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Looking for Class 11 Maths NCERT solutions for Trigonometry Miscellaneous Exercise? You’re at the right place! This section provides comprehensive and step-by-step solutions to all the questions from the Miscellaneous Exercise of Chapter 3 – Trigonometric Functions. These solutions are crafted as per the latest NCERT guidelines, helping you revise all important concepts like trigonometric identities, functions, signs in different quadrants, and general solutions of trigonometric equations. Whether you’re solving problems from the Class 11 Maths Ch 3 Miscellaneous Exercise or referring to the NCERT Exemplar Class 11 Maths, these solutions will guide you toward complete conceptual clarity. Strengthen your understanding of Class 11 Maths Trigonometric topics and get exam-ready today!
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Miscellaneous Exercise
1. Prove that: \( 2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}=0 \)
Answer
L.H.S.
\(=2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}\)
\(=2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+2 \cos \left(\frac{\frac{3 \pi}{13}+\frac{5 \pi}{13}}{2}\right) \cos \left(\frac{\frac{3 \pi}{13}-\frac{5 \pi}{13}}{2}\right)\)
\(=\left[\cos x+\cos y=2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x+y}{2}\right)\right]\)
\(=2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+2 \cos \frac{4 \pi}{13} \cos \left(\frac{-\pi}{13}\right)\)
\(=2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+2 \cos \frac{4 \pi}{13} \cos \frac{\pi}{13}\)
\(=2 \cos \frac{\pi}{13}\left[\cos \frac{9 \pi}{13}+\cos \frac{4 \pi}{13}\right]\)
\(=2 \cos \frac{\pi}{13}\left[2 \cos \left(\frac{\frac{9 \pi}{13}+\frac{4 \pi}{2}}{2}\right) \cos \left(\frac{\frac{9 \pi}{13}-\frac{4 \pi}{23}}{2}\right)\right]\)
\(=2 \cos \frac{\pi}{13}\left[2 \cos \frac{\pi}{2} \cos \frac{5 \pi}{26}\right]\)
\(=2 \cos \times \frac{\pi}{13} \times 2 \times 0 \times \cos \frac{5 \pi}{26}\)
\(=0\)
Hence, L.H.S \( = \) R.H.S\(\ldots\)Proved
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2. Prove that \( (\sin 3 x+\sin x) \sin x+(\cos 3 x-\cos x) \cos x=0 \)
Answer
L.H.S.
\(=(\sin 3 x+\sin x) \sin x+(\cos 3 x-\cos x) \cos x\)
\(=\sin 3 x \sin x+\sin ^{2} x+\cos 3 x \cos x-\cos ^{2} x\)
\(=\cos 3 x \cos x+\sin 3 x \sin x-\left(\cos ^{2} x-\sin ^{2} x\right)\)
\(=\cos (3 x-x)-\cos 2 x[\cos (\mathrm{A}-\mathrm{B})=\cos \mathrm{A} \operatorname{Cos} \mathrm{B}+\sin A \sin B] \)
\(=\cos 2 x-\cos 2 x\)
\(=0\)
\(=\text { R.H.S. }\)
3. Prove that -
\((\cos x+\cos y)^{2}+(\sin x-\sin y)^{2}=4 \cos ^{2}\left(\frac{x+y}{2}\right)\)
\((\cos x+\cos y)^{2}+(\sin x-\sin y)^{2}=4 \cos ^{2}\left(\frac{x+y}{2}\right)\)
Answer
\(\text { L.H.S. }=(\cos x+\cos y)^{2}+(\sin x-\sin y)^{2}\)
\(=\cos ^{2} x+\cos ^{2} y+2 \cos x \cos y+\sin ^{2} x+\sin ^{2} y-2 \sin x \sin y\)
\(=\left(\cos ^{2} x+\sin ^{2} x\right)+\left(\cos ^{2} y+\sin ^{2} y\right)+2(\cos x \cos y-\sin x \sin y)\)
\(=1+1+2 \cos (x+y)[\cos (\mathrm{A}+\mathrm{B})=(\cos \mathrm{A} \cos \mathrm{B}-\sin \mathrm{A} \sin \mathrm{B}]\)
\(=2+2 \cos (x+y)\)
\(=2[1+\cos (x+y)]\)
\(=2\left[1+2 \cos ^{2}\left(\frac{x+y}{2}\right)-1\right]\left[\cos 2 \mathrm{~A}=2 \cos ^{2} \mathrm{~A}-1\right]\)
\(=4 \cos ^{2}\left(\frac{x+y}{2}\right)\)
\(=\text { R.H.S }\)
Hence, L.H.S \( = \) R.H.S\(\ldots\)Proved
4. Prove that -
\((\cos x-\cos y)^{2}+(\sin x-\sin y)^{2}=4 \sin ^{2}\left(\frac{x-y}{2}\right)\)
\((\cos x-\cos y)^{2}+(\sin x-\sin y)^{2}=4 \sin ^{2}\left(\frac{x-y}{2}\right)\)
Answer
\(\text { L.H.S. }=(\cos x-\cos y)^{2}+(\sin x-\sin y)^{2}\)
\(=\cos ^{2} x+\cos ^{2} y-2 \cos x \cos y+\sin ^{2} x+\sin ^{2} y-2 \sin x \sin y\)
\(=\left(\cos ^{2} x+\sin ^{2} x\right)+\left(\cos ^{2} y+\sin ^{2} y\right)-2(\cos x \cos y+\sin x \sin y)\)
\(=1+1-2 \cos (x-y)[\cos (\mathrm{A}-\mathrm{B})=(\cos \mathrm{A} \cos \mathrm{B}+\sin \mathrm{A} \sin \mathrm{B}]\)
\(=2[1+\cos (x+y)]\)
\(=2\left[1-\left\{1-2 \sin ^{2}\left(\frac{x-y}{2}\right)\right\}\right]\left[\cos 2 \mathrm{~A}=1-\sin ^{2} \mathrm{~A}\right]\)
\(=4 \sin ^{2}\left(\frac{x-y}{2}\right)\)
\(=\text { R.H.S }\)
Hence, L.H.S \( = \) R.H.S...Proved
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5. Prove that: \( \sin x+\sin 3 x+\sin 5 x+\sin 7 x=4 \cos x \cos 2 x \sin 4 x \)
Answer
\(\text { L.H.S }=\sin x+\sin 3 x+\sin 5 x+\sin 7 x\)
\(=(\sin x+\sin 5 x)+(\sin 3 x+\sin 7 x)\)
\(=2 \sin \left(\frac{x+5 x}{2}\right) \cdot \cos \left(\frac{x-5 x}{2}\right)+2 \sin \left(\frac{3 x+7 x}{2}\right) \cos \left(\frac{3 x-7 x}{2}\right)\)
\(=2 \sin 3 x \cos (-2 x)+2 \sin 5 x \cos (-2 x)\)
\(=2 \sin 3 x \cos 2 x+2 \sin 5 x \cos 2 x\)
\(=2 \cos 2 x[\sin 3 x+\sin 5 x]\)
\(=2 \cos 2 x\left[2 \sin \left(\frac{3 x+5 x}{2}\right) \cdot \cos \left(\frac{3 x-5 x}{2}\right)\right]\)
\(=2 \cos 2 x[2 \sin 4 x \cdot \cos (-x)]\)
\(=4 \cos 2 x \sin 4 x \cos x\)
\(=\text { R.H.S }\)
Hence, L.H.S = R.H.S...Proved
6. Prove that: \( \frac{(\sin 7 x+\sin 5 x)+(\sin \theta x+\sin 3 x)}{(\cos 7 x+\cos 5 x)+(\cos 9 x+\cos 3 x)}=\tan 6 x \)
Answer
It is known that
\( \sin \mathrm{A}+\sin \mathrm{B}=2 \sin \left(\frac{A+B}{2}\right) . \operatorname{Cos}\left(\frac{A-B}{2}\right), \cos \mathrm{A}+\cos \mathrm{B}=2 \cos \left(\frac{A+B}{2}\right) \). \( \operatorname{Cos}\left(\frac{A-B}{2}\right) \)
L.H.S \( =\frac{(\sin 7 x+\sin 5 x)+(\sin \theta x+\sin n x)}{(\cos 7 x+\cos 5 x)+(\cos 9 x+\sin 3 x)} \)
\( =\frac{\left[2 \sin \left(\frac{7 x+5 x}{2}\right) \cdot \cos \left(\frac{7 x-5 x}{2}\right)\right]+\left[2 \sin \left(\frac{9 x+3 x}{2}\right) \cdot \cos \left(\frac{9 x-3 x}{2}\right)\right]}{\left[2 \cos \left(\frac{7 x+5 x}{2}\right) \cdot \cos \left(\frac{7 x-5 x}{2}\right)\right]+\left[2 \cos \left(\frac{9 x+3 x}{2}\right) \cdot \cos \left(\frac{9 x-3 x}{2}\right)\right]} \)
\( =\frac{[2 \sin 6 x \cdot \cos x]+[2 \sin \operatorname{n} 6 x \cdot \cos 3 x]}{[2 \cos 6 x \cdot \cos x]+[2 \cos 6 x \cdot \cos 3 x]} \)
\( =\frac{2 \sin 6 x[\cos x+\cos 3 x]}{2 \cos 6 x[\cos x+\cos 3 x]} \)
\( =\tan 6 \mathrm{x} \)
\( = \) R.H.S
Hence, L.H.S = R.H.S...Proved
7. Find \( \sin \frac{x}{2}, \cos \frac{x}{2}2 \) and \( \tan \frac{x}{2} \), if \( \tan x=-\frac{4}{3}, x \) in quadrant II
Answer
Here, \( x \) is in quadrant II.
\(=\text { i.e., } \frac{\pi}{2} < x < \pi\)
\(=\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}\)
Therefore, \( \sin \frac{x}{2}, \cos \frac{x}{2} \) and \( \tan \frac{x}{2} \) are lies in first quadrant.
It is given that \( \tan x=-\frac{4}{3} \)
\( \operatorname{Sec}^{2} x=1+\tan ^{2} x=1+\left(-\frac{4}{3}\right)^{2}=1+\frac{16}{9}=\frac{25}{9} \)
\( =\cos ^{2} x=\frac{9}{25} \)
\( =\cos x= \pm \frac{3}{5} \)
As \(x\) is in quadrant II, \( \cos x \) is negative.
\( \cos x=\frac{-3}{5} \)
Now, \( \cos x=2 \cos ^{2} \frac{x}{2}-1 \)
\( =\frac{-3}{5}=2 \cos ^{2} \frac{x}{2}-1 \)
\( =2 \cos ^{2} \frac{x}{2}=1-\frac{3}{5} \)
\( =2 \cos ^{2} \frac{x}{2}=\frac{2}{5} \)
\( =\cos ^{2} \frac{x}{2}=\frac{1}{5} \)
\( =\cos \frac{x}{2}=\frac{1}{\sqrt{5}}\quad [\cos \frac{x}{2} \) is positive \( ] \)
\( =\cos \frac{x}{2}=\frac{\sqrt{5}}{5} \)
\( \operatorname{Sin}^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}=1 \)
\( =\sin ^{2} \frac{x}{2}=1-\frac{1}{5}=\frac{4}{5} \)
\( =\sin ^{2} \frac{x}{2}+\left(\frac{1}{\sqrt{5}}\right)^{2}=1 \)
\( =\sin \frac{x}{2}=\frac{2}{\sqrt{5}}\quad [\sin \frac{x}{2} \) is positive \( ] \)
\( =\sin \frac{x}{2}=\frac{2 \sqrt{5}}{5} \)
\( \tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}=\frac{\left(\frac{2}{\sqrt{5}}\right)}{\left(\frac{1}{\sqrt{5}}\right)}=2 \)
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 .
8. Find, \( \sin \frac{x}{2}, \cos \frac{x}{2} \),and \( \tan \frac{x}{2} \) for \( \cos x=-\frac{1}{3}, x \) in quadrant III
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, where \( \sin \frac{x}{2} \) as is positive.
It is given that \( \cos x=-\frac{1}{3} \)
\( \operatorname{Cos} x=1-2 \sin ^{2} \frac{x}{2} \)
\( \Rightarrow \sin ^{2} \frac{x}{2}=\frac{1-\cos x}{2} \)
\( \Rightarrow \sin ^{2} \frac{x}{2}=\frac{1-\left(-\frac{1}{3}\right)}{2}=\frac{\left(1+\frac{1}{3}\right)}{2}=\frac{\frac{4}{3}}{2}=\frac{2}{3} \)
\( \Rightarrow \sin \frac{x}{2}=\frac{\sqrt{2}}{\sqrt{3}}\quad [\sin \frac{x}{2} \) is positive \( ] \)
\( \Rightarrow \sin \frac{x}{2}=\frac{\sqrt{2}}{\sqrt{2}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{6}}{3} \)
Now,
\( \operatorname{Cos} x=2 \cos ^{2} \frac{x}{2}-1 \)
\(\Rightarrow \cos ^{2} \frac{x}{2}=\frac{1+\cos x}{2}=\frac{1+\left(-\frac{1}{3}\right)}{2}=\frac{\left(\frac{3-1}{3}\right)}{2}=\frac{\left(\frac{2}{3}\right)}{2}=\frac{1}{2}\)
\(\Rightarrow \cos ^{2} \frac{x}{2}=-\frac{1}{\sqrt{3}}\quad [\cos \frac{x}{2} \text { is negative }]\)
\(\Rightarrow \cos \frac{x}{2}=-\frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{-\sqrt{3}}{3} \tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}=\frac{\left(\frac{\sqrt{2}}{\sqrt{3}}\right)}{\left(\frac{-1}{\sqrt{3}}\right)}=-\sqrt{2}\)
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 \( -\sqrt{2} \).
