Class 11 Maths Chapter 3 – Trigonometric Functions Examples

Eg 1. Convert $40^\circ20'$ into radian measure.

Sol. We know that $180^\circ=\pi$ radian.

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

Using $1^\circ=60'$ ,

40^\circ20'=\left(40+\frac{20}{60}\right)^\circ=\left(40\frac13\right)^\circ=\left(\frac{121}{3}\right)^\circ

Substituting in the formula,

\begin{array}{l} \displaystyle 40^\circ20'=\frac{121}{3}\times\frac{\pi}{180}\text{ radian} \\[0.8em] \hspace{2.8em} \displaystyle =\frac{121\pi}{540}\text{ radian} \end{array}

∴ $40^\circ20'=\frac{121\pi}{540}$ radian.

Eg 2. Convert 6 radians into degree measure.

Sol. We know that $\pi$ radian $=180^\circ$ .

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

\begin{array}{l} \displaystyle 6\text{ radians}=\frac{180}{\pi}\times6\text{ degree} \end{array}

Using $\pi=\frac{22}{7}$ ,

\begin{array}{l} \displaystyle =\frac{180\times7\times6}{22}\text{ degree} \end{array}

\begin{array}{l} \displaystyle =\frac{3780}{11}\text{ degree}=343\frac{7}{11}\text{ degree} \end{array}

Using $1^\circ=60'$ ,

\begin{array}{l} \displaystyle =343^\circ+\frac{7}{11}\times60' \end{array}

\begin{array}{l} \displaystyle =343^\circ+38\frac{2}{11}' \end{array}

Using $1'=60''$ ,

\begin{array}{l} \displaystyle =343^\circ+38'+\frac{2}{11}\times60'' \end{array}

\begin{array}{l} \displaystyle =343^\circ+38'+10.9'' \\[0.4em] \displaystyle \approx343^\circ38'11'' \end{array}

∴ $6$ radians $=343^\circ38'11''$ approximately.

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

Sol. Here, $l=37.4\text{ cm}$ and

\theta=60^\circ=60\times\frac{\pi}{180}=\frac{\pi}{3}\text{ radian}

We know that

\textcolor{#1B5E20}{ r=\frac{l}{\theta} }

Substituting the values,

\begin{array}{l} \displaystyle r=\frac{37.4}{\dfrac{\pi}{3}}=\frac{37.4\times3}{\pi} \\[1.8em] \displaystyle =\frac{37.4\times3\times7}{22} \\[0.8em] \displaystyle =35.7\text{ cm} \end{array}

∴ The radius of the circle is $35.7\text{ cm}$ .

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

Sol. Here, $r=1.5\text{ cm}$ .

In 60 minutes, the minute hand of a watch completes one revolution, i.e., it moves $2\pi$ radians or $360^\circ$ .

∴ In 40 minutes, it moves

\theta=\frac{2\pi}{60}\times40=\frac{4\pi}{3}\text{ radians}

Let $l$ be the distance moved by the tip of the minute hand. Then,

\textcolor{#1B5E20}{ l=r\theta }

Substituting the values,

\begin{array}{l} \displaystyle l=1.5\times\frac{4\pi}{3}\text{ cm}=2\pi\text{ cm} \\[0.8em] \displaystyle =2\times3.14\text{ cm} \\[0.4em] \displaystyle =6.28\text{ cm} \end{array}

∴ The tip of the minute hand moves $6.28\text{ cm}$ .

Eg 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.

Sol. Let $r_1$ and $r_2$ be the radii of the two circles and $l$ be the length of each arc.

Converting the angles into radians,

\begin{array}{l} \displaystyle \theta_1=65^\circ=65\times\frac{\pi}{180}=\frac{13\pi}{36}\text{ radian} \\[0.8em] \displaystyle \theta_2=110^\circ=110\times\frac{\pi}{180}=\frac{22\pi}{36}\text{ radian} \end{array}

We know that $l=r\theta$ . Since the arcs are of the same length,

r_1\theta_1=r_2\theta_2

Substituting the values,

\begin{array}{l} \displaystyle \frac{13\pi}{36}\times r_1=\frac{22\pi}{36}\times r_2 \\[0.8em] \displaystyle \frac{r_1}{r_2}=\frac{22}{13} \end{array}

∴ $r_1:r_2=22:13$ .

Eg 6. If $\cos x=-\frac35$ , $x$ lies in the third quadrant, find the values of other five trigonometric functions.

Sol. Given, $\cos x=-\frac35$ .

\sec x=\frac{1}{\cos x}=\frac{1}{-\dfrac35}=-\frac53

We know that

\textcolor{#1B5E20}{ \sin^2x+\cos^2x=1 }

Substituting the value of $\cos x$ ,

\displaystyle \sin^2x=1-\left(-\frac35\right)^2=1-\frac{9}{25}=\frac{16}{25}

\displaystyle \sin x=\pm\frac45

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

\begin{array}{l} \displaystyle \sin x=-\frac45 \\[0.4em] \displaystyle \text{cosec}\,x=\frac{1}{\sin x}=\frac{1}{-\dfrac45}=-\frac54 \end{array}

Now,

\begin{array}{l} \displaystyle \tan x=\frac{\sin x}{\cos x}=\frac{-\dfrac45}{-\dfrac35}=\frac43 \\[2em] \displaystyle \cot x=\frac{\cos x}{\sin x}=\frac{-\dfrac35}{-\dfrac45}=\frac34 \end{array}

∴ $\sin x=-\frac45$ , $\text{cosec}\,x=-\frac54$ , $\sec x=-\frac53$ , $\tan x=\frac43$ and $\cot x=\frac34$ .

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

Sol. Given, $\cot x=-\frac{5}{12}$ .

\tan x=\frac{1}{\cot x}=\frac{1}{-\dfrac{5}{12}}=-\frac{12}{5}

We know that

\textcolor{#1B5E20}{ \sec^2x=1+\tan^2x }

Substituting the value of $\tan x$ ,

\displaystyle \sec^2x=1+\left(-\frac{12}{5}\right)^2=1+\frac{144}{25}=\frac{169}{25}

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

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

\begin{array}{l} \displaystyle \sec x=-\frac{13}{5} \\[0.8em] \displaystyle \cos x=\frac{1}{\sec x}=\frac{1}{-\dfrac{13}{5}}=-\frac{5}{13} \end{array}

Now,

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

\displaystyle \text{cosec}\,x=\frac{1}{\sin x}=\frac{1}{\dfrac{12}{13}}=\frac{13}{12}

∴ $\tan x=-\frac{12}{5}$ , $\sec x=-\frac{13}{5}$ , $\cos x=-\frac{5}{13}$ , $\sin x=\frac{12}{13}$ and $\text{cosec}\,x=\frac{13}{12}$ .

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

Sol. We know that the values of $\sin x$ repeat after an interval of $2\pi$ .

Writing $\frac{31\pi}{3}=10\pi+\frac{\pi}{3}$ ,

\begin{array}{l} \displaystyle \sin\frac{31\pi}{3}=\sin\left(10\pi+\frac{\pi}{3}\right) \\[0.8em] \hspace{3em} \displaystyle =\sin\frac{\pi}{3}=\frac{\sqrt3}{2} \end{array}

Eg 9. Find the value of $\cos(-1710^\circ)$ .

Sol. We know that the values of $\cos x$ repeat after an interval of $360^\circ$ .

Writing $-1710^\circ=-5\times360^\circ+90^\circ$ ,

\begin{array}{l} \displaystyle \cos(-1710^\circ)=\cos(-5\times360^\circ+90^\circ) \\[0.4em] \hspace{0.8em} \displaystyle =\cos90^\circ \\[0.4em] \hspace{0.8em} \displaystyle =0 \end{array}

Eg 10. Prove that $3\sin\frac{\pi}{6}\sec\frac{\pi}{3}-4\sin\frac{5\pi}{6}\cot\frac{\pi}{4}=1$

Sol. Using $\sin(\pi-x)=\sin x$ ,

\sin\frac{5\pi}{6}=\sin\left(\pi-\frac{\pi}{6}\right)=\sin\frac{\pi}{6}=\frac12

Also, $\sin\frac{\pi}{6}=\frac12$ , $\sec\frac{\pi}{3}=2$ and $\cot\frac{\pi}{4}=1$ .

\text{LHS}=3\sin\frac{\pi}{6}\sec\frac{\pi}{3}-4\sin\frac{5\pi}{6}\cot\frac{\pi}{4}

Substituting the values,

\begin{array}{l} \displaystyle =3\times\frac12\times2-4\times\frac12\times1 \\[0.4em] \displaystyle =3-2 \\[0.4em] \displaystyle =1=\text{RHS} \end{array}

Hence proved.

Eg 11. Find the value of $\sin15^\circ$ .

Sol. Writing $15^\circ=45^\circ-30^\circ$ ,

\begin{array}{l} \displaystyle \sin15^\circ=\sin(45^\circ-30^\circ) \end{array}

Using the identity

\textcolor{#1B5E20}{ \begin{array}{l} \displaystyle \sin(x-y)=\sin x\cos y-\cos x\sin y \end{array} }

\begin{array}{l} \hspace{0.8em} \displaystyle =\sin45^\circ\cos30^\circ-\cos45^\circ\sin30^\circ \end{array}

Substituting the values,

\begin{array}{l} \displaystyle =\frac{1}{\sqrt2}\times\frac{\sqrt3}{2}-\frac{1}{\sqrt2}\times\frac12 \\[0.8em] \displaystyle =\frac{\sqrt3-1}{2\sqrt2} \end{array}

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

Sol. Writing $\frac{13\pi}{12}=\pi+\frac{\pi}{12}$ ,

\begin{array}{l} \displaystyle \tan\frac{13\pi}{12}=\tan\left(\pi+\frac{\pi}{12}\right) \end{array}

Using $\tan(\pi+x)=\tan x$ ,

\begin{array}{l} \hspace{0.8em} \displaystyle =\tan\frac{\pi}{12} \end{array}

Writing $\frac{\pi}{12}=\frac{\pi}{4}-\frac{\pi}{6}$ ,

\begin{array}{l} \displaystyle \tan\frac{\pi}{12}=\tan\left(\frac{\pi}{4}-\frac{\pi}{6}\right) \end{array}

Using the identity

\textcolor{#1B5E20}{ \begin{array}{l} \displaystyle \tan(x-y)=\frac{\tan x-\tan y}{1+\tan x\tan y} \end{array} }

\begin{array}{l} \hspace{0.8em} \displaystyle =\frac{\tan\dfrac{\pi}{4}-\tan\dfrac{\pi}{6}}{1+\tan\dfrac{\pi}{4}\tan\dfrac{\pi}{6}} \end{array}

Substituting the values,

\begin{array}{l} \hspace{2.6em} \displaystyle =\frac{1-\dfrac{1}{\sqrt3}}{1+\dfrac{1}{\sqrt3}} \\[1.2em] \hspace{2.6em} \displaystyle =\frac{\sqrt3-1}{\sqrt3+1} \end{array}

Rationalising the denominator,

\begin{array}{l} \hspace{0.8em} \displaystyle =\frac{\sqrt3-1}{\sqrt3+1}\times\frac{\sqrt3-1}{\sqrt3-1} \\[0.8em] \hspace{0.8em} \displaystyle =\frac{(\sqrt3-1)^2}{(\sqrt3+1)(\sqrt3-1)} \\[0.8em] \hspace{0.8em} \displaystyle =\frac{4-2\sqrt3}{2} \\[0.8em] \hspace{0.8em} \displaystyle =2-\sqrt3 \end{array}

Eg 13. Prove that $\frac{\sin(x+y)}{\sin(x-y)}=\frac{\tan x+\tan y}{\tan x-\tan y}$ .

Sol.

\begin{array}{l} \displaystyle \text{LHS}=\frac{\sin(x+y)}{\sin(x-y)} \end{array}

Using $\sin(x\pm y)=\sin x\cos y\pm\cos x\sin y$ ,

\begin{array}{l} \hspace{0.8em} =\frac{\sin x\cos y+\cos x\sin y}{\sin x\cos y-\cos x\sin y} \end{array}

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

\begin{array}{l} \hspace{0.8em} \displaystyle =\frac{\dfrac{\sin x}{\cos x}+\dfrac{\sin y}{\cos y}}{\dfrac{\sin x}{\cos x}-\dfrac{\sin y}{\cos y}} \\[0.8em] \hspace{0.8em} \displaystyle =\frac{\tan x+\tan y}{\tan x-\tan y}=\text{RHS} \end{array}

Hence proved.

Eg 14. Show that $\tan3x\tan2x\tan x=\tan3x-\tan2x-\tan x$

Sol. We know that $3x=2x+x$ .

Using $\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$ ,

\tan3x=\tan(2x+x)=\frac{\tan2x+\tan x}{1-\tan2x\tan x}

Cross-multiplying,

\begin{array}{l} \displaystyle \tan3x(1-\tan2x\tan x)=\tan2x+\tan x \\[0.4em] \hspace{0.8em} \displaystyle \Rightarrow\tan3x-\tan3x\tan2x\tan x=\tan2x+\tan x \\[0.4em] \hspace{0.8em} \displaystyle \Rightarrow\tan3x\tan2x\tan x=\tan3x-\tan2x-\tan x \end{array}

Hence proved.

Eg 15. Prove that $\cos\left(\frac{\pi}{4}+x\right)+\cos\left(\frac{\pi}{4}-x\right)=\sqrt2\cos x$

Sol. Taking $A=\frac{\pi}{4}+x$ and $B=\frac{\pi}{4}-x$ ,

\begin{array}{l} \displaystyle \frac{A+B}{2}=\frac{\pi}{4} \\[0.8em] \displaystyle \frac{A-B}{2}=x \end{array}

Using $\cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$ ,

\begin{array}{l} \displaystyle \text{LHS}=\cos\left(\frac{\pi}{4}+x\right)+\cos\left(\frac{\pi}{4}-x\right) \\[0.8em] \hspace{0.8em} \displaystyle =2\cos\frac{\pi}{4}\cos x \end{array}

\begin{array}{l} \hspace{0.8em} \displaystyle =2\times\frac{1}{\sqrt2}\cos x \\[0.8em] \hspace{0.8em} \displaystyle =\sqrt2\cos x=\text{RHS} \end{array}

Hence proved.

Eg 16. Prove that $\frac{\cos7x+\cos5x}{\sin7x-\sin5x}=\cot x$

Sol.

\begin{array}{l} \displaystyle \text{LHS}=\frac{\cos7x+\cos5x}{\sin7x-\sin5x} \end{array}

Using $\cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$

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

\begin{array}{l} \hspace{1.8em} \displaystyle =\frac{2\cos\dfrac{7x+5x}{2}\cos\dfrac{7x-5x}{2}}{2\cos\dfrac{7x+5x}{2}\sin\dfrac{7x-5x}{2}} \end{array}

\begin{array}{l} \hspace{1.8em} \displaystyle =\frac{2\cos6x\cos x}{2\cos6x\sin x} \\[1em] \hspace{1.8em} \displaystyle =\frac{\cos x}{\sin x} \\[0.8em] \hspace{1.8em} \displaystyle =\cot x=\text{RHS} \end{array}

Hence proved.

Eg 17. Prove that $\frac{\sin5x-2\sin3x+\sin x}{\cos5x-\cos x}=\tan x$

Sol.

\begin{array}{l} \displaystyle \text{LHS}=\frac{\sin5x-2\sin3x+\sin x}{\cos5x-\cos x} \\[1em] \hspace{0.8em} \displaystyle =\frac{(\sin5x+\sin x)-2\sin3x}{\cos5x-\cos x} \end{array}

Using $\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$

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

\begin{array}{l} \hspace{1em} \displaystyle =\frac{2\sin\dfrac{5x+x}{2}\cos\dfrac{5x-x}{2}-2\sin3x}{-2\sin\dfrac{5x+x}{2}\sin\dfrac{5x-x}{2}} \end{array}

\begin{array}{l} \hspace{1em} \displaystyle =\frac{2\sin3x\cos2x-2\sin3x}{-2\sin3x\sin2x} \\[1em] \hspace{1em} \displaystyle =\frac{2\sin3x(\cos2x-1)}{-2\sin3x\sin2x} \\[1em] \hspace{1em} \displaystyle =\frac{1-\cos2x}{\sin2x} \end{array}

Using $1-\cos2x=2\sin^2x$ and $\sin2x=2\sin x\cos x$ ,

\begin{array}{l} \hspace{1em} \displaystyle =\frac{2\sin^2x}{2\sin x\cos x} \end{array}

\begin{array}{l} \hspace{1em} \displaystyle =\frac{\sin x}{\cos x} \\[0.8em] \hspace{1em} \displaystyle =\tan x=\text{RHS} \end{array}

Hence proved.

Eg 18. If $\sin x=\frac35$ , $\cos y=-\frac{12}{13}$ , where $x$ and $y$ both lie in second quadrant, find the value of $\sin(x+y)$ .

Sol. We know that

\textcolor{#1B5E20}{ \sin(x+y)=\sin x\cos y+\cos x\sin y\qquad ...(i) }

Using $\cos^2x=1-\sin^2x$ ,

\begin{array}{l} \displaystyle \cos^2x=1-\left(\frac35\right)^2=1-\frac{9}{25}=\frac{16}{25} \end{array}

\begin{array}{l} \displaystyle \cos x=\pm\frac45 \end{array}

Since $x$ lies in the second quadrant, $\cos x$ is negative, so $\cos x=-\frac45$ .

Using $\sin^2y=1-\cos^2y$ ,

\begin{array}{l} \displaystyle \sin^2y=1-\left(-\frac{12}{13}\right)^2=1-\frac{144}{169}=\frac{25}{169} \end{array}

\begin{array}{l} \displaystyle \sin y=\pm\frac{5}{13} \end{array}

Since $y$ lies in the second quadrant, $\sin y$ is positive, so $\sin y=\frac{5}{13}$ .

Substituting the values of $\sin x$ , $\cos y$ , $\cos x$ and $\sin y$ ,

Using (i),

\begin{array}{l} \displaystyle \sin(x+y)=\frac35\times\left(-\frac{12}{13}\right)+\left(-\frac45\right)\times\frac{5}{13} \end{array}

\begin{array}{l} \hspace{3.6em} \displaystyle =-\frac{36}{65}-\frac{20}{65} \\[1em] \hspace{3.6em} \displaystyle =-\frac{56}{65} \end{array}

Eg 19. Prove that $\cos2x\cos\frac{x}{2}-\cos3x\cos\frac{9x}{2}=\sin5x\sin\frac{5x}{2}$ .

Sol. Using $2\cos A\cos B=\cos(A+B)+\cos(A-B)$ ,

\displaystyle 2\cos2x\cos\frac{x}{2}=\cos\left(2x+\frac{x}{2}\right)+\cos\left(2x-\frac{x}{2}\right)

\displaystyle =\cos\frac{5x}{2}+\cos\frac{3x}{2}

Similarly,

\displaystyle 2\cos\frac{9x}{2}\cos3x=\cos\left(\frac{9x}{2}+3x\right)+\cos\left(\frac{9x}{2}-3x\right)

\displaystyle =\cos\frac{15x}{2}+\cos\frac{3x}{2}

Substituting these expressions in the LHS,

\displaystyle \text{LHS}=\cos2x\cos\frac{x}{2}-\cos3x\cos\frac{9x}{2}

\displaystyle =\frac12\left[2\cos2x\cos\frac{x}{2}-2\cos\frac{9x}{2}\cos3x\right]

\displaystyle =\frac12\left[\left(\cos\frac{5x}{2}+\cos\frac{3x}{2}\right)-\left(\cos\frac{15x}{2}+\cos\frac{3x}{2}\right)\right]

\begin{array}{l} \hspace{0.8em} \displaystyle =\frac12\left[\cos\frac{5x}{2}-\cos\frac{15x}{2}\right] \end{array}

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

\begin{array}{l} \displaystyle =\frac12\left[-2\sin\frac{\dfrac{5x}{2}+\dfrac{15x}{2}}{2}\sin\frac{\dfrac{5x}{2}-\dfrac{15x}{2}}{2}\right] \end{array}

\begin{array}{l} \displaystyle =-\sin5x\sin\left(-\frac{5x}{2}\right) \end{array}

Using $\sin(-\theta)=-\sin\theta$ ,

=\sin5x\sin\frac{5x}{2}=\text{RHS}

Hence proved.

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

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

We know that

\textcolor{#1B5E20}{ \tan2x=\frac{2\tan x}{1-\tan^2x} }

Substituting $x=\frac{\pi}{8}$ ,

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

Let $y=\tan\frac{\pi}{8}$ . Since $\tan\frac{\pi}{4}=1$ ,

\begin{array}{l} \displaystyle 1=\frac{2y}{1-y^2} \end{array}

Cross-multiplying,

\begin{array}{l} \displaystyle 1-y^2=2y \\[0.4em] \hspace{0.8em} \displaystyle \Rightarrow y^2+2y-1=0 \end{array}

Using the quadratic formula $y=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ ,

\begin{array}{l} \displaystyle y=\frac{-2\pm\sqrt{(-2)^2-4(1)(-1)}}{2(1)} \end{array}

\begin{array}{l} \hspace{0.8em} \displaystyle =\frac{-2\pm\sqrt{4+4}}{2}=\frac{-2\pm2\sqrt2}{2} \\[0.8em] \hspace{0.8em} \displaystyle =-1\pm\sqrt2 \end{array}

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

\tan\frac{\pi}{8}=\sqrt2-1

Eg 21. If $\tan x=\frac34$ , $\pi<x<\frac{3\pi}{2}$ , find the value of $\sin\frac{x}{2}$ , $\cos\frac{x}{2}$ and $\tan\frac{x}{2}$ .

Sol. Since $\pi<x<\frac{3\pi}{2}$ , $x$ lies in the third quadrant, so $\cos x$ is negative.

Also,

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

So, $\frac{x}{2}$ lies in the second quadrant; $\sin\frac{x}{2}$ is positive and $\cos\frac{x}{2}$ is negative.

We know that

\textcolor{#1B5E20}{ \sec^2x=1+\tan^2x }

Substituting the value of $\tan x$ ,

\displaystyle \sec^2x=1+\left(\frac34\right)^2=1+\frac{9}{16}=\frac{25}{16}

\displaystyle \cos^2x=\frac{1}{\sec^2x}=\frac{1}{\dfrac{25}{16}}=\frac{16}{25}

Since $\cos x$ is negative, $\cos x=-\frac45$ .

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

\begin{array}{l} \displaystyle 2\sin^2\frac{x}{2}=1-\left(-\frac45\right)=1+\frac45=\frac95 \end{array}

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

\displaystyle \sin\frac{x}{2}=\frac{3}{\sqrt{10}}=\frac{3\sqrt{10}}{10}

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

\begin{array}{l} \displaystyle 2\cos^2\frac{x}{2}=1+\left(-\frac45\right)=1-\frac45=\frac15 \end{array}

\displaystyle \cos^2\frac{x}{2}=\frac{1}{10}

\displaystyle \cos\frac{x}{2}=-\frac{1}{\sqrt{10}}=-\frac{\sqrt{10}}{10}

Therefore,

\tan\frac{x}{2}=\frac{\sin\dfrac{x}{2}}{\cos\dfrac{x}{2}}=\frac{\dfrac{3}{\sqrt{10}}}{-\dfrac{1}{\sqrt{10}}}=-3

∴ $\sin\frac{x}{2}=\frac{3\sqrt{10}}{10}$ , $\cos\frac{x}{2}=-\frac{\sqrt{10}}{10}$ and $\tan\frac{x}{2}=-3$ .

Eg 22. Prove that $\cos^2x+\cos^2\left(x+\frac{\pi}{3}\right)+\cos^2\left(x-\frac{\pi}{3}\right)=\frac32$ .

Sol.

\begin{array}{l} \displaystyle \text{LHS}=\cos^2x+\cos^2\left(x+\frac{\pi}{3}\right)+\cos^2\left(x-\frac{\pi}{3}\right) \end{array}

Using $\cos^2\theta=\frac{1+\cos2\theta}{2}$ ,

\begin{array}{l} \hspace{0.8em} \displaystyle =\frac{1+\cos2x}{2}+\frac{1+\cos\left(2x+\dfrac{2\pi}{3}\right)}{2}+\frac{1+\cos\left(2x-\dfrac{2\pi}{3}\right)}{2} \end{array}

\begin{array}{l} \displaystyle =\frac12\left[3+\cos2x+\cos\left(2x+\frac{2\pi}{3}\right)\right.\allowbreak+\left.\cos\left(2x-\frac{2\pi}{3}\right)\right] \end{array}

Using $\cos(A+B)+\cos(A-B)=2\cos A\cos B$ ,

\begin{array}{l} \displaystyle =\frac12\left[3+\cos2x+2\cos2x\cos\frac{2\pi}{3}\right] \end{array}

Using $\cos(\pi-\theta)=-\cos\theta$ , we have $\cos\frac{2\pi}{3}=-\cos\frac{\pi}{3}=-\frac12$ .

\displaystyle =\frac12\left[3+\cos2x+2\cos2x\times\left(-\frac12\right)\right]

\displaystyle =\frac12\left[3+\cos2x-\cos2x\right]

\displaystyle =\frac32=\text{RHS}

Hence proved.