Miscellaneous Example (Revised) - Chapter 3 Trigonometric Functions class 11 ncert solutions Maths - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Chapter 3: Trigonometric Functions | NCERT Solutions for Class 11 Maths

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

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

Therefore $\cos x= \pm \frac{4}{5}$.
Since $x$ lies in second quadrant, $\cos x$ is negative.
Hence
$
\cos x=-\frac{4}{5}
$

Now
$
\sin ^2 y=1-\cos ^2 y=1-\frac{144}{169}=\frac{25}{169}
$
i.e.
$
\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\left(-\frac{12}{13}\right)+\left(-\frac{4}{5}\right) \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

$
\begin{aligned}
\text { L.H.S. } & =\frac{1}{2}\left[2 \cos 2 x \cos \frac{x}{2}-2 \cos \frac{9 x}{2} \cos 3 x\right] \\
= & \frac{1}{2}\left[\cos \left(2 x+\frac{x}{2}\right)+\cos \left(2 x-\frac{x}{2}\right)-\cos \left(\frac{9 x}{2}+3 x\right)-\cos \left(\frac{9 x}{2}-3 x\right)\right] \\
= & \frac{1}{2}\left[\cos \frac{5 x}{2}+\cos \frac{3 x}{2}-\cos \frac{15 x}{2}-\cos \frac{3 x}{2}\right]=\frac{1}{2}\left[\cos \frac{5 x}{2}-\cos \frac{15 x}{2}\right] \\
= & \frac{1}{2}\left[-2 \sin \left\{\frac{\frac{5 x}{2}+\frac{15 x}{2}}{2}\right\} \sin \left\{\frac{\frac{5 x}{2}-\frac{15 x}{2}}{2}\right\}\right] \\
= & -\sin 5 x \sin \left(-\frac{5 x}{2}\right)=\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 $\quad \tan \frac{\pi}{4}=\frac{2 \tan \frac{\pi}{8}}{1-\tan ^2 \frac{\pi}{8}}$

Let $y=\tan \frac{\pi}{8}$. Then $1=\frac{2 y}{1-y^2}$
or
$
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} \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
$
\cos ^2 \frac{x}{2}=\frac{1}{10}
$
or
$
\cos \frac{x}{2}=-\frac{1}{\sqrt{10}} \text { (Why?) }
$

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

Example 22

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

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