Miscellaneous Exercise (Revised) - Chapter 3 - Trigonometric Functions - Ncert Solutions class 11 - Maths
Updated On 26-08-2025 By Lithanya
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Chapter 3: Trigonometric Functions | NCERT Solutions for Class 11 Maths
Prove that:
Ex 3.4 Question 1.
$2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}=0$
Answer.
L.H.S. $=2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}$
$=\cos \left(\frac{9 \pi}{13}+\frac{\pi}{13}\right)+\cos \left(\frac{9 \pi}{13}-\frac{\pi}{13}\right)+\cos \left(\frac{3 \pi}{13}\right)+\cos \left(\frac{5 \pi}{13}\right)$
$=\cos \frac{10 \pi}{13}+\cos \frac{8 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}$
$=\cos \left(\pi-\frac{3 \pi}{13}\right)+\cos \left(\pi-\frac{5 \pi}{13}\right)+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}$
$=-\cos \frac{3 \pi}{13}-\cos \frac{5 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}=0=$ R.H.S.
Ex 3.4 Question 2.
$(\sin 3 x+\sin x) \sin x+(\cos 3 x-\cos x) \cos x=0$
Answer.
L.H.S. $=(\sin 3 x+\sin x) \sin x+(\cos 3 x-\cos x) \cos x$
$=\left[2 \sin \left(\frac{3 x+x}{2}\right) \cos \left(\frac{3 x-x}{2}\right)\right] \sin x+\left[-2 \sin \left(\frac{3 x+x}{2}\right) \sin \left(\frac{3 x-x}{2}\right)\right] \cos x$
$\begin{aligned}
& =[2 \sin 2 x \cos x] \sin x+[-2 \sin 2 x \sin x] \cos x \\
& =2 \sin 2 x \cdot \sin x \cdot \cos x-2 \sin 2 x \cdot \sin x \cdot \cos x=0=R H S
\end{aligned}$
Ex 3.4 Question 3.
$(\cos x+\cos y)^2+(\sin x-\sin y)^2=4 \cos ^2 \frac{x+y}{2}$
Answer:
$
\begin{aligned}
& \text {L.H.S. }=(\cos x+\cos y)^2+(\sin x-\sin y)^2 \\
& =\left[2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)\right]^2+\left[2 \cos \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)\right]^2 \\
& =4 \cos ^2\left(\frac{x+y}{2}\right) \cos ^2\left(\frac{x-y}{2}\right)+4 \cos ^2\left(\frac{x+y}{2}\right) \sin ^2\left(\frac{x-y}{2}\right) \\
& =4 \cos ^2\left(\frac{x+y}{2}\right)\left[\cos ^2\left(\frac{x-y}{2}\right)+\sin ^2\left(\frac{x-y}{2}\right)\right] \\
& =4 \cos ^2\left(\frac{x+y}{2}\right)=\text { R.H.S. }
\end{aligned}
$
Ex 3.4 Question 4.
$(\cos x-\cos y)^2+(\sin x-\sin y)^2=4 \sin ^2 \frac{x-y}{2}$
Answer:
$
\begin{aligned}
& \text {L.H.S. }=(\cos x-\cos y)^2+(\sin x-\sin y)^2 \\
& =\left[-2 \sin \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)\right]^2+\left[2 \cos \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)\right]^2 \\
& =4 \sin ^2\left(\frac{x+y}{2}\right) \sin ^2\left(\frac{x-y}{2}\right)+4 \cos ^2\left(\frac{x+y}{2}\right) \sin ^2\left(\frac{x-y}{2}\right) \\
& =4 \sin ^2\left(\frac{x-y}{2}\right)\left[\sin ^2\left(\frac{x+y}{2}\right)+\cos ^2\left(\frac{x+y}{2}\right)\right] \\
& =4 \sin ^2\left(\frac{x-y}{2}\right)=\text { R.H.S. }
\end{aligned}
$
Ex 3.4 Question 5.
$\sin x+\sin 3 x+\sin 5 x+\sin 7 x=4 \cos x \cos 2 x \sin 4 x$
Answer:
$\begin{aligned}
& \text {L.H.S. }=\sin x+\sin 3 x+\sin 5 x+\sin 7 x \\
& =(\sin 7 x+\sin x)+(\sin 5 x+\sin 3 x) \\
& =\left[2 \sin \left(\frac{7 x+x}{2}\right) \cos \left(\frac{7 x-x}{2}\right)\right]+\left[2 \sin \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)\right] \\
& =[2 \sin 4 x \cos 3 x]+[2 \sin 4 x \cos x] \\
& =2 \sin 4 x[\cos 3 x+\cos x] \\
& =2 \sin 4 x\left[2 \cos \left(\frac{3 x+x}{2}\right) \cos \left(\frac{3 x-x}{2}\right)\right] \\
& =2 \sin 4 x[2 \cos 2 x \cos x] \\
& =4 \cos x \cos 2 x \sin 4 x=\text { R.H.S. }
\end{aligned}$
Ex 3.4 Question 6.
$\text {} \frac{(\sin 7 x+\sin 5 x)+(\sin 9 x+\sin 3 x)}{(\cos 7 x+\cos 5 x)+(\cos 9 x+\cos 3 x)}=\tan 6 x$
Answer:
$\begin{aligned}
& \text {L.H.S. }=\frac{(\sin 7 x+\sin 5 x)+(\sin 9 x+\sin 3 x)}{(\cos 7 x+\cos 5 x)+(\cos 9 x+\cos 3 x)} \\
& =\frac{\left[2 \sin \left(\frac{7 x+5 x}{2}\right) \cos \left(\frac{7 x-5 x}{2}\right)\right]+\left[2 \sin \left(\frac{9 x+3 x}{2}\right) \cos \left(\frac{9 x-3 x}{2}\right)\right]}{\left.\left.2 \cos \left(\frac{7 x+5 x}{2}\right) \cos \left(\frac{7 x-5 x}{2}\right)\right]+2 \cos \left(\frac{9 x+3 x}{2}\right) \cos \left(\frac{9 x-3 x}{2}\right)\right]} \\
& =\frac{2 \sin 6 x \cos x+2 \sin 6 x \cos 3 x}{2 \cos 6 x \cos x+2 \cos 6 x \cos 3 x} \\
& =\frac{2 \sin 6 x(\cos x+\cos 3 x)}{2 \cos 6 x(\cos x+\cos 3 x)} \\
& =\tan 6 x=\text { R.H.S. }
\end{aligned}$
Ex 3.4 Question 7.
$\sin 3 x+\sin 2 x-\sin x=4 \sin x \cos \frac{x}{2} \cos \frac{3 x}{2}$
Answer:
$
\begin{aligned}
& \text {L.H.S. }=\sin 3 x+\sin 2 x-\sin x \\
& =(\sin 3 x-\sin x)+\sin 2 x \\
& =\left[2 \cos \left(\frac{3 x+x}{2}\right) \sin \left(\frac{3 x-x}{2}\right)\right]+2 \sin x \cos x \\
& =2 \cos 2 x \sin x+2 \sin x \cos x \\
& =2 \sin x[\cos 2 x+\cos x] \\
& =2 \sin x\left[2 \cos \left(\frac{2 x+x}{2}\right) \cos \left(\frac{2 x-x}{2}\right)\right] \\
& =2 \sin x\left[2 \cos \left(\frac{3 x}{2}\right) \cos \left(\frac{x}{2}\right)\right] \\
& =4 \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:
Ex 3.4 Question 8.
$\tan x=\frac{-4}{3}=x$ in quadrant II.
Answer.
Given: $\tan x=\frac{-4}{3} x$ in quadrant II.
$
\begin{aligned}
& \therefore \sec ^2 x=1+\tan ^2 x \\
& \Rightarrow \sec ^2 x=1+\left(\frac{-4}{3}\right)^2=1+\frac{16}{9}=\frac{25}{9}
\end{aligned}
$
$
\Rightarrow \sec x= \pm \frac{5}{3} \cos x= \pm \frac{3}{5}=\frac{-3}{5}
$
[ $x$ lies in II quadrant]
Also $\frac{\pi}{2} $\therefore \frac{x}{2}$ lies in first quadrant.
$\therefore \sin \frac{x}{2}, \cos \frac{x}{2}, \tan \frac{x}{2}$ are positive.
Now, $\cos \frac{x}{2}=\sqrt{\frac{1+\cos x}{2}}=\sqrt{\frac{1-\frac{3}{5}}{2}}=\sqrt{\frac{1}{5}} \times \frac{\sqrt{5}}{\sqrt{5}}=\frac{\sqrt{5}}{5}$
$\sin \frac{x}{2}=\sqrt{\frac{1-\cos x}{2}}=\sqrt{\frac{1+\frac{3}{5}}{2}}=\sqrt{\frac{4}{5}}=\frac{2}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}}=\frac{2 \sqrt{5}}{5}$
$
\tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}=\frac{\frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}}=2
$
Ex 3.4 Question 9.
$\cos x=\frac{-1}{3}, x$ in quadrant III.
Answer.
Given: $\cos x=\frac{-1}{3}, x$ in quadrant III.
Now, $\pi $\therefore \frac{x}{2}$ lies in second quadrant.
$\therefore \sin \frac{x}{2}$ are positive and $\cos \frac{x}{2}, \tan \frac{x}{2}$ are negative.
Now, $\cos \frac{x}{2}=\sqrt{\frac{1+\cos x}{2}}=-\sqrt{\frac{1-\frac{1}{3}}{2}}=-\frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{-\sqrt{3}}{3}$
$
\begin{aligned}
& \sin \frac{x}{2}=\sqrt{\frac{1-\cos x}{2}}=\sqrt{\frac{1+\frac{1}{3}}{2}}=\sqrt{\frac{2}{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{6}}{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}
\end{aligned}
$
Ex 3.4 Question 10.
$\sin x=\frac{1}{4}, x$ in quadrant II.
Answer.
Given: $\sin x=\frac{1}{4}, x$ in quadrant II.
$
\therefore \cos ^2 x=1-\sin ^2 x
$
$
\begin{aligned}
& \Rightarrow \cos ^2 x=1-\left(\frac{1}{4}\right)^2=1-\frac{1}{16}=\frac{15}{16} \\
& \Rightarrow \cos x= \pm \frac{\sqrt{15}}{4} \cos x=-\frac{\sqrt{15}}{4}=\frac{-3}{5}
\end{aligned}
$
[ $x$ lies in II quadrant]
Also $\frac{\pi}{2} $\therefore \frac{x}{2}$ lies in first quadrant.
$\therefore \sin \frac{x}{2}, \cos \frac{x}{2}, \tan \frac{x}{2}$ are positive.
$
\begin{aligned}
& \sin \frac{x}{2}=\sqrt{\frac{1-\cos x}{2}}=\sqrt{\frac{1+\frac{\sqrt{15}}{4}}{2}}=\sqrt{\frac{4+\sqrt{15}}{8}}=\frac{\sqrt{8+2 \sqrt{15}}}{4} \\
& \tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}=\frac{\sqrt{4+\sqrt{15}}}{\sqrt{4-\sqrt{15}}} \times \frac{\sqrt{4+\sqrt{15}}}{\sqrt{4+\sqrt{15}}}=\frac{4+\sqrt{15}}{\sqrt{16-15}}=4+\sqrt{15}
\end{aligned}
$
