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Looking for Class 11 Maths NCERT Solutions for Trigonometry Miscellaneous Exercise? You’ve come to the right place! This section provides complete and step-by-step solutions for all the questions from the Miscellaneous Exercise of Chapter 3 – Trigonometric Functions. These solutions are crafted as per the latest NCERT guidelines and cover a mix of concepts like trigonometric identities, angle conversions, function values, and general solutions of equations. Whether you’re working through the Class 11 Maths Ch 3 Miscellaneous Exercise solutions or practicing from the NCERT Exemplar Class 11 Maths, these answers will help reinforce your understanding of Class 11 Maths Trigonometric concepts. View or download the full NCERT Solution for Class 11 Maths Chapter 3 and enhance your problem-solving skills 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 \( -\sqrt{2} \).
