$ \begin{aligned} \cos 2 x & =\cos ^{2} x-\sin ^{2} x \\ & =\cos ^{2} x-(1-\cos ^{2} x)=2 \cos ^{2} x-1 \end{aligned} $

Again, $\quad \cos 2 x=\cos ^{2} x-\sin ^{2} x$

$ =1-\sin ^{2} x-\sin ^{2} x=1-2 \sin ^{2} x . $

We have

$ \cos 2 x=\cos ^{2} x-\sin ^{2} x=\frac{\cos ^{2} x-\sin ^{2} x}{\cos ^{2} x+\sin ^{2} x} $

Dividing numerator and denominator by $\cos ^{2} x$, we get

$ \cos 2 x=\frac{1-\tan ^{2} x}{1+\tan ^{2} x}, x \neq n \pi+\frac{\pi}{2} \text{, where } n \text{ is an integer } $

15. $\sin 2 x=2 \sin x \cos x=\frac{2 \tan x}{1+\tan ^{2} x} \quad x \neq n \pi+\frac{\pi}{2}$, where $n$ is an integer

We have

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

Replacing $y$ by $x$, we get $\sin 2 x=2 \sin x \cos x$.

Again

$ \sin 2 x=\frac{2 \sin x \cos x}{\cos ^{2} x+\sin ^{2} x} $

Dividing each term by $\cos ^{2} x$, we get

$ \sin 2 x=\frac{2 \tan x}{1+\tan ^{2} x} $

16. $\tan 2 x=\frac{2 \tan x}{1-\tan ^{2} x}$ if $2 x \neq n \pi+\frac{\pi}{2}$, where $n$ is an integer

We know that

$ \tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y} $

Replacing $y$ by $x$, we get $\tan 2 x=\frac{2 \tan x}{1-\tan ^{2} x}$

17. $\sin 3 x=3 \sin x-4 \sin ^{3} x$

We have,

$ \begin{aligned} \sin 3 x & =\sin (2 x+x) \\ & =\sin 2 x \cos x+\cos 2 x \sin x \\ & =2 \sin x \cos x \cos x+(1-2 \sin ^{2} x) \sin x \\ & =2 \sin x(1-\sin ^{2} x)+\sin x-2 \sin ^{3} x \\ & =2 \sin x-2 \sin ^{3} x+\sin x-2 \sin ^{3} x \\ & =3 \sin x-4 \sin ^{3} x \end{aligned} $

18. $\cos 3 x=4 \cos ^{3} x-3 \cos x$

We have,

$ \begin{aligned} \cos 3 x & =\cos (2 x+x) \\ & =\cos 2 x \cos x-\sin 2 x \sin x \\ & =(2 \cos ^{2} x-1) \cos x-2 \sin x \cos x \sin x \\ & =(2 \cos ^{2} x-1) \cos x-2 \cos x(1-\cos ^{2} x) \\ & =2 \cos ^{3} x-\cos x-2 \cos x+2 \cos ^{3} x \\ & =4 \cos ^{3} x-3 \cos x . \end{aligned} $

19. $\tan 3 x=\frac{3 \tan x-\tan ^{3} x}{1-3 \tan ^{2} x}$ if $3 x \neq n \pi+\frac{\pi}{2}$, where $n$ is an integer

We have $\tan 3 x=\tan (2 x+x)$

$ =\frac{\tan 2 x+\tan x}{1-\tan 2 x \tan x}=\frac{\frac{2 \tan x}{1-\tan ^{2} x}+\tan x}{1-\frac{2 \tan x \cdot \tan x}{1-\tan ^{2} x}} $

$ =\frac{2 \tan x+\tan x-\tan ^{3} x}{1-\tan ^{2} x-2 \tan ^{2} x}=\frac{3 \tan x-\tan ^{3} x}{1-3 \tan ^{2} x} $

20.

$ \text{ (i) } \cos x+\cos y=2 \cos \frac{x+y}{2} \cos \frac{x-y}{2} $

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

(iii) $\sin x+\sin y=2 \sin \frac{x+y}{2} \cos \frac{x-y}{2}$

(iv) $\sin x-\sin y=2 \cos \frac{x+y}{2} \sin \frac{x-y}{2}$

We know that

$\cos (x+y)=\cos x \cos y-\sin x \sin y \quad \quad \quad \quad \ldots (1)$

and $\quad \cos (x-y)=\cos x \cos y+\sin x \sin y \quad \quad \quad \quad \ldots (2)$

Adding and subtracting (1) and (2), we get

$ \cos (x+y)+\cos (x-y)=2 \cos x \cos y \quad \quad \quad \quad \ldots (3) $

and $\quad \cos (x+y)-\cos (x-y)=-2 \sin x \sin y \quad \quad \quad \quad \ldots (4)$

Further $\sin (x+y)=\sin x \cos y+\cos x \sin y \quad \quad \quad \quad \ldots (5)$

and $\quad \sin (x-y)=\sin x \cos y-\cos x \sin y \quad \quad \quad \quad \ldots (6)$

Adding and subtracting (5) and (6), we get

$ \begin{aligned} & \sin (x+y)+\sin (x-y)=2 \sin x \cos y \quad \quad \quad \quad \ldots (7) \\ & \sin (x+y)-\sin (x-y)=2 \cos x \sin y \quad \quad \quad \quad \ldots (8) \end{aligned} $

Let $x+y=\theta$ and $x-y=\phi$. Therefore

$ x=(\frac{\theta+\phi}{2}) \text{ and } y=(\frac{\theta-\phi}{2}) $

Substituting the values of $x$ and $y$ in (3), (4), (7) and (8), we get

$ \begin{aligned} & \cos \theta+\cos \phi=2 \cos (\frac{\theta+\phi}{2}) \cos (\frac{\theta-\phi}{2}) \\ & \cos \theta-\cos \phi=-2 \sin (\frac{\theta+\phi}{2}) \sin (\frac{\theta-\phi}{2}) \\ & \sin \theta+\sin \phi=2 \sin (\frac{\theta+\phi}{2}) \cos (\frac{\theta-\phi}{2}) \end{aligned} $

$ \sin \theta-\sin \phi=2 \cos (\frac{\theta+\phi}{2}) \sin (\frac{\theta-\phi}{2}) $

Since $\theta$ and $\phi$ can take any real values, we can replace $\theta$ by $x$ and $\phi$ by $y$.

Thus, we get

$ \begin{aligned} & \cos x+\cos y=2 \cos \frac{x+y}{2} \cos \frac{x-y}{2} ; \cos x-\cos y=-2 \sin \frac{x+y}{2} \sin \frac{x-y}{2}, \\ & \sin x+\sin y=2 \sin \frac{x+y}{2} \cos \frac{x-y}{2} ; \sin x-\sin y=2 \cos \frac{x+y}{2} \sin \frac{x-y}{2} . \end{aligned} $

Remark As a part of identities given in 20, we can prove the following results:

21. (i) $2 \cos x \cos y=\cos (x+y)+\cos (x-y)$

(ii) $-2 \sin x \sin y=\cos (x+y)-\cos (x-y)$

(iii) $2 \sin x \cos y=\sin (x+y)+\sin (x-y)$

(iv) $2 \cos x \sin y=\sin (x+y)-\sin (x-y)$.

Example 10 Prove that

$ 3 \sin \frac{\pi}{6} \sec \frac{\pi}{3}-4 \sin \frac{5 \pi}{6} \cot \frac{\pi}{4}=1 $

Solution We have

$ \begin{aligned} \text{ L.H.S. } & =3 \sin \frac{\pi}{6} \sec \frac{\pi}{3}-4 \sin \frac{5 \pi}{6} \cot \frac{\pi}{4} \\ & =3 \times \frac{1}{2} \times 2-4 \sin (\pi-\frac{\pi}{6}) \times 1=3-4 \sin \frac{\pi}{6} \\ & =3-4 \times \frac{1}{2}=1=\text{ R.H.S. } \end{aligned} $

Example 11 Find the value of $\sin 15^{\circ}$.

Solution We have

$ \begin{aligned} \sin 15^{\circ} & =\sin (45^{\circ}-30^{\circ}) \\ & =\sin 45^{\circ} \cos 30^{\circ}-\cos 45^{\circ} \sin 30^{\circ} \\ & =\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \times \frac{1}{2}=\frac{\sqrt{3}-1}{2 \sqrt{2}} . \end{aligned} $

Example 12 Find the value of $\tan \frac{13 \pi}{12}$.

Solution We have

$ \begin{aligned} \tan \frac{13 \pi}{12} & =\tan (\pi+\frac{\pi}{12})=\tan \frac{\pi}{12}=\tan (\frac{\pi}{4}-\frac{\pi}{6}) \\ & =\frac{\tan \frac{\pi}{4}-\tan \frac{\pi}{6}}{1+\tan \frac{\pi}{4} \tan \frac{\pi}{6}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}=2-\sqrt{3} \end{aligned} $

Example 13 Prove that

$ \frac{\sin (x+y)}{\sin (x-y)}=\frac{\tan x+\tan y}{\tan x-\tan y} . $

Solution We have

$ \text{ L.H.S. }=\frac{\sin (x+y)}{\sin (x-y)}=\frac{\sin x \cos y+\cos x \sin y}{\sin x \cos y-\cos x \sin y} $

Dividing the numerator and denominator by $\cos x \cos y$, we get

$ \frac{\sin (x+y)}{\sin (x-y)}=\frac{\tan x+\tan y}{\tan x-\tan y} $

Example 14 Show that

$ \tan 3 x \tan 2 x \tan x=\tan 3 x-\tan 2 x-\tan x $

Solution We know that $3 x=2 x+x$

Therefore, $\tan 3 x=\tan (2 x+x)$

or $\quad \tan 3 x=\frac{\tan 2 x+\tan x}{1-\tan 2 x \tan x}$

or

$ \tan 3 x-\tan 3 x \tan 2 x \tan x=\tan 2 x+\tan x $

or

$ \tan 3 x-\tan 2 x-\tan x=\tan 3 x \tan 2 x \tan x $

or

$ \tan 3 x \tan 2 x \tan x=\tan 3 x-\tan 2 x-\tan x . $

Example 15 Prove that

$ \cos (\frac{\pi}{4}+x)+\cos (\frac{\pi}{4}-x)=\sqrt{2} \cos x $

Solution Using the Identity 20(i), we have

$ \begin{aligned} \text{ L.H.S. } & =\cos (\frac{\pi}{4}+x)+\cos (\frac{\pi}{4}-x) \\ & =2 \cos (\frac{\frac{\pi}{4}+x+\frac{\pi}{4}-x}{2}) \cos (\frac{\frac{\pi}{4}+x-(\frac{\pi}{4}-x)}{2}) \\ & =2 \cos \frac{\pi}{4} \cos x=2 \times \frac{1}{\sqrt{2}} \cos x=\sqrt{2} \cos x=\text{ R.H.S. } \end{aligned} $

Example 16 Prove that $\frac{\cos 7 x+\cos 5 x}{\sin 7 x-\sin 5 x}=\cot x$

Solution Using the Identities 20 (i) and 20 (iv), we get

$ \text{ L.H.S. }=\frac{2 \cos \frac{7 x+5 x}{2} \cos \frac{7 x-5 x}{2}}{2 \cos \frac{7 x+5 x}{2} \sin \frac{7 x-5 x}{2}}=\frac{\cos x}{\sin x}=\cot x=\text{ R.H.S. } $

Example 17 Prove that $=\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x$

Solution We have

$ \begin{aligned} \text{ L.H.S. } & =\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\frac{\sin 5 x+\sin x-2 \sin 3 x}{\cos 5 x-\cos x} \\ & =\frac{2 \sin 3 x \cos 2 x-2 \sin 3 x}{-2 \sin 3 x \sin 2 x}=-\frac{\sin 3 x(\cos 2 x-1)}{\sin 3 x \sin 2 x} \\ & =\frac{1-\cos 2 x}{\sin 2 x}=\frac{2 \sin ^{2} x}{2 \sin x \cos x}=\tan x=\text{ R.H.S. } \end{aligned} $

EXERCISE 3.3

Prove that:

1. $\sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}=-\frac{1}{2}$

Show Answer

Answer :

L.H.S. $=\sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}$

$=(\frac{1}{2})^{2}+(\frac{1}{2})^{2}-(1)^{2}$

$=\frac{1}{4}+\frac{1}{4}-1=-\frac{1}{2}$

$=$ R.H.S.

2. $2 \sin ^{2} \frac{\pi}{6}+cosec^{2} \frac{7 \pi}{6} \cos ^{2} \frac{\pi}{3}=\frac{3}{2}$

Show Answer

Answer :

L.H.S. $=2 \sin ^{2} \frac{\pi}{6}+cosec 2 \frac{7 \pi}{6} \cos ^{2} \frac{\pi}{3}$

$ \begin{aligned} & =2(\frac{1}{2})^{2}+cosec^{2}(\pi+\frac{\pi}{6})(\frac{1}{2})^{2} \\ & =2 \times \frac{1}{4}+(-cosec \frac{\pi}{6})^{2}(\frac{1}{4}) \\ & =\frac{1}{2}+(-2)^{2}(\frac{1}{4}) \\ & =\frac{1}{2}+\frac{4}{4}=\frac{1}{2}+1=\frac{3}{2} \\ & =\text{ R.H.S. } \end{aligned} $

3. $\cot ^{2} \frac{\pi}{6}+cosec \frac{5 \pi}{6}+3 \tan ^{2} \frac{\pi}{6}=6$

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Answer :

L.H.S. $=\cot ^{2} \frac{\pi}{6}+cosec \frac{5 \pi}{6}+3 \tan ^{2} \frac{\pi}{6}$

$=(\sqrt{3})^{2}+cosec(\pi-\frac{\pi}{6})+3(\frac{1}{\sqrt{3}})^{2}$

$=3+cosec \frac{\pi}{6}+3 \times \frac{1}{3}$

$=3+2+1=6$

$=$ R.H.S

4. $2 \sin ^{2} \frac{3 \pi}{4}+2 \cos ^{2} \frac{\pi}{4}+2 \sec ^{2} \frac{\pi}{3}=10$

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Answer :

L.H.S $=2 \sin ^{2} \frac{3 \pi}{4}+2 \cos ^{2} \frac{\pi}{4}+2 \sec ^{2} \frac{\pi}{3}$ $=2{\sin (\pi-\frac{\pi}{4})}^{2}+2(\frac{1}{\sqrt{2}})^{2}+2(2)^{2}$

$=2{\sin \frac{\pi}{4}}^{2}+2 \times \frac{1}{2}+8$

$=2(\frac{1}{\sqrt{2}})^{2}+1+8$

$=1+1+8$

$=10$

$=$ R.H.S

5. Find the value of:

(i) $\sin 75^{\circ}$

(ii) $\tan 15^{\circ}$

Show Answer

Answer :

(i) $\sin 75^{\circ}=\sin (45^{\circ}+30^{\circ})$

$=\sin 45^{\circ} \cos 30^{\circ}+\cos 45^{\circ} \sin 30^{\circ}$

$[\sin (x+y)=\sin x \cos y+\cos x \sin y]$

$=(\frac{1}{\sqrt{2}})(\frac{\sqrt{3}}{2})+(\frac{1}{\sqrt{2}})(\frac{1}{2})$

$=\frac{\sqrt{3}}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}=\frac{\sqrt{3}+1}{2 \sqrt{2}}$

(ii) $\tan 15^{\circ}=\tan (45^{\circ} \hat{a} \in^{\prime \prime} 30^{\circ})$

$ \begin{aligned} & =\frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ} \tan 30^{\circ}} \quad[\tan (x-y)=\frac{\tan x-\tan y}{1+\tan x \tan y}] \\ & =\frac{1-\frac{1}{\sqrt{3}}}{1+1(\frac{1}{\sqrt{3}})}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}} \\ & =\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{(\sqrt{3}-1)^{2}}{(\sqrt{3}+1)(\sqrt{3}-1)}=\frac{3+1-2 \sqrt{3}}{(\sqrt{3})^{2}-(1)^{2}} \\ & =\frac{4-2 \sqrt{3}}{3-1}=2-\sqrt{3} \end{aligned} $

Prove the following:

6. $\cos (\frac{\pi}{4}-x) \cos (\frac{\pi}{4}-y)-\sin (\frac{\pi}{4}-x) \sin (\frac{\pi}{4}-y)=\sin (x+y)$

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Answer :

$ \begin{aligned} & \cos (\frac{\pi}{4}-x) \cos (\frac{\pi}{4}-y)-\sin (\frac{\pi}{4}-x) \sin (\frac{\pi}{4}-y) \\ &= \frac{1}{2}[2 \cos (\frac{\pi}{4}-x) \cos (\frac{\pi}{4}-y)]+\frac{1}{2}[-2 \sin (\frac{\pi}{4}-x) \sin (\frac{\pi}{4}-y)] \\ &= \frac{1}{2}[\cos {(\frac{\pi}{4}-x)+(\frac{\pi}{4}-y)}+\cos {(\frac{\pi}{4}-x)-(\frac{\pi}{4}-y)}] \\ &+\frac{1}{2}[\cos {(\frac{\pi}{4}-x)+(\frac{\pi}{4}-y)}-\cos {(\frac{\pi}{4}-x)-(\frac{\pi}{4}-y)}] \\ & {[\because 2 \cos A \cos B=\cos (A+B)+\cos (A-B)] } \\ &-2 \sin A \sin B=\cos (A+B)-\cos (A-B)] \\ &= 2 \times \frac{1}{2}[\cos {(\frac{\pi}{4}-x)+(\frac{\pi}{4}-y)}] \\ &= \cos [\frac{\pi}{2}-(x+y)] \\ &= \sin (x+y) \\ &= \text{ R.H.S } \end{aligned} $

7. $\frac{\tan (\frac{\pi}{4}+x)}{\tan (\frac{\pi}{4}-x)}=(\frac{1+\tan x}{1-\tan x})^{2} \quad$

Show Answer

Answer :

It is known that

$ \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B} \text{ and } \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B} $

$ \frac{\tan (\frac{\pi}{4}+x)}{\tan (\frac{\pi}{4}-x)}=\frac{(\frac{\tan \frac{\pi}{4}+\tan x}{1-\tan \frac{\pi}{4} \tan x})}{(\frac{\tan \frac{\pi}{4}-\tan x}{1+\tan \frac{\pi}{4} \tan x})}=\frac{(\frac{1+\tan x}{1-\tan x})}{(\frac{1-\tan x}{1+\tan x})}=(\frac{1+\tan x}{1-\tan x})^{2}=\text{ R.H.S. } $

8. $\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos (\frac{\pi}{2}+x)}=\cot ^{2} x$

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Answer :

$ \begin{aligned} \text{ L.H.S. } & =\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos (\frac{\pi}{2}+x)} \\ & =\frac{[-\cos x][\cos x]}{(\sin x)(-\sin x)} \\ & =\frac{-\cos ^{2} x}{-\sin ^{2} x} \\ & =\cot ^{2} x \\ & =\text{ R.H.S. } \end{aligned} $

9. $\cos (\frac{3 \pi}{2}+x) \cos (2 \pi+x)[\cot (\frac{3 \pi}{2}-x)+\cot (2 \pi+x)]=1$

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Answer :

$ \begin{aligned} & \text{ L.H.S. }=\cos (\frac{3 \pi}{2}+x) \cos (2 \pi+x)[\cot (\frac{3 \pi}{2}-x)+\cot (2 \pi+x)] \\ & =\sin x \cos x[\tan x+\cot x] \\ & =\sin x \cos x(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}) \\ & =(\sin x \cos x)[\frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos x}] \\ & =1=\text{ R.H.S. } \end{aligned} $

10. $\sin (n+1) x \sin (n+2) x+\cos (n+1) x \cos (n+2) x=\cos x$

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Answer :

L.H.S. $=\sin (n+1) x \sin (n+2) x+\cos (n+1) x \cos (n+2) x$

$=\frac{1}{2}[2 \sin (n+1) x \sin (n+2) x+2 \cos (n+1) x \cos (n+2) x]$

$=\frac{1}{2} \begin{cases} \cos {(n+1) x-(n+2) x}-\cos {(n+1) x+(n+2) x} \\ +\cos {(n+1) x+(n+2) x}+\cos {(n+1) x-(n+2) x} \end{cases} $

$ \begin{cases} \because-2 \sin A \sin B=\cos (A+B)-\cos (A-B) \\ 2 \cos A \cos B=\cos (A+B)+\cos (A-B) \end{cases} $

$=\frac{1}{2} \times 2 \cos {(n+1) x-(n+2) x}$

$=\cos (-x)=\cos x=$ R.H.S.

11. $\cos (\frac{3 \pi}{4}+x)-\cos (\frac{3 \pi}{4}-x)=-\sqrt{2} \sin x$

Show Answer

Answer :

It is known that

$ \cos A-\cos B=-2 \sin (\frac{A+B}{2}) \cdot \sin (\frac{A-B}{2}) $

$\therefore$ L.H.S. $=\cos (\frac{3 \pi}{4}+x)-\cos (\frac{3 \pi}{4}-x)$

$=-2 \sin {\frac{(\frac{3 \pi}{4}+x)+(\frac{3 \pi}{4}-x)}{2}} \cdot \sin {\frac{(\frac{3 \pi}{4}+x)-(\frac{3 \pi}{4}-x)}{2}}$

$=-2 \sin (\frac{3 \pi}{4}) \sin x$

$=-2 \sin (\pi-\frac{\pi}{4}) \sin x$

$=-2 \sin \frac{\pi}{4} \sin x$

$=-2 \times \frac{1}{\sqrt{2}} \times \sin x$

$=-\sqrt{2} \sin x$

$=$ R.H.S.

12. $\sin ^{2} 6 x-\sin ^{2} 4 x=\sin 2 x \sin 10 x \quad$

Show Answer

Answer :

It is known

that $\sin A+\sin B=2 \sin (\frac{A+B}{2}) \cos (\frac{A-B}{2}), \sin A-\sin B=2 \cos (\frac{A+B}{2}) \sin (\frac{A-B}{2})$

$\therefore$ L.H.S. $=\sin ^{2} 6 x$ -$\sin ^{2} 4 x$

$=(\sin 6 x+\sin 4 x)(\sin 6 x$ -$\sin$

$4 x)$

$=[2 \sin (\frac{6 x+4 x}{2}) \cos (\frac{6 x-4 x}{2})][2 \cos (\frac{6 x+4 x}{2}) \cdot \sin (\frac{6 x-4 x}{2})]$

$ \begin{aligned} & =(2 \sin 5 x \cos x)(2 \cos 5 x \sin x) \\ & =(2 \sin 5 x \cos 5 x)(2 \sin x \cos x) \\ & =\sin 10 x \sin 2 x \\ & =\text{ R.H.S. } \end{aligned} $

13. $\sin 2 x+2 \sin 4 x+\sin 6 x=4 \cos ^{2} x \sin 4 x$

Show Answer

Answer :

L.H.S. $=\sin 2 x+2 \sin 4 x+\sin 6 x$

$=[\sin 2 x+\sin 6 x]+2 \sin 4 x$

$=[2 \sin (\frac{2 x+6 x}{2}) \cos (\frac{2 x-6 x}{2})]+2 \sin 4 x$

$[\because \sin A+\sin B=2 \sin (\frac{A+B}{2}) \cos (\frac{A-B}{2})]$

$=2 \sin 4 x \cos ( 2 x)+2 \sin 4 x$

$ \begin{aligned} & =2 \sin 4 x \cos 2 x+2 \sin 4 x \\ & =2 \sin 4 x(\cos 2 x+1) \\ & =2 \sin 4 x(2 \cos ^{2} x - 1+1) \\ & =2 \sin 4 x(2 \cos ^{2} x) \\ & =4 \cos ^{2} x \sin 4 x \\ & =\text{ R.H.S. } \end{aligned} $

14. $\cot 4 x(\sin 5 x+\sin 3 x)=\cot x(\sin 5 x-\sin 3 x)$

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Answer :

L.H.S $=\cot 4 x(\sin 5 x+\sin 3 x)$

$=\frac{\cos 4 x}{\sin 4 x}[2 \sin (\frac{5 x+3 x}{2}) \cos (\frac{5 x-3 x}{2})]$

$[\because \sin A+\sin B=2 \sin (\frac{A+B}{2}) \cos (\frac{A-B}{2})]$

$=(\frac{\cos 4 x}{\sin 4 x})[2 \sin 4 x \cos x]$

$=2 \cos 4 x \cos x$

R.H.S. $=\cot x(\sin 5 x$ - $\sin 3 x)$

$=\frac{\cos x}{\sin x}[2 \cos (\frac{5 x+3 x}{2}) \sin (\frac{5 x-3 x}{2})]$

$[\because \sin A-\sin B=2 \cos (\frac{A+B}{2}) \sin (\frac{A-B}{2})]$

$=\frac{\cos x}{\sin x}[2 \cos 4 x \sin x]$

$=2 \cos 4 x \cdot \cos x$

L.H.S. = R.H.S.

15. $\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}=-\frac{\sin 2 x}{\cos 10 x}$

Show Answer

Answer :

It is known that

$ \begin{aligned} & \cos A-\cos B=-2 \sin (\frac{A+B}{2}) \sin (\frac{A-B}{2}), \sin A-\sin B=2 \cos (\frac{A+B}{2}) \sin (\frac{A-B}{2}) \\ & \therefore \text{ L.H.S }=\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x} \\ & =\frac{-2 \sin (\frac{9 x+5 x}{2}) \cdot \sin (\frac{9 x-5 x}{2})}{2 \cos (\frac{17 x+3 x}{2}) \cdot \sin (\frac{17 x-3 x}{2})} \\ & =\frac{-2 \sin 7 x \cdot \sin 2 x}{2 \cos 10 x \cdot \sin 7 x} \\ & =-\frac{\sin 2 x}{\cos 10 x} \\ & =\text{ R.H.S. } \end{aligned} $

16. $\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}=\tan 4 x$

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Answer :

It is known that

$ \sin A+\sin B=2 \sin (\frac{A+B}{2}) \cos (\frac{A-B}{2}), \cos A+\cos B=2 \cos (\frac{A+B}{2}) \cos (\frac{A-B}{2}) $

$\therefore$ L.H.S. $=\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}$

$=\frac{2 \sin (\frac{5 x+3 x}{2}) \cdot \cos (\frac{5 x-3 x}{2})}{2 \cos (\frac{5 x+3 x}{2}) \cdot \cos (\frac{5 x-3 x}{2})}$

$=\frac{2 \sin 4 x \cdot \cos x}{2 \cos 4 x \cdot \cos x}$

$=\frac{\sin 4 x}{\cos 4 x}$

$=\tan 4 x=$ R.H.S.

17. $\frac{\sin x-\sin y}{\cos x+\cos y}=\tan \frac{x-y}{2}$

Show Answer

Answer :

It is known that

$\sin A-\sin B=2 \cos (\frac{A+B}{2}) \sin (\frac{A-B}{2}), \cos A+\cos B=2 \cos (\frac{A+B}{2}) \cos (\frac{A-B}{2})$

$\therefore$ L.H.S. $=\frac{\sin x-\sin y}{\cos x+\cos y}$

$=\frac{2 \cos (\frac{x+y}{2}) \cdot \sin (\frac{x-y}{2})}{2 \cos (\frac{x+y}{2}) \cdot \cos (\frac{x-y}{2})}$

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

$=\tan (\frac{x-y}{2})=$ R.H.S.

18. $\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}=\tan 2 x$

Show Answer

Answer :

It is known that

$\sin A+\sin B=2 \sin (\frac{A+B}{2}) \cos (\frac{A-B}{2}), \cos A+\cos B=2 \cos (\frac{A+B}{2}) \cos (\frac{A-B}{2})$

$\therefore$ L.H.S. $=\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}$

$ \begin{aligned} & =\frac{2 \sin (\frac{x+3 x}{2}) \cos (\frac{x-3 x}{2})}{2 \cos (\frac{x+3 x}{2}) \cos (\frac{x-3 x}{2})} \\ & =\frac{\sin 2 x}{\cos 2 x} \\ & =\tan 2 x \\ & =\text{ R.H.S } \end{aligned} $

19. $\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}=2 \sin x$

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Answer :

It is known that

$ \sin A-\sin B=2 \cos (\frac{A+B}{2}) \sin (\frac{A-B}{2}), \cos ^{2} A-\sin ^{2} A=\cos 2 A $

$\therefore$ L.H.S. $=\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}$

$=\frac{2 \cos (\frac{x+3 x}{2}) \sin (\frac{x-3 x}{2})}{-\cos 2 x}$

$=\frac{2 \cos 2 x \sin (-x)}{-\cos 2 x}$

$=-2 \times(-\sin x)$

$=2 \sin x=$ R.H.S.

20. $\frac{\cos 4 x+\cos 3 x+\cos 2 x}{\sin 4 x+\sin 3 x+\sin 2 x}=\cot 3 x$

Show Answer

Answer :

L.H.S. $=\frac{\cos 4 x+\cos 3 x+\cos 2 x}{\sin 4 x+\sin 3 x+\sin 2 x}$

$ \begin{aligned} & =\frac{(\cos 4 x+\cos 2 x)+\cos 3 x}{(\sin 4 x+\sin 2 x)+\sin 3 x} \\ & =\frac{2 \cos (\frac{4 x+2 x}{2}) \cos (\frac{4 x-2 x}{2})+\cos 3 x}{2 \sin (\frac{4 x+2 x}{2}) \cos (\frac{4 x-2 x}{2})+\sin 3 x} \\ & {[\because \cos A+\cos B=2 \cos (\frac{A+B}{2}) \cos (\frac{A-B}{2}), \sin A+\sin B=2 \sin (\frac{A+B}{2}) \cos (\frac{A-B}{2})]} \\ & =\frac{2 \cos 3 x \cos x+\cos 3 x}{2 \sin 3 x \cos x+\sin 3 x} \\ & =\frac{\cos 3 x(2 \cos x+1)}{\sin 3 x(2 \cos x+1)} \\ & =\cot 3 x=\text{ R.H.S. } \end{aligned} $

21. $\cot x \cot 2 x-\cot 2 x \cot 3 x-\cot 3 x \cot x=1$

Show Answer

Answer :

L.H.S. $=\cot x \cot 2 x$ -$\cot 2 x \cot 3 x$ -$\cot 3 x \cot x$

$=\cot x \cot 2 x$ -$\cot 3 x(\cot 2 x+\cot x)$

$=\cot x \cot 2 x - \cot (2 x+x)(\cot 2 x+\cot x)$

$=\cot x \cot 2 x-[\frac{\cot 2 x \cot x-1}{\cot x+\cot 2 x}](\cot 2 x+\cot x)$

$[\because \cot (A+B)=\frac{\cot A \cot B-1}{\cot A+\cot B}]$

$=\cot x \cot 2 x$ -( $\cot 2 x \cot x$ -1$)$

$=1=$ R.H.S.

22. $\tan 4 x=\frac{4 \tan x(1-\tan ^{2} x)}{1-6 \tan ^{2} x+\tan ^{4} x} \quad$ 24. $\cos 4 x=1-8 \sin ^{2} x \cos ^{2} x$

Show Answer

Answer :

$ \begin{aligned} & \text{ It is known that } \tan 2 A=\frac{2 \tan A}{1-\tan ^{2} A} \\ & \therefore \text{ L.H.S. }=\tan 4 x=\tan 2(2 x) \\ & =\frac{2 \tan 2 x}{1-\tan ^{2}(2 x)} \\ & =\frac{2(\frac{2 \tan x}{1-\tan ^{2} x})}{1-(\frac{2 \tan x}{1-\tan ^{2} x})^{2}} \\ & =\frac{(\frac{4 \tan x}{1-\tan ^{2} x})}{[1-\frac{4 \tan ^{2} x}{(1-\tan ^{2} x)^{2}}]} \\ & =\frac{(\frac{4 \tan x}{1-\tan ^{2} x})}{[\frac{(1-\tan ^{2} x)^{2}-4 \tan ^{2} x}{(1-\tan ^{2} x)^{2}}]} \\ & =\frac{4 \tan x(1-\tan ^{2} x)}{(1-\tan ^{2} x)^{2}-4 \tan ^{2} x} \\ & =\frac{4 \tan x(1-\tan ^{2} x)}{1+\tan ^{4} x-2 \tan ^{2} x-4 \tan ^{2} x} \\ & =\frac{4 \tan x(1-\tan ^{2} x)}{1-6 \tan ^{2} x+\tan ^{4} x}=\text{ R.H.S. } \end{aligned} $

23. $\cos 6 x=32 \cos ^{6} x-48 \cos ^{4} x+18 \cos ^{2} x-1$

Show Answer

Answer :

L.H.S. $=\cos 6 x$

$=\cos 3(2 x)$

$=4 \cos ^{3} 2 x-3 \cos 2 x[\cos 3 A=4 \cos ^{3} A-3 \cos A]$

$=4[(2 \cos ^{2} x-1)^{3}-3(2 \cos ^{2} x-1)[\cos 2 x=2 \cos ^{2} x-1].$

$=4[(2 \cos ^{2} x)^{3}-(1)^{3}-3(2 \cos ^{2} x)^{2}+3(2 \cos ^{2} x)]-6 \cos ^{2} x+3$

$=4[8 \cos ^{6} x-1-12 \cos ^{4} x+6 \cos ^{2} x]-6 \cos ^{2} x+3$

$=32 \cos ^{6} x-4-48 \cos ^{4} x+24 \cos ^{2} x-6 \cos ^{2} x+3$

$=32 \cos ^{6} x-48 \cos ^{4} x+18 \cos ^{2} x-1$

$=$ R.H.S.

Miscellaneous Examples

Example 18 If $\sin x=\frac{3}{5}, \cos y=-\frac{12}{13}$, where $x$ and $y$ both lie in second quadrant, find the value of $\sin (x+y)$.

Solution We know that

$ \begin{aligned} & \sin (x+y)=\sin x \cos y+\cos x \sin y \\ & \cos ^{2} x=1-\sin ^{2} x=1-\frac{9}{25}=\frac{16}{25} \end{aligned} $

Therefore $\cos x= \pm \frac{4}{5}$.

Since $x$ lies in second quadrant, $\cos x$ is negative.

Hence $\quad \cos x=-\frac{4}{5}$

Now $\quad \sin ^{2} y=1-\cos ^{2} y=1-\frac{144}{169}=\frac{25}{169}$

i.e. $\quad \sin y= \pm \frac{5}{13}$.

Since $y$ lies in second quadrant, hence $\sin y$ is positive. Therefore, $\sin y=\frac{5}{13}$. Substituting the values of $\sin x, \sin y, \cos x$ and $\cos y$ in (1), we get

$ \sin (x+y)=\frac{3}{5} \times(-\frac{12}{13})+(-\frac{4}{5}) \times \frac{5}{13}=-\frac{36}{65}-\frac{20}{65}=-\frac{56}{65} $

Example 19 Prove that

$ \cos 2 x \cos \frac{x}{2}-\cos 3 x \cos \frac{9 x}{2}=\sin 5 x \sin \frac{5 x}{2} $

Solution We have

L.H.S. $=\frac{1}{2}[2 \cos 2 x \cos \frac{x}{2}-2 \cos \frac{9 x}{2} \cos 3 x]$

$ \begin{aligned} & =\frac{1}{2}[\cos (2 x+\frac{x}{2})+\cos (2 x-\frac{x}{2})-\cos (\frac{9 x}{2}+3 x)-\cos (\frac{9 x}{2}-3 x)] \\ & =\frac{1}{2}[\cos \frac{5 x}{2}+\cos \frac{3 x}{2}-\cos \frac{15 x}{2}-\cos \frac{3 x}{2}]=\frac{1}{2}[\cos \frac{5 x}{2}-\cos \frac{15 x}{2}] \\ & =\frac{1}{2}[-2 \sin {\frac{\frac{5 x}{2}+\frac{15 x}{2}}{2}} \sin {\frac{\frac{5 x}{2}-\frac{15 x}{2}}{2}}] \\ & =-\sin 5 x \sin (-\frac{5 x}{2})=\sin 5 x \sin \frac{5 x}{2}=\text{ R.H.S. } \end{aligned} $

Example 20 Find the value of $\tan \frac{\pi}{8}$.

Solution Let $x=\frac{\pi}{8}$. Then $2 x=\frac{\pi}{4}$.

Now $\quad \tan 2 x=\frac{2 \tan x}{1-\tan ^{2} x}$

or

$ \tan \frac{\pi}{4}=\frac{2 \tan \frac{\pi}{8}}{1-\tan ^{2} \frac{\pi}{8}} $

Let $y=\tan \frac{\pi}{8}$. Then $1=\frac{2 y}{1-y^{2}}$

or $\quad y^{2}+2 y-1=0$

Therefore

$ y=\frac{-2 \pm 2 \sqrt{2}}{2}=-1 \pm \sqrt{2} $

Since $\frac{\pi}{8}$ lies in the first quadrant, $y=\tan \frac{\pi}{8}$ is positve. Hence

$ \tan \frac{\pi}{8}=\sqrt{2}-1 $

Example 21 If $\tan x=\frac{3}{4}, \pi<x<\frac{3 \pi}{2}$, find the value of $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$.

Solution Since $\pi<x<\frac{3 \pi}{2}, \cos x$ is negative.

Also

$ \frac{\pi}{2}<\frac{x}{2}<\frac{3 \pi}{4} $

Therefore, $\sin \frac{x}{2}$ is positive and $\cos \frac{x}{2}$ is negative.

Now

$ \sec ^{2} x=1+\tan ^{2} x=1+\frac{9}{16}=\frac{25}{16} $

Therefore

$ \cos ^{2} x=\frac{16}{25} \quad \text{ or } \cos x=-\frac{4}{5} \quad \text{ (Why?) } $

Now

$ 2 \sin ^{2} \frac{x}{2}=1-\cos x=1+\frac{4}{5}=\frac{9}{5} \text{. } $

Therefore

$ \sin ^{2} \frac{x}{2}=\frac{9}{10} $

or

$ \sin \frac{x}{2}=\frac{3}{\sqrt{10}} \quad \text{ (Why?) } $

Again

$ 2 \cos ^{2} \frac{x}{2}=1+\cos x=1-\frac{4}{5}=\frac{1}{5} $

Therefore $\quad \cos ^{2} \frac{x}{2}=\frac{1}{10}$

or

$ \cos \frac{x}{2}=-\frac{1}{\sqrt{10}}(\text{ Why? }) $

Hence $\quad \tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}=\frac{3}{\sqrt{10}} \times(\frac{-\sqrt{10}}{1})=-3$.

Example 22 Prove that $\cos ^{2} x+\cos ^{2}(x+\frac{\pi}{3})+\cos ^{2}(x-\frac{\pi}{3})=\frac{3}{2}$.

Solution We have

$ \begin{aligned} \text{ L.H.S. } & =\frac{1+\cos 2 x}{2}+\frac{1+\cos (2 x+\frac{2 \pi}{3})}{2}+\frac{1+\cos (2 x-\frac{2 \pi}{3})}{2} . \\ & =\frac{1}{2}[3+\cos 2 x+\cos (2 x+\frac{2 \pi}{3})+\cos (2 x-\frac{2 \pi}{3})] \\ & =\frac{1}{2}[3+\cos 2 x+2 \cos 2 x \cos \frac{2 \pi}{3}] \\ & =\frac{1}{2}[3+\cos 2 x+2 \cos 2 x \cos (\pi-\frac{\pi}{3})] \\ & =\frac{1}{2}[3+\cos 2 x-2 \cos 2 x \cos \frac{\pi}{3}] \\ & =\frac{1}{2}[3+\cos 2 x-\cos 2 x]=\frac{3}{2}=\text{ R.H.S. } \end{aligned} $

Miscellaneous Exercise on Chapter 3

Prove that:

1. $2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}=0$

Show Answer

Answer :

L.H.S.

$=2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}$

$=2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+2 \cos (\frac{\frac{3 \pi}{13}+\frac{5 \pi}{13}}{2}) \cos (\frac{\frac{3 \pi}{13}-\frac{5 \pi}{13}}{2})[\cos x+\cos y=2 \cos (\frac{x+y}{2}) \cos (\frac{x-y}{2})]$

$=2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+2 \cos \frac{4 \pi}{13} \cos (\frac{-\pi}{13})$

$=2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+2 \cos \frac{4 \pi}{13} \cos \frac{\pi}{13}$

$=2 \cos \frac{\pi}{13}[\cos \frac{9 \pi}{13}+\cos \frac{4 \pi}{13}]$

$=2 \cos \frac{\pi}{13}[2 \cos (\frac{\frac{9 \pi}{13}+\frac{4 \pi}{13}}{2}) \cos (\frac{\frac{9 \pi}{13}-\frac{4 \pi}{13}}{2})]$

$=2 \cos \frac{\pi}{13}[2 \cos \frac{\pi}{2} \cos \frac{5 \pi}{26}]$

$=2 \cos \frac{\pi}{13} \times 2 \times 0 \times \cos \frac{5 \pi}{26}$

$=0=$ R.H.S

2. $(\sin 3 x+\sin x) \sin x+(\cos 3 x-\cos x) \cos x=0$

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Answer :

L.H.S.

$=(\sin 3 x+\sin x) \sin x+(\cos 3 x$ ấ" $\cos x) \cos x$

$=\sin 3 x \sin x+\sin ^{2} x+\cos 3 x \cos x-\cos ^{2} x$

$=\cos 3 x \cos x+\sin 3 x \sin x-(\cos ^{2} x-\sin ^{2} x)$

$=\cos (3 x-x)-\cos 2 x \quad[\cos (A-B)=\cos A \cos B+\sin A \sin B]$

$=\cos 2 x-\cos 2 x$

$=0$

$=$ RH.S.

3. $(\cos x+\cos y)^{2}+(\sin x-\sin y)^{2}=4 \cos ^{2} \frac{x+y}{2}$

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Answer :

L.H.S. $=(\cos x+\cos y)^{2}+(\sin x-\sin y)^{2}$

$=\cos ^{2} x+\cos ^{2} y+2 \cos x \cos y+\sin ^{2} x+\sin ^{2} y-2 \sin x \sin y$

$=(\cos ^{2} x+\sin ^{2} x)+(\cos ^{2} y+\sin ^{2} y)+2(\cos x \cos y-\sin x \sin y)$

$=1+1+2 \cos (x+y) \quad[\cos (A+B)=(\cos A \cos B-\sin A \sin B)]$

$=2+2 \cos (x+y)$

$=2[1+\cos (x+y)]$

$=2[1+2 \cos ^{2}(\frac{x+y}{2})-1] \quad[\cos 2 A=2 \cos ^{2} A-1]$

$=4 \cos ^{2}(\frac{x+y}{2})=$ R.H.S.

4. $(\cos x-\cos y)^{2}+(\sin x-\sin y)^{2}=4 \sin ^{2} \frac{x-y}{2}$

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Answer :

L.H.S. $=(\cos x-\cos y)^{2}+(\sin x-\sin y)^{2}$

$ \begin{matrix} =\cos ^{2} x+\cos ^{2} y-2 \cos x \cos y+\sin ^{2} x+\sin ^{2} y-2 \sin x \sin y \\ =(\cos ^{2} x+\sin ^{2} x)+(\cos ^{2} y+\sin ^{2} y)-2[\cos x \cos y+\sin x \sin y] \\ =1+1-2[\cos (x-y)] & \\ {[\cos (A-B)=\cos A \cos B+\sin A \sin B]} \\ =2[1-\cos (x-y)] & \\ =2[1-{1-2 \sin ^{2}(\frac{x-y}{2})}] & \\ {[\cos 2 A=1-2 \sin ^{2} A]} \\ =4 \sin ^{2}(\frac{x-y}{2})=\text{ R.H.S. } \end{matrix} $

5. $\sin x+\sin 3 x+\sin 5 x+\sin 7 x=4 \cos x \cos 2 x \sin 4 x$

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Answer :

It is known that

$ \sin A+\sin B=2 \sin (\frac{A+B}{2}) \cdot \cos (\frac{A-B}{2}) $

$\therefore$ L.H.S. $=\sin x+\sin 3 x+\sin 5 x+\sin 7 x$

$=(\sin x+\sin 5 x)+(\sin 3 x+\sin 7 x)$

$=2 \sin (\frac{x+5 x}{2}) \cdot \cos (\frac{x-5 x}{2})+2 \sin (\frac{3 x+7 x}{2}) \cos (\frac{3 x-7 x}{2})$

$=2 \sin 3 x \cos (-2 x)+2 \sin 5 x \cos (-2 x)$

$=2 \sin 3 x \cos 2 x+2 \sin 5 x \cos 2 x$

$=2 \cos 2 x[\sin 3 x+\sin 5 x]$

$=2 \cos 2 x[2 \sin (\frac{3 x+5 x}{2}) \cdot \cos (\frac{3 x-5 x}{2})]$

$=2 \cos 2 x[2 \sin 4 x \cdot \cos (-x)]$

$=4 \cos 2 x \sin 4 x \cos x=$ R.H.S.

6. $\frac{(\sin 7 x+\sin 5 x)+(\sin 9 x+\sin 3 x)}{(\cos 7 x+\cos 5 x)+(\cos 9 x+\cos 3 x)}=\tan 6 x$

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Answer :

It is known that

$\sin A+\sin B=2 \sin (\frac{A+B}{2}) \cdot \cos (\frac{A-B}{2}), \cos A+\cos B=2 \cos (\frac{A+B}{2}) \cdot \cos (\frac{A-B}{2})$

$=\frac{(\sin 7 x+\sin 5 x)+(\sin 9 x+\sin 3 x)}{(\cos 7 x+\cos 5 x)+(\cos 9 x+\cos 3 x)}$

$=\frac{[2 \sin (\frac{7 x+5 x}{2}) \cdot \cos (\frac{7 x-5 x}{2})]+[2 \sin (\frac{9 x+3 x}{2}) \cdot \cos (\frac{9 x-3 x}{2})]}{[2 \cos (\frac{7 x+5 x}{2}) \cdot \cos (\frac{7 x-5 x}{2})]+[2 \cos (\frac{9 x+3 x}{2}) \cdot \cos (\frac{9 x-3 x}{2})]}$

$=\frac{[2 \sin 6 x \cdot \cos x]+[2 \sin 6 x \cdot \cos 3 x]}{[2 \cos 6 x \cdot \cos x]+[2 \cos 6 x \cdot \cos 3 x]}$

$=\frac{2 \sin 6 x[\cos x+\cos 3 x]}{2 \cos 6 x[\cos x+\cos 3 x]}$

$=\tan 6 x$

$=$ R.H.S.

7. $\sin 3 x+\sin 2 x-\sin x=4 \sin x \cos \frac{x}{2} \cos \frac{3 x}{2}$

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Answer :

L.H.S. $=\sin 3 x+\sin 2 x-\sin x$

$ \begin{aligned} & =\sin 3 x+(\sin 2 x-\sin x) \\ & =\sin 3 x+[2 \cos (\frac{2 x+x}{2}) \sin (\frac{2 x-x}{2})] \quad[\sin A-\sin B=2 \cos (\frac{A+B}{2}) \sin (\frac{A-B}{2})] \\ & =\sin 3 x+[2 \cos (\frac{3 x}{2}) \sin (\frac{x}{2})] \\ & =\sin 3 x+2 \cos \frac{3 x}{2} \sin \frac{x}{2} \\ & =2 \sin \frac{3 x}{2} \cdot \cos \frac{3 x}{2}+2 \cos \frac{3 x}{2} \sin \frac{x}{2} \quad[\sin 2 A=2 \sin A \cdot \cos B] \\ & =2 \cos (\frac{3 x}{2})[\sin (\frac{3 x}{2})+\sin (\frac{x}{2})] \\ & =2 \cos (\frac{3 x}{2})[2 \sin {\frac{(\frac{3 x}{2})+(\frac{x}{2})}{2}} \cos {\frac{(\frac{3 x}{2})-(\frac{x}{2})}{2}}][\sin A+\sin B=2 \sin (\frac{A+B}{2}) \cos (\frac{A-B}{2})] \\ & .=2 \cos (\frac{3 x}{2}) \cdot 2 \sin x \cos (\frac{x}{2})] \cos (\frac{3 x}{2})=\text{ R.H.S. } \end{aligned} $

Find $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$ in each of the following:

8. $\tan x=-\frac{4}{3}, x$ in quadrant II 9. $\cos x=-\frac{1}{3}, x$ in quadrant III

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Answer :

Here, $x$ is in quadrant II.

i.e., $\frac{\pi}{2}<x<\pi$

$\Rightarrow \frac{\pi}{4}<\frac{x}{2}<\frac{\pi}{2}$

Therefore, $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$ are all positive.

It is given that $\tan x=-\frac{4}{3}$.

$\sec ^{2} x=1+\tan ^{2} x=1+(\frac{-4}{3})^{2}=1+\frac{16}{9}=\frac{25}{9}$

$\therefore \cos ^{2} x=\frac{9}{25}$

$\Rightarrow \cos x= \pm \frac{3}{5}$

As $x$ is in quadrant II, $\cos x$ is negative.

$ \therefore \quad \cos x=\frac{-3}{5} $

Now, $\cos x=2 \cos ^{2} \frac{x}{2}-1$

$\Rightarrow \frac{-3}{5}=2 \cos ^{2} \frac{x}{2}-1$

$\Rightarrow 2 \cos ^{2} \frac{x}{2}=1-\frac{3}{5}$

$\Rightarrow 2 \cos ^{2} \frac{x}{2}=\frac{2}{5}$

$\Rightarrow \cos ^{2} \frac{x}{2}=\frac{1}{5}$

$\Rightarrow \cos \frac{x}{2}=\frac{1}{\sqrt{5}} \quad[\because \cos \frac{x}{2}.$ is positive $]$

$\therefore \cos \frac{x}{2}=\frac{\sqrt{5}}{5}$

$\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}=1$

$\Rightarrow \sin ^{2} \frac{x}{2}+(\frac{1}{\sqrt{5}})^{2}=1$

$\Rightarrow \sin ^{2} \frac{x}{2}=1-\frac{1}{5}=\frac{4}{5}$

$\Rightarrow \sin \frac{x}{2}=\frac{2}{\sqrt{5}}$

$[\because \sin \frac{x}{2}.$ is positive $]$

$\therefore \sin \frac{x}{2}=\frac{2 \sqrt{5}}{5}$ $\tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}=\frac{(\frac{2}{\sqrt{5}})}{(\frac{1}{\sqrt{5}})}=2$

Thus, the respective values of $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$ are $\frac{2 \sqrt{5}}{5}, \frac{\sqrt{5}}{5}$, and 2

9. Find $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$ for $\cos x=-\frac{1}{3}, x$

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Answer :

Here, $x$ is in quadrant III.

i.e., $\pi<x<\frac{3 \pi}{2}$

$\Rightarrow \frac{\pi}{2}<\frac{x}{2}<\frac{3 \pi}{4}$

Therefore, $\cos \frac{x}{2}$ and $\tan \frac{x}{2}$ are negative, whereas $\sin \frac{x}{2}$ is positive.

It is given that $\cos x=-\frac{1}{3}$.

$\cos x=1-2 \sin ^{2} \frac{x}{2}$

$\Rightarrow \sin ^{2} \frac{x}{2}=\frac{1-\cos x}{2}$

$\Rightarrow \sin ^{2} \frac{x}{2}=\frac{1-(-\frac{1}{3})}{2}=\frac{(1+\frac{1}{3})}{2}=\frac{\frac{4}{3}}{2}=\frac{2}{3}$

$\Rightarrow \sin \frac{x}{2}=\frac{\sqrt{2}}{\sqrt{3}} \quad[\because \sin \frac{x}{2}.$ is positive $]$

$\therefore \sin \frac{x}{2}=\frac{\sqrt{2}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{6}}{3}$

Now,

$\cos x=2 \cos ^{2} \frac{x}{2}-1$ $\Rightarrow \cos ^{2} \frac{x}{2}=\frac{1+\cos x}{2}=\frac{1+(-\frac{1}{3})}{2}=\frac{(\frac{3-1}{3})}{2}=\frac{(\frac{2}{3})}{2}=\frac{1}{3}$

$\Rightarrow \cos \frac{x}{2}=-\frac{1}{\sqrt{3}} \quad[\because \cos \frac{x}{2}.$ is negative $]$

$\therefore \cos \frac{x}{2}=-\frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{-\sqrt{3}}{3}$

$\tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}=\frac{(\frac{\sqrt{2}}{\sqrt{3}})}{(\frac{-1}{\sqrt{3}})}=-\sqrt{2}$

Thus, the respective values of $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$ are $\frac{\sqrt{6}}{3}, \frac{-\sqrt{3}}{3}$, and $-\sqrt{2}$

10. $\sin x=\frac{1}{4}, x$ in quadrant II

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Answer :

Here, $x$ is in quadrant II.

i.e., $\frac{\pi}{2}<x<\pi$

$\Rightarrow \frac{\pi}{4}<\frac{x}{2}<\frac{\pi}{2}$

Therefore, $\sin \frac{x}{2}, \cos \frac{x}{2}$, and $\tan \frac{x}{2}$ are all positive.

It is given that $\sin x=\frac{1}{4}$.

$\cos ^{2} x=1-\sin ^{2} x=1-(\frac{1}{4})^{2}=1-\frac{1}{16}=\frac{15}{16}$

$\Rightarrow \cos x=-\frac{\sqrt{15}}{4}$

[cos $x$ is negative in quadrant II]

$ \begin{aligned} & \sin ^{2} \frac{x}{2}=\frac{1-\cos x}{2}=\frac{1-(-\frac{\sqrt{15}}{4})}{2}=\frac{4+\sqrt{15}}{8} \\ & \Rightarrow \sin \frac{x}{2}=\sqrt{\frac{4+\sqrt{15}}{8}} \quad[\because \sin \frac{x}{2} \text{ is positive }] \\ & =\sqrt{\frac{4+\sqrt{15}}{8} \times \frac{2}{2}} \\ & =\sqrt{\frac{8+2 \sqrt{15}}{16}} \\ & =\frac{\sqrt{8+2 \sqrt{15}}}{4} \\ & \cos ^{2} \frac{x}{2}=\frac{1+\cos x}{2}=\frac{1+(-\frac{\sqrt{15}}{4})}{2}=\frac{4-\sqrt{15}}{8} \\ & \Rightarrow \cos \frac{x}{2}=\sqrt{\frac{4-\sqrt{15}}{8}} \quad[\because \cos \frac{x}{2} \text{ is positive }] \\ & =\sqrt{\frac{4-\sqrt{15}}{8} \times \frac{2}{2}} \\ & =\sqrt{\frac{8-2 \sqrt{15}}{16}} \\ & =\frac{\sqrt{8-2 \sqrt{15}}}{4} \\ & \tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}=\frac{(\frac{\sqrt{8+2 \sqrt{15}}}{4})}{(\frac{\sqrt{8-2 \sqrt{15}}}{4})}=\frac{\sqrt{8+2 \sqrt{15}}}{\sqrt{8-2 \sqrt{15}}} \\ & =\sqrt{\frac{8+2 \sqrt{15}}{8-2 \sqrt{15}} \times \frac{8+2 \sqrt{15}}{8+2 \sqrt{15}}} \\ & =\sqrt{\frac{(8+2 \sqrt{15})^{2}}{64-60}}=\frac{8+2 \sqrt{15}}{2}=4+\sqrt{15} \end{aligned} $

Thus, the respective values of $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$ are $\frac{\sqrt{8+2 \sqrt{15}}}{4}, \frac{\sqrt{8-2 \sqrt{15}}}{4}$, and $4+\sqrt{15}$

Summary

If in a circle of radius $r$, an arc of length $l$ subtends an angle of $\theta$ radians, then $l=r \theta$

Radian measure $=\frac{\pi}{180} \times$ Degree measure

Degree measure $=\frac{180}{\pi} \times$ Radian measure

$\cos ^{2} x+\sin ^{2} x=1$

$1+\tan ^{2} x=\sec ^{2} x$

$1+\cot ^{2} x=cosec^{2} x$

$ \cos (2 n \pi+x)=\cos x$

$ \sin (2 n \pi+x)=\sin x$

$ \sin (-x)=-\sin x$

$ \cos (-x)=\cos x$

$ \cos (x+y)=\cos x \cos y-\sin x \sin y$

$\cos (x-y)=\cos x \cos y+\sin x \sin y$

$\cos (\frac{\pi}{2}-x)=\sin x$ $\sin (\frac{\pi}{2}-x)=\cos x$

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

$ \sin (x-y)=\sin x \cos y-\cos x \sin y$

$\cos (\frac{\pi}{2}+x)=-\sin x$

$ \sin (\frac{\pi}{2}+x)=\cos x $

$\cos (\pi-x)=-\cos x$

$ \sin (\pi-x)=\sin x $

$\cos (\pi+x)=-\cos x$

$\sin (\pi+x)=-\sin x$

$\cos (2 \pi-x)=\cos x$

$ \sin (2 \pi-x)=-\sin x $

If none of the angles $x, y$ and $(x \pm y)$ is an odd multiple of $\frac{\pi}{2}$, then

$\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y}$

$\tan (x-y)=\frac{\tan x-\tan y}{1+\tan x \tan y}$

If none of the angles $x, y$ and $(x \pm y)$ is a multiple of $\pi$, then

$ \cot (x+y)=\frac{\cot x \cot y-1}{\cot y+\cot x} $

$\cot (x-y)=\frac{\cot x \cot y+1}{\cot y-\cot x}$

$\cos 2 x=\cos ^{2} x-\sin ^{2} x=2 \cos ^{2} x-1=1-2 \sin ^{2} x=\frac{1-\tan ^{2} x}{1+\tan ^{2} x}$

$ \sin 2 x=2 \sin x \cos x=\frac{2 \tan x}{1+\tan ^{2} x}$

$ \tan 2 x=\frac{2 \tan x}{1-\tan ^{2} x}$

$\sin 3 x=3 \sin x-4 \sin ^{3} x$

$ \cos 3 x=4 \cos ^{3} x-3 \cos x$ $\tan 3 x=\frac{3 \tan x-\tan ^{3} x}{1-3 \tan ^{2} x}$

(i) $\cos x+\cos y=2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}$

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

(iii) $\sin x+\sin y=2 \sin \frac{x+y}{2} \cos \frac{x-y}{2}$

(iv) $\sin x-\sin y=2 \cos \frac{x+y}{2} \sin \frac{x-y}{2}$

(i) $2 \cos x \cos y=\cos (x+y)+\cos (x-y)$

(ii) $-2 \sin x \sin y=\cos (x+y)-\cos (x-y)$

(iii) $2 \sin x \cos y=\sin (x+y)+\sin (x-y)$

(iv) $2 \cos x \sin y=\sin (x+y)-\sin (x-y)$.

Historical Note

The study of trigonometry was first started in India. The ancient Indian Mathematicians, Aryabhatta (476), Brahmagupta (598), Bhaskara I (600) and Bhaskara II (1114) got important results. All this knowledge first went from India to middle-east and from there to Europe. The Greeks had also started the study of trigonometry but their approach was so clumsy that when the Indian approach became known, it was immediately adopted throughout the world.

In India, the predecessor of the modern trigonometric functions, known as the sine of an angle, and the introduction of the sine function represents the main contribution of the siddhantas (Sanskrit astronomical works) to the history of mathematics.

Bhaskara I (about 600) gave formulae to find the values of sine functions for angles more than $90^{\circ}$. A sixteenth century Malayalam work Yuktibhasa (period) contains a proof for the expansion of $\sin (A+B)$. Exact expression for sines or cosines of $18^{\circ}, 36^{\circ}, 54^{\circ}, 72^{\circ}$, etc., are given by Bhaskara II.

The symbols $\sin ^{-1} x, \cos ^{-1} x$, etc., for arc $\sin x$, arc $\cos x$, etc., were suggested by the astronomer Sir John F.W. Hersehel (1813) The names of Thales (about 600 B.C.) is invariably associated with height and distance problems. He is credited with the determination of the height of a great pyramid in Egypt by measuring shadows of the pyramid and an auxiliary staff (or gnomon) of known height, and comparing the ratios:

$ \frac{H}{S}=\frac{h}{s}=\tan (\text{ sun’s altitude }) $

Thales is also said to have calculated the distance of a ship at sea through the proportionality of sides of similar triangles. Problems on height and distance using the similarity property are also found in ancient Indian works.