Answer :

$\text{L.H.S.}=2 \cos \dfrac{\pi}{13} \cos \dfrac{9 \pi}{13}+\cos \dfrac{3 \pi}{13}+\cos \dfrac{5 \pi}{13}$

$\hspace{1.1cm}=2 \cos \dfrac{\pi}{13} \cos \dfrac{9 \pi}{13}+2 \cos \left(\dfrac{\dfrac{3 \pi}{13}+\dfrac{5 \pi}{13}}{2}\right) \cos \left(\dfrac{\dfrac{3 \pi}{13}-\dfrac{5 \pi}{13}}{2}\right)\qquad\bigg[\because \ \cos x+\cos y=2 \cos \left(\dfrac{x+y}{2}\right) \cos \left(\dfrac{x-y}{2}\right)\bigg]$

$\hspace{1.1cm}=2 \cos \dfrac{\pi}{13} \cos \dfrac{9 \pi}{13}+2 \cos \dfrac{4 \pi}{13} \cos \left(\dfrac{-\pi}{13}\right)$

$\hspace{1.1cm}=2 \cos \dfrac{\pi}{13} \cos \dfrac{9 \pi}{13}+2 \cos \dfrac{4 \pi}{13} \cos \dfrac{\pi}{13}$

$\hspace{1.1cm}=2 \cos \dfrac{\pi}{13}\bigg[\cos \dfrac{9 \pi}{13}+\cos \dfrac{4 \pi}{13}\bigg]$

$\hspace{1.1cm}=2 \cos \dfrac{\pi}{13}\left[2 \cos \left(\dfrac{\dfrac{9 \pi}{13}+\dfrac{4 \pi}{13}}{2}\right) \cos \left(\dfrac{\dfrac{9 \pi}{13}-\dfrac{4 \pi}{13}}{2}\right)\right]$

$\hspace{1.1cm}=2 \cos \dfrac{\pi}{13}\bigg[2 \cos \dfrac{\pi}{2} \cos \dfrac{5 \pi}{26}\bigg]$

$\hspace{1.1cm}=2 \cos \dfrac{\pi}{13} \times 2 \times 0 \times \cos \dfrac{5 \pi}{26}$

$\hspace{1.1cm}=0= \text{R.H.S.}$

Answer :

$\text{L.H.S.}=\left(\sin 3 x+\sin x\right) \sin x+\left(\cos 3 x - \cos x\right) \cos x$

$\hspace{1.1cm}=\sin 3 x \sin x+\sin ^{2} x+\cos 3 x \cos x-\cos ^{2} x$

$\hspace{1.1cm}=\cos 3 x \cos x+\sin 3 x \sin x-\left(\cos ^{2} x-\sin ^{2} x\right)$

$\hspace{1.1cm}=\cos \left(3 x-x\right)-\cos 2 x \quad\bigg[\because \ \cos \left(A-B\right)=\cos A \cos B+\sin A \sin B\bigg]$

$\hspace{1.1cm}=\cos 2 x-\cos 2 x=0$

$\hspace{1.1cm}= \text{R.H.S.}$

Answer :

$\text{L.H.S.}=\left(\cos x+\cos y\right)^{2}+\left(\sin x-\sin y\right)^{2}$

$\hspace{1.1cm}=\cos ^{2} x+\cos ^{2} y+2 \cos x \cos y+\sin ^{2} x+\sin ^{2} y-2 \sin x \sin y$

$\hspace{1.1cm}=\left(\cos ^{2} x+\sin ^{2} x\right)+\left(\cos ^{2} y+\sin ^{2} y\right)+2\left(\cos x \cos y-\sin x \sin y\right)$

$\hspace{1.1cm}=1+1+2 \cos \left(x+y\right) \quad\bigg[\because \ \cos \left(A+B\right)=\left(\cos A \cos B-\sin A \sin B\right)\bigg]$

$\hspace{1.1cm}=2+2 \cos \left(x+y\right)$

$\hspace{1.1cm}=2\bigg[1+\cos \left(x+y\right)\bigg]$

$\hspace{1.1cm}=2\bigg[1+2 \cos ^{2}\left(\dfrac{x+y}{2}\right)-1\bigg] \quad\bigg[\because \ \cos 2 A=2 \cos ^{2} A-1\bigg]$

$\hspace{1.1cm}=4 \cos ^{2}\left(\dfrac{x+y}{2}\right)$

$\hspace{1.1cm}=\text{R.H.S.}$

Answer :

$\text{L.H.S.}=\left(\cos x-\cos y\right)^{2}+\left(\sin x-\sin y\right)^{2}$

$\hspace{1.1cm}=\cos ^{2} x+\cos ^{2} y-2 \cos x \cos y+\sin ^{2} x+\sin ^{2} y-2 \sin x \sin y $

$\hspace{1.1cm}=\left(\cos ^{2} x+\sin ^{2} x\right)+\left(\cos ^{2} y+\sin ^{2} y\right)-2\bigg[\cos x \cos y+\sin x \sin y\bigg]$

$\hspace{1.1cm}=1+1-2\bigg[\cos \left(x-y\right)\bigg] \qquad {\bigg[\because \ \cos \left(A-B\right)=\cos A \cos B+\sin A \sin B\bigg]} $

$\hspace{1.1cm}=2\bigg[1-\cos \left(x-y\right)\bigg] $

$\hspace{1.1cm}=2\bigg[1-\left \lbrace 1-2 \sin ^{2}\left(\dfrac{x-y}{2}\right)\right \rbrace \bigg] $ $\qquad {\bigg[\because \ \cos 2 A=1-2 \sin ^{2} A\bigg]} $

$\hspace{1.1cm}=4 \sin ^{2}\left(\dfrac{x-y}{2}\right)$

$\hspace{1.1cm}=\text{ R.H.S. }$

Answer :

It is known that

$ \sin A+\sin B=2 \sin \left(\dfrac{A+B}{2}\right) \cdot \cos \left(\dfrac{A-B}{2}\right) $

$\therefore \ \ $ $\text{L.H.S.}$ $=\sin x+\sin 3 x+\sin 5 x+\sin 7 x$

$\hspace{1.6cm}=\left(\sin x+\sin 5 x\right)+\left(\sin 3 x+\sin 7 x\right)$

$\hspace{1.6cm}=2 \sin \left(\dfrac{x+5 x}{2}\right) \cdot \cos \left(\dfrac{x-5 x}{2}\right)+2 \sin \left(\dfrac{3 x+7 x}{2}\right) \cos \left(\dfrac{3 x-7 x}{2}\right)$

$\hspace{1.6cm}=2 \sin 3 x \cos \left(-2 x\right)+2 \sin 5 x \cos \left(-2 x\right)$

$\hspace{1.6cm}=2 \sin 3 x \cos 2 x+2 \sin 5 x \cos 2 x$

$\hspace{1.6cm}=2 \cos 2 x\bigg[\sin 3 x+\sin 5 x\bigg]$

$\hspace{1.6cm}=2 \cos 2 x\bigg[2 \sin \left(\dfrac{3 x+5 x}{2}\right) \cdot \cos \left(\dfrac{3 x-5 x}{2}\right)\bigg]$

$\hspace{1.6cm}=2 \cos 2 x\bigg[2 \sin 4 x \cdot \cos \left(-x\right)\bigg]$

$\hspace{1.6cm}=4 \cos 2 x \sin 4 x \cos x$

$\hspace{1.6cm}=\text{R.H.S.}$

Answer :

It is known that

$\sin A+\sin B=2 \sin \left(\dfrac{A+B}{2}\right) \cdot \cos \left(\dfrac{A-B}{2}\right), \cos A+\cos B=2 \cos \left(\dfrac{A+B}{2}\right) \cdot \cos \left(\dfrac{A-B}{2}\right)$

$\hspace{2cm}=\dfrac{\left(\sin 7 x+\sin 5 x\right)+\left(\sin 9 x+\sin 3 x\right)}{\left(\cos 7 x+\cos 5 x\right)+\left(\cos 9 x+\cos 3 x\right)}$

$\hspace{2cm}=\dfrac{\bigg[2 \sin \left(\dfrac{7 x+5 x}{2}\right) \cdot \cos \left(\dfrac{7 x-5 x}{2}\right)\bigg]+\bigg[2 \sin \left(\dfrac{9 x+3 x}{2}\right) \cdot \cos \left(\dfrac{9 x-3 x}{2}\right)\bigg]}{\bigg[2 \cos \left(\dfrac{7 x+5 x}{2}\right) \cdot \cos \left(\dfrac{7 x-5 x}{2}\right)\bigg]+\bigg[2 \cos \left(\dfrac{9 x+3 x}{2}\right) \cdot \cos \left(\dfrac{9 x-3 x}{2}\right)\bigg]}$

$\hspace{2cm}=\dfrac{\bigg[2 \sin 6 x \cdot \cos x\bigg]+\bigg[2 \sin 6 x \cdot \cos 3 x\bigg]}{\bigg[2 \cos 6 x \cdot \cos x\bigg]+\bigg[2 \cos 6 x \cdot \cos 3 x\bigg]}$

$\hspace{2cm}=\dfrac{2 \sin 6 x\bigg[\cos x+\cos 3 x\bigg]}{2 \cos 6 x\bigg[\cos x+\cos 3 x\bigg]}$

$\hspace{2cm}=\tan 6 x$

$\hspace{2cm}=\text{ R.H.S.}$

Answer :

$ \begin{aligned} \text{L.H.S.} & =\sin 3 x+\sin 2 x-\sin x \\ \\ & =\sin 3 x+\left(\sin 2 x-\sin x\right) \\ \\ & =\sin 3 x+\bigg[2 \cos \left(\dfrac{2 x+x}{2}\right) \sin \left(\dfrac{2 x-x}{2}\right)\bigg] \quad\bigg[\because \ \sin A-\sin B=2 \cos \left(\dfrac{A+B}{2}\right) \sin \left(\dfrac{A-B}{2}\right)\bigg] \\ \\ & =\sin 3 x+\bigg[2 \cos \left(\dfrac{3 x}{2}\right) \sin \left(\dfrac{x}{2}\right)\bigg] \\ \\ & =\sin 3 x+2 \cos \dfrac{3 x}{2} \sin \dfrac{x}{2} \\ \\ & =2 \sin \dfrac{3 x}{2} \cdot \cos \dfrac{3 x}{2}+2 \cos \dfrac{3 x}{2} \sin \dfrac{x}{2} \quad\bigg[\because \ \sin 2 A=2 \sin A \cdot \cos B\bigg] \\ \\ & =2 \cos \left(\dfrac{3 x}{2}\right)\bigg[\sin \left(\dfrac{3 x}{2}\right)+\sin \left(\dfrac{x}{2}\right)\bigg] \\ \\ & =2 \cos \left(\dfrac{3 x}{2}\right)\left[2 \sin \left \lbrace \dfrac{\left(\dfrac{3 x}{2}\right)+\left(\dfrac{x}{2}\right)}{2}\right \rbrace \cos \left\lbrace \dfrac{\left(\dfrac{3 x}{2}\right)-\left(\dfrac{x}{2}\right)}{2}\right \rbrace \right]\qquad \bigg[\because \ \sin A+\sin B=2 \sin \left(\dfrac{A+B}{2}\right) \cos \left(\dfrac{A-B}{2}\right)\bigg] \\ \\ & =2 \cos \left(\dfrac{3 x}{2}\right) \left[ 2 \sin x \cos \left(\dfrac{x}{2}\right)\right] =\text{ R.H.S. } \end{aligned} $

Answer :

Here, $x$ is in quadrant II.

i.e., $\dfrac{\pi}{2}<x<\pi$

$\Rightarrow \dfrac{\pi}{4}<\dfrac{x}{2}<\dfrac{\pi}{2}$

Therefore, $\sin \dfrac{x}{2}, \cos \dfrac{x}{2}$ and $\tan \dfrac{x}{2}$ are all positive.

It is given that $\tan x=-\dfrac{4}{3}$

$\sec ^{2} x=1+\tan ^{2} x=1+\left(\dfrac{-4}{3}\right)^{2}=1+\dfrac{16}{9}=\dfrac{25}{9}$

$\therefore \ \ \cos ^{2} x=\dfrac{9}{25}$

$\Rightarrow \cos x= \pm \dfrac{3}{5}$

As $x$ is in quadrant II, $\cos x$ is negative.

$ \therefore \quad \cos x=\dfrac{-3}{5} $

Now, $\cos x=2 \cos ^{2} \dfrac{x}{2}-1$

$\Rightarrow \dfrac{-3}{5}=2 \cos ^{2} \dfrac{x}{2}-1$

$\Rightarrow 2 \cos ^{2} \dfrac{x}{2}=1-\dfrac{3}{5}$

$\Rightarrow 2 \cos ^{2} \dfrac{x}{2}=\dfrac{2}{5}$

$\Rightarrow \cos ^{2} \dfrac{x}{2}=\dfrac{1}{5}$

$\Rightarrow \cos \dfrac{x}{2}=\dfrac{1}{\sqrt{5}} \quad\bigg[\because \cos \dfrac{x}{2}.$ is positive $\bigg]$

$\therefore \ \ \cos \dfrac{x}{2}=\dfrac{\sqrt{5}}{5}$

$\Rightarrow \sin ^{2} \dfrac{x}{2}+\cos ^{2} \dfrac{x}{2}=1$

$\Rightarrow \sin ^{2} \dfrac{x}{2}+\left(\dfrac{1}{\sqrt{5}}\right)^{2}=1$

$\Rightarrow \sin ^{2} \dfrac{x}{2}=1-\dfrac{1}{5}=\dfrac{4}{5}$

$\Rightarrow \sin \dfrac{x}{2}=\dfrac{2}{\sqrt{5}}$ $\qquad\bigg[\because \ \ \sin \dfrac{x}{2}.$ is positive $\bigg]$

$\therefore \ \ \sin \dfrac{x}{2}=\dfrac{2 \sqrt{5}}{5}$

$ \quad \ \tan \dfrac{x}{2}=\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}=\dfrac{\left(\dfrac{2}{\sqrt{5}}\right)}{\left(\dfrac{1}{\sqrt{5}}\right)}=2$

Thus, the respective values of $\sin \dfrac{x}{2}, \cos \dfrac{x}{2}$ and $\tan \dfrac{x}{2}$ are $\dfrac{2 \sqrt{5}}{5}, \dfrac{\sqrt{5}}{5}$, and 2

Answer :

Here, $x$ is in quadrant III.

i.e., $\pi<x<\dfrac{3 \pi}{2}$

$\Rightarrow \dfrac{\pi}{2}<\dfrac{x}{2}<\dfrac{3 \pi}{4}$

Therefore, $ \ \cos \dfrac{x}{2}$ and $\tan \dfrac{x}{2}$ are negative, whereas $\sin \dfrac{x}{2}$ is positive.

It is given that $\cos x=-\dfrac{1}{3}$

$\cos x=1-2 \sin ^{2} \dfrac{x}{2}$

$\Rightarrow \sin ^{2} \dfrac{x}{2}=\dfrac{1-\cos x}{2}$

$\Rightarrow \sin ^{2} \dfrac{x}{2}=\dfrac{1-\left(-\dfrac{1}{3}\right)}{2}=\dfrac{\left(1+\dfrac{1}{3}\right)}{2}=\dfrac{\dfrac{4}{3}}{2}=\dfrac{2}{3}$

$\Rightarrow \sin \dfrac{x}{2}=\dfrac{\sqrt{2}}{\sqrt{3}} \quad\bigg[\because \ \ \sin \dfrac{x}{2}$ is positive $\bigg]$

$\therefore \ \ \sin \dfrac{x}{2}=\dfrac{\sqrt{2}}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{\sqrt{6}}{3}$

Now,

$\cos x=2 \cos ^{2} \dfrac{x}{2}-1$

$\Rightarrow \ \cos ^{2} \dfrac{x}{2}=\dfrac{1+\cos x}{2}=\dfrac{1+\left(-\dfrac{1}{3}\right)}{2}=\dfrac{\left(\dfrac{3-1}{3}\right)}{2}=\dfrac{\left(\dfrac{2}{3}\right)}{2}=\dfrac{1}{3}$

$\Rightarrow \cos \dfrac{x}{2}=-\dfrac{1}{\sqrt{3}} \qquad\bigg[\because \ \ \cos \dfrac{x}{2}$ is negative $\bigg]$

$\therefore \ \ \cos \dfrac{x}{2}=-\dfrac{1}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{-\sqrt{3}}{3}$

$\tan \dfrac{x}{2}=\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}=\dfrac{\left(\dfrac{\sqrt{2}}{\sqrt{3}}\right)}{\left(\dfrac{-1}{\sqrt{3}}\right)}=-\sqrt{2}$

Thus, the respective values of $\sin \dfrac{x}{2}, \cos \dfrac{x}{2}$ and $\tan \dfrac{x}{2}$ are $\dfrac{\sqrt{6}}{3}, \dfrac{-\sqrt{3}}{3}$, and $-\sqrt{2}$

Answer :

Here, $x$ is in quadrant II.

i.e., $\dfrac{\pi}{2}<x<\pi$

$\Rightarrow \dfrac{\pi}{4}<\dfrac{x}{2}<\dfrac{\pi}{2}$

Therefore, $\sin \dfrac{x}{2}, \cos \dfrac{x}{2}$, and $\tan \dfrac{x}{2}$ are all positive.

It is given that $\sin x=\dfrac{1}{4}$.

$ \ \cos ^{2} x=1-\sin ^{2} x=1-\left(\dfrac{1}{4}\right)^{2}=1-\dfrac{1}{16}=\dfrac{15}{16}$

$\quad\cos x=-\dfrac{\sqrt{15}}{4}$ $\qquad\bigg[\because \ \cos x$ is negative in quadrant II $\bigg]$

$ \begin{aligned} \sin ^{2} \dfrac{x}{2} & =\dfrac{1-\cos x}{2}=\dfrac{1-\left(-\dfrac{\sqrt{15}}{4}\right)}{2}=\dfrac{4+\sqrt{15}}{8} \\ \\ \sin \dfrac{x}{2} & =\sqrt{\dfrac{4+\sqrt{15}}{8}} \quad\bigg[\because \sin \dfrac{x}{2} \text{ is positive }\bigg] \\ \\ & =\sqrt{\dfrac{4+\sqrt{15}}{8} \times \dfrac{2}{2}} \\ \\ & =\sqrt{\dfrac{8+2 \sqrt{15}}{16}} =\dfrac{\sqrt{8+2 \sqrt{15}}}{4} \\ \\ \cos ^{2} \dfrac{x}{2} & =\dfrac{1+\cos x}{2}=\dfrac{1+\left(-\dfrac{\sqrt{15}}{4}\right)}{2}=\dfrac{4-\sqrt{15}}{8} \\ \\ \cos \dfrac{x}{2} & =\sqrt{\dfrac{4-\sqrt{15}}{8}} \qquad\bigg[\because \cos \dfrac{x}{2} \text{ is positive }\bigg] \\ \\ & =\sqrt{\dfrac{4-\sqrt{15}}{8} \times \dfrac{2}{2}} \\ \\ & =\sqrt{\dfrac{8-2 \sqrt{15}}{16}} =\dfrac{\sqrt{8-2 \sqrt{15}}}{4} \\ \\ \tan \dfrac{x}{2} & =\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}=\dfrac{\left(\dfrac{\sqrt{8+2 \sqrt{15}}}{4}\right)}{\left(\dfrac{\sqrt{8-2 \sqrt{15}}}{4}\right)}=\dfrac{\sqrt{8+2 \sqrt{15}}}{\sqrt{8-2 \sqrt{15}}} \\ \\ & =\sqrt{\dfrac{8+2 \sqrt{15}}{8-2 \sqrt{15}} \times \dfrac{8+2 \sqrt{15}}{8+2 \sqrt{15}}} \\ \\ & =\sqrt{\dfrac{\left(8+2 \sqrt{15}\right)^{2}}{64-60}}=\dfrac{8+2 \sqrt{15}}{2}=4+\sqrt{15} \end{aligned} $

Thus, the respective values of $\sin \dfrac{x}{2}, \cos \dfrac{x}{2}$ and $\tan \dfrac{x}{2}$ are $\dfrac{\sqrt{8+2 \sqrt{15}}}{4}, \dfrac{\sqrt{8-2 \sqrt{15}}}{4}$, and $4+\sqrt{15}$