Answer :

L.H.S.

$=2 \cos \dfrac{\pi}{13} \cos \dfrac{9 \pi}{13}+\cos \dfrac{3 \pi}{13}+\cos \dfrac{5 \pi}{13}$

$=2 \cos \dfrac{\pi}{13} \cos \dfrac{9 \pi}{13}+2 \cos (\dfrac{\dfrac{3 \pi}{13}+\dfrac{5 \pi}{13}}{2}) \cos (\dfrac{\dfrac{3 \pi}{13}-\dfrac{5 \pi}{13}}{2})[\cos x+\cos y=2 \cos (\dfrac{x+y}{2}) \cos (\dfrac{x-y}{2})]$

$=2 \cos \dfrac{\pi}{13} \cos \dfrac{9 \pi}{13}+2 \cos \dfrac{4 \pi}{13} \cos (\dfrac{-\pi}{13})$

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

$=2 \cos \dfrac{\pi}{13}[\cos \dfrac{9 \pi}{13}+\cos \dfrac{4 \pi}{13}]$

$=2 \cos \dfrac{\pi}{13}[2 \cos (\dfrac{\dfrac{9 \pi}{13}+\dfrac{4 \pi}{13}}{2}) \cos (\dfrac{\dfrac{9 \pi}{13}-\dfrac{4 \pi}{13}}{2})]$

$=2 \cos \dfrac{\pi}{13}[2 \cos \dfrac{\pi}{2} \cos \dfrac{5 \pi}{26}]$

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

$=0=$ R.H.S

Answer :

L.H.S.

$=(\sin 3 x+\sin x) \sin x+(\cos 3 x$ ấ" $\cos x) \cos x$

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

$=\cos 3 x \cos x+\sin 3 x \sin x-(\cos ^{2} x-\sin ^{2} x)$

$=\cos (3 x-x)-\cos 2 x \quad[\cos (A-B)=\cos A \cos B+\sin A \sin B]$

$=\cos 2 x-\cos 2 x$

$=0$

$=$ RH.S.

Answer :

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

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

$=(\cos ^{2} x+\sin ^{2} x)+(\cos ^{2} y+\sin ^{2} y)+2(\cos x \cos y-\sin x \sin y)$

$=1+1+2 \cos (x+y) \quad[\cos (A+B)=(\cos A \cos B-\sin A \sin B)]$

$=2+2 \cos (x+y)$

$=2[1+\cos (x+y)]$

$=2[1+2 \cos ^{2}(\dfrac{x+y}{2})-1] \quad[\cos 2 A=2 \cos ^{2} A-1]$

$=4 \cos ^{2}(\dfrac{x+y}{2})=$ R.H.S.

Answer :

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

$ \begin{matrix} =\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] \\ =1+1-2[\cos (x-y)] & \\ {[\cos (A-B)=\cos A \cos B+\sin A \sin B]} \\ =2[1-\cos (x-y)] & \\ =2[1-\{1-2 \sin ^{2}(\dfrac{x-y}{2})\}] & \\ {[\cos 2 A=1-2 \sin ^{2} A]} \\ =4 \sin ^{2}(\dfrac{x-y}{2})=\text{ R.H.S. } \end{matrix} $

Answer :

It is known that

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

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

$=(\sin x+\sin 5 x)+(\sin 3 x+\sin 7 x)$

$=2 \sin (\dfrac{x+5 x}{2}) \cdot \cos (\dfrac{x-5 x}{2})+2 \sin (\dfrac{3 x+7 x}{2}) \cos (\dfrac{3 x-7 x}{2})$

$=2 \sin 3 x \cos (-2 x)+2 \sin 5 x \cos (-2 x)$

$=2 \sin 3 x \cos 2 x+2 \sin 5 x \cos 2 x$

$=2 \cos 2 x[\sin 3 x+\sin 5 x]$

$=2 \cos 2 x[2 \sin (\dfrac{3 x+5 x}{2}) \cdot \cos (\dfrac{3 x-5 x}{2})]$

$=2 \cos 2 x[2 \sin 4 x \cdot \cos (-x)]$

$=4 \cos 2 x \sin 4 x \cos x=$ R.H.S.

Answer :

It is known that

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

$=\dfrac{(\sin 7 x+\sin 5 x)+(\sin 9 x+\sin 3 x)}{(\cos 7 x+\cos 5 x)+(\cos 9 x+\cos 3 x)}$

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

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

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

$=\tan 6 x$

$=$ R.H.S.

Answer :

L.H.S. $=\sin 3 x+\sin 2 x-\sin x$

$ \begin{aligned} & =\sin 3 x+(\sin 2 x-\sin x) \\ & =\sin 3 x+[2 \cos (\dfrac{2 x+x}{2}) \sin (\dfrac{2 x-x}{2})] \quad[\sin A-\sin B=2 \cos (\dfrac{A+B}{2}) \sin (\dfrac{A-B}{2})] \\ & =\sin 3 x+[2 \cos (\dfrac{3 x}{2}) \sin (\dfrac{x}{2})] \\ & =\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[\sin 2 A=2 \sin A \cdot \cos B] \\ & =2 \cos (\dfrac{3 x}{2})[\sin (\dfrac{3 x}{2})+\sin (\dfrac{x}{2})] \\ & =2 \cos (\dfrac{3 x}{2})[2 \sin \{\dfrac{(\dfrac{3 x}{2})+(\dfrac{x}{2})}{2}\} \cos \{\dfrac{(\dfrac{3 x}{2})-(\dfrac{x}{2})}{2}\}][\sin A+\sin B=2 \sin (\dfrac{A+B}{2}) \cos (\dfrac{A-B}{2})] \\ & .=2 \cos (\dfrac{3 x}{2}) \cdot 2 \sin x \cos (\dfrac{x}{2})] \cos (\dfrac{3 x}{2})=\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 $\tan x=-\dfrac{4}{3}$.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$[\because \sin \dfrac{x}{2}.$ is positive $]$

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

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 $\cos x=-\dfrac{1}{3}$.

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

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

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

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

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

Now,

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

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

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

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

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 $\sin x=\dfrac{1}{4}$.

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

$\Rightarrow \cos x=-\dfrac{\sqrt{15}}{4}$

[cos $x$ is negative in quadrant II]

$ \begin{aligned} & \sin ^{2} \dfrac{x}{2}=\dfrac{1-\cos x}{2}=\dfrac{1-(-\dfrac{\sqrt{15}}{4})}{2}=\dfrac{4+\sqrt{15}}{8} \\ & \Rightarrow \sin \dfrac{x}{2}=\sqrt{\dfrac{4+\sqrt{15}}{8}} \quad[\because \sin \dfrac{x}{2} \text{ is positive }] \\ & =\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+(-\dfrac{\sqrt{15}}{4})}{2}=\dfrac{4-\sqrt{15}}{8} \\ & \Rightarrow \cos \dfrac{x}{2}=\sqrt{\dfrac{4-\sqrt{15}}{8}} \quad[\because \cos \dfrac{x}{2} \text{ is positive }] \\ & =\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{(\dfrac{\sqrt{8+2 \sqrt{15}}}{4})}{(\dfrac{\sqrt{8-2 \sqrt{15}}}{4})}=\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{(8+2 \sqrt{15})^{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 $4+\sqrt{15}$