NCERT Solutions Class 11 Maths Chapter-3 (Trigonometric Functions)Miscellaneous Exercise

NCERT Solutions Class 11 Maths from class 11th Students will get the answers of Chapter-3 (Trigonometric Functions)Miscellaneous Exercise . This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.

We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

NCERT Question-Answer

Class 11 Mathematics

Chapter-3 (Trigonometric Functions)Miscellaneous Exercise

Questions and answers given in practice

Chapter-3 (Trigonometric Functions)

Miscellaneous Exercise

Q1. Prove that: $2\cos \frac{\pi }{13}\cos \frac{9\pi }{13}+\cos \frac{3\pi }{13}+\cos \frac{5\pi }{13}=0$

Answer. $\begin{array}{c}\text{L.H.S.}\\ =2\cos \frac{\pi }{13}\cos \frac{9\pi }{13}+\cos \frac{5\pi }{13}+\cos \frac{5\pi }{13}\\ =2\cos \frac{\pi }{13}\cos \frac{9\pi }{13}+2\cos \left(\frac{\frac{3\pi }{13}+\frac{5\pi }{13}}{2}\right)\cos \left(\frac{\frac{3\pi }{13}-\frac{5\pi }{13}}{2}\right)[\cos x+\cos y=2\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)]\end{array}$ $\begin{array}{c}=2\cos \frac{\pi }{13}\cos \frac{9\pi }{13}+2\cos \frac{4\pi }{13}\cos \left(\frac{-\pi }{13}\right)\\ =2\cos \frac{\pi }{13}\cos \frac{9\pi }{13}+2\cos \frac{4\pi }{13}\cos \frac{\pi }{13}\\ =2\cos \frac{\pi }{13}[\cos \frac{9\pi }{13}+\cos \frac{4\pi }{13}]\end{array}$ $\begin{array}{c}=2\cos \frac{\pi }{13}\left[2\cos (\frac{9\pi }{13}+\frac{4\pi }{13})\cos \left(\frac{\frac{9\pi }{13}-\frac{4\pi }{13}}{2}\right)\right]\\ =2\cos \frac{\pi }{13}\left[2\cos \frac{\pi }{2}\cos \frac{5\pi }{26}\right]\\ =2\cos \frac{\pi }{13}\times 2\times 0\times \cos \frac{5\pi }{26}\\ =0=R.H.S\end{array}$

Q2. Prove that: (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0

Answer. $\begin{array}{c}\text{L.H.S.}\\ =(\sin 3x+\sin x)\sin x+(\cos 3x-\cos x)\cos x\end{array}$ $\begin{array}{c}=\sin 3x\sin x+{\sin }^{2}x+\cos 3x\cos x-{\cos }^{2}x\\ =\cos 3x\cos x+\sin 3x\sin x-({\cos }^{2}x-{\sin }^{2}x)\\ =\cos (3x-x)-\cos 2x\left[\cos \right(A-B)=\cos A\cos B+\sin A\sin B]\\ =\cos 2x-\cos 2x\\ =0\\ =R.H.S\end{array}$

Q3. Prove that: $(\cos x+\cos y{)}^2+(\sin x-\sin y{)}^2=4{\cos }^2\frac{x+y}{2}$

Answer. $\begin{array}{c}\text{L.H.S.}=(\cos x+\cos y{)}^2+(\sin x-\sin y{)}^2\\ ={\cos }^{2}x+{\cos }^{2}y+2\cos x\cos y+{\sin }^{2}x+{\sin }^{2}y-2\sin x\sin y\\ =({\cos }^{2}x+{\sin }^{2}x)+({\cos }^{2}y+{\sin }^{2}y)+2(\cos x\cos y-\sin x\sin y)\end{array}$ $\begin{array}{c}=1+1+2\cos (x+y)\\ =2+2\cos (x+y)\end{array}\left[\cos \right(A+B)=(\cos A\cos B-\sin A\sin B\left)\right]$ $\begin{array}{cc}& =2[1+\cos (x+y\left)\right]\\ & =2[1+2{\cos }^2\left(\frac{x+y}{2}\right)-1][\cos 2A=2{\cos }^{2}A-1]\\ & =4{\cos }^2\left(\frac{x+y}{2}\right)=R.H.S\end{array}$

Q4. Prove that: $(\cos x-\cos y{)}^2+(\sin x-\sin y{)}^2=4{\sin }^2\frac{x-y}{2}$

Answer. L.H.S $=(\cos x-\cos y{)}^2+(\sin x-\sin y{)}^2$ $\begin{array}{c}={\cos }^{2}x+{\cos }^{2}y-2\cos x\cos y+{\sin }^{2}x+{\sin }^{2}y-2\sin x\sin y\\ =({\cos }^{2}x+{\sin }^{2}x)+({\cos }^{2}y+{\sin }^{2}y)-2[\cos x\cos y+\sin x\sin y]\end{array}$ $=1+1-2\left[\cos \right(x-y\left)\right]\left[\cos \right(A-B)=\cos A\cos B+\sin A\sin B]$ $=2[1-\cos (x-y\left)\right]$ $=2[1-\{1-2{\sin }^2\left(\frac{x-y}{2}\right)\}][\cos 2A=1-2{\sin }^{2}A]$ $=4{\sin }^2\left(\frac{x-y}{2}\right)=R.H.S$

Q5. Prove that: sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x

Answer. It is known that $\sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cdot \cos \left(\frac{A-B}{2}\right)$ $\begin{array}{c}\text{L.H.S.}=\sin x+\sin 3x+\sin 5x+\sin 7x\\ =(\sin x+\sin 5x)+(\sin 3x+\sin 7x)\\ =2\sin \left(\frac{x+5x}{2}\right)\cdot \cos \left(\frac{x-5x}{2}\right)+2\sin \left(\frac{3x+7x}{2}\right)\cos \left(\frac{3x-7x}{2}\right)\end{array}$ $\begin{array}{c}=2\sin 3x\cos (-2x)+2\sin 5x\cos (-2x)\\ =2\sin 3x\cos 2x+2\sin 5x\cos 2x\\ =2\cos 2x[\sin 3x+\sin 5x]\\ =2\cos 2x[2\sin \left(\frac{3x+5x}{2}\right)\cdot \cos \left(\frac{3x-5x}{2}\right)]\end{array}$ $\begin{array}{c}=2\cos 2x[2\sin 4x\cdot \cos (-x\left)\right]\\ =4\cos 2x\sin 4x\cos x=\text{R.H.S.}\end{array}$

Q6. Prove that: $\frac{(\sin 7x+\sin 5x)+(\sin 9x+\sin 3x)}{(\cos 7x+\cos 5x)+(\cos 9x+\cos 3x)}=\tan 6x$

Answer. It is known that $\sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cdot \cos \left(\frac{A-B}{2}\right),\cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cdot \cos \left(\frac{A-B}{2}\right)$ L.H.S $=\frac{(\sin 7x+\sin 5x)+(\sin 9x+\sin 3x)}{(\cos 7x+\cos 5x)+(\cos 9x+\cos 3x)}$ $=\frac{[2\sin \left(\frac{7x+5x}{2}\right)\cdot \cos \left(\frac{7x-5x}{2}\right)]+[2\sin \left(\frac{9x+3x}{2}\right)\cdot \cos \left(\frac{9x-3x}{2}\right)]}{[2\cos \left(\frac{7x+5x}{2}\right)-\cos \left(\frac{7x-5x}{2}\right)]+[2\cos \left(\frac{9x+3x}{2}\right)\cdot \cos \left(\frac{9x-3x}{2}\right)]}$ $\begin{array}{c}=[2\sin 6x\cdot \cos x]+[2\sin 6x\cdot \cos 3x]\\ =[2\cos 6x\cdot \cos x]+[2\cos 6x\cdot \cos 3x]\\ =\frac{2\sin 6x[\cos x+\cos 3x]}{2\cos 6x[\cos x+\cos 3x]}\\ =\tan 6x\\ =\text{R.H.S.}\end{array}$

Q7. Prove that: $\sin 3x+\sin 2x-\sin x=4\sin x\cos \frac{x}{2}\cos \frac{3x}{2}$

Answer. L.H.S $=\sin 3x+\sin 2x-\sin x$ $\begin{array}{c}=\sin 3x+(\sin 2x-\sin x)\\ =\sin 3x+\left[2\cos \left(\frac{2x+x}{2}\right)\sin \left(\frac{2x-x}{2}\right)\right][\sin A-\sin B=2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)]\end{array}$ $\begin{array}{c}=\sin 3x+\left[2\cos \left(\frac{3x}{2}\right)\sin \left(\frac{x}{2}\right)\right]\\ =\sin 3x+2\cos \frac{3x}{2}\sin \frac{x}{2}\end{array}$ $=2\sin \frac{3x}{2}\cdot \cos \frac{3x}{2}+2\cos \frac{3x}{2}\sin \frac{x}{2}[\sin 2A=2\sin A\cdot \cos B]$ $\begin{array}{c}=2\cos \left(\frac{3x}{2}\right)[\sin \left(\frac{3x}{2}\right)+\sin \left(\frac{x}{2}\right)]\\ =2\cos \left(\frac{3x}{2}\right)\left[2\sin \left\{\frac{\left(\frac{3x}{2}\right)+\left(\frac{x}{2}\right)}{2}\right\}\cos \left\{\frac{\left(\frac{3x}{2}\right)-\left(\frac{x}{2}\right)}{2}\right\}\right][[\sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)]\end{array}$ $\begin{array}{c}=2\cos \left(\frac{3x}{2}\right)\cdot 2\sin x\cos \left(\frac{x}{2}\right)\\ =4\sin x\cos \left(\frac{x}{2}\right)\cos \left(\frac{3x}{2}\right)=R.H.S\end{array}$

Q8. Find $\sin \frac{x}{2},\cos \frac{x}{2}\text{and}\tan \frac{x}{2}$ in ${\tan }^{\prime }x=-\frac{4}{3},x\text{in quadrant}II$

Answer. Here, x is in quadrant II. $\begin{array}{c}\frac{\pi }{2}<x<\pi \\ \Rightarrow \frac{\pi }{4}<\frac{x}{2}<\frac{\pi }{2}\end{array}$ Therefore, $\sin \frac{x}{2},\cos \frac{x}{2}\text{and}\tan \frac{x}{2}$ are all positive. It is given that $\tan x=-\frac{4}{3}$ $\begin{array}{c}{\sec }^{2}x=1+{\tan }^{2}x=1+{\left(\frac{-4}{3}\right)}^2=1+\frac{16}{9}=\frac{25}{9}\\ \therefore {\cos }^{2}x=\frac{9}{25}\\ \Rightarrow \cos x=\pm \frac{3}{5}\end{array}$ As x is in quadrant II, cos x is negative. $\begin{array}{c}\cos x=\frac{-3}{5}\\ \text{Now,}\cos x=2{\cos }^2\frac{x}{2}-1\\ \Rightarrow \frac{-3}{5}=2{\cos }^2\frac{x}{2}-1\\ \Rightarrow 2{\cos }^2\frac{x}{2}=1-\frac{3}{5}\end{array}$ $\begin{array}{c}\Rightarrow 2{\cos }^2\frac{x}{2}=\frac{2}{5}\\ \Rightarrow {\cos }^2\frac{x}{2}=\frac{1}{5}\end{array}$ $\Rightarrow \cos \frac{x}{2}=\frac{1}{\sqrt{5}}\because \cos \frac{x}{2}\text{is positive}]$ $\begin{array}{c}\therefore \cos \frac{x}{2}=\frac{\sqrt{5}}{5}\\ {\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}=1\\ \Rightarrow {\sin }^2\frac{x}{2}+{\left(\frac{1}{\sqrt{5}}\right)}^2=1\\ \Rightarrow {\sin }^2\frac{x}{2}=1-\frac{1}{5}=\frac{4}{5}\end{array}$ $\Rightarrow \sin \frac{x}{2}=\frac{2}{\sqrt{5}}[\because \sin \frac{x}{2}\text{is positive}]$ $\begin{array}{c}\therefore \sin \frac{x}{2}=\frac{2\sqrt{5}}{5}\\ \tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}=\frac{\left(\frac{2}{\sqrt{5}}\right)}{\left(\frac{1}{\sqrt{5}}\right)}=2\end{array}$ Thus, the respective values of $\sin \frac{x}{2},\cos \frac{x}{2}\text{and}\tan \frac{x}{2}are\frac{2\sqrt{5}}{5},\frac{\sqrt{5}}{5},\text{and}2$

Q9. Find $\sin \frac{x}{2},\cos \frac{x}{2}\text{and}\tan \frac{x}{2}$ in $\cos x=-\frac{1}{3},x$ in quadrant III.

Answer. Here, x is in quadrant III. $\begin{array}{c}\text{i.e.,}\pi <x<\frac{3\pi }{2}\\ \Rightarrow \frac{\pi }{2}<\frac{x}{2}<\frac{3\pi }{4}\end{array}$ Therefore, $\cos \frac{x}{2}and\tan \frac{x}{2}\text{are negative, whereas}$ $\sin \frac{x}{2}$ is positive. It is given that $\cos x=-\frac{1}{3}$. $\begin{array}{c}\cos x=1-2{\sin }^2\frac{x}{2}\\ \Rightarrow {\sin }^2\frac{x}{2}=\frac{1-\cos x}{2}\\ \Rightarrow {\sin }^2\frac{x}{2}=\frac{1-(-\frac{1}{3})}{2}=\frac{(1+\frac{1}{3})}{2}=\frac{3}{2}=\frac{2}{3}\end{array}$ $\Rightarrow \sin \frac{x}{2}=\frac{\sqrt{2}}{\sqrt{3}}[\because \sin \frac{x}{2}\text{is positive}]$ $\begin{array}{c}\therefore \sin \frac{x}{2}=\frac{\sqrt{2}}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{6}}{3}\\ \cos x=2{\cos }^2\frac{x}{2}-1\end{array}$ $\begin{array}{c}\Rightarrow {\cos }^2\frac{x}{2}=\frac{1+\cos x}{2}=\frac{1+(-\frac{1}{3})}{2}=\frac{\left(\frac{3-1}{3}\right)}{2}=\frac{\left(\frac{2}{3}\right)}{2}=\frac{1}{3}\\ \Rightarrow \cos \frac{x}{2}=-\frac{1}{\sqrt{3}}\end{array}$ $\begin{array}{c}\therefore \cos \frac{x}{2}=-\frac{1}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{-\sqrt{3}}{3}\\ \tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}=\frac{\left(\frac{\sqrt{2}}{\sqrt{3}}\right)}{\left(\frac{-1}{\sqrt{3}}\right)}=-\sqrt{2}\end{array}$ Thus, the respective values of $\sin \frac{x}{2},\cos \frac{x}{2}\text{and}\tan \frac{x}{2}\text{are}\frac{\sqrt{6}}{3},\frac{-\sqrt{3}}{3},\text{and}-\sqrt{2}$.

Q10. Find $\sin \frac{x}{2},\cos \frac{x}{2}\text{and}\tan \frac{x}{2}$ in $\sin x=\frac{1}{4},x$ is in quadrant II.

Answer. Here, x is in quadrant II. $\begin{array}{c}\text{i.e.,}\frac{\pi }{2}<x<\pi \\ \Rightarrow \frac{\pi }{4}<\frac{x}{2}<\frac{\pi }{2}\end{array}$ Therefore, $\sin \frac{x}{2},\cos \frac{x}{2}$ and $\tan \frac{x}{2}$ are all positive. It is given that $\sin x=\frac{1}{4}$ $\begin{array}{c}{\cos }^{2}x=1-{\sin }^{2}x=1-{\left(\frac{1}{4}\right)}^2=1-\frac{1}{16}=\frac{15}{16}\\ \Rightarrow \cos x=-\frac{\sqrt{15}}{4}\left[\cos x\text{is negative in quadrant II}\right]\end{array}$ ${\sin }^2\frac{x}{2}=\frac{1-\cos x}{2}=\frac{1-(-\frac{\sqrt{15}}{4})}{2}=\frac{4+\sqrt{15}}{8}$ $\Rightarrow \sin \frac{x}{2}=\sqrt{\frac{4+\sqrt{15}}{8}}[\because \sin \frac{x}{2}\text{is positive}]$ $\begin{array}{c}=\sqrt{\frac{4+\sqrt{15}}{8}\times \frac{2}{2}}\\ =\sqrt{\frac{8+2\sqrt{15}}{16}}\\ =\frac{\sqrt{8+2\sqrt{15}}}{4}\end{array}$ ${\cos }^2\frac{x}{2}=\frac{1+\cos x}{2}=\frac{1+(-\frac{\sqrt{15}}{4})}{2}=\frac{4-\sqrt{15}}{8}$ $\Rightarrow \cos \frac{x}{2}=\sqrt{\frac{4-\sqrt{15}}{8}}[\because \cos \frac{x}{2}\text{is positive}]$ $\begin{array}{c}=\sqrt{\frac{4-\sqrt{15}}{8}\times \frac{2}{2}}\\ =\sqrt{\frac{8-2\sqrt{15}}{16}}\\ =\frac{\sqrt{8-2\sqrt{15}}}{4}\end{array}$ $\tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}=\frac{(\frac{\sqrt{8+2\sqrt{15}}}{4}}{\frac{\sqrt{8-2\sqrt{15}}}{4}})=\frac{\sqrt{8+2\sqrt{15}}}{\sqrt{8-2\sqrt{15}}}$ $\begin{array}{c}=\sqrt{\frac{8+2\sqrt{15}}{8-2\sqrt{15}}\times \frac{8+2\sqrt{15}}{8+2\sqrt{15}}}\\ =\sqrt{\frac{(8+2\sqrt{15}{)}^2}{64-60}}=\frac{8+2\sqrt{15}}{2}=4+\sqrt{15}\end{array}$ Thus, the respective values of $\sin \frac{x}{2},\cos \frac{x}{2}\text{and}\tan \frac{x}{2}\text{are}\frac{\sqrt{8+2\sqrt{15}}}{4},\frac{\sqrt{8-2\sqrt{15}}}{4}$, and $4+\sqrt{15}.$