Prove that
1. $2\cos \frac{\pi}{13}\cos \frac{9\pi}{13}$ $+\cos\frac{3\pi}{13}+\cos\frac{5\pi}{13}=0$
L.H.S.,
= $2\cos \frac{\pi}{13}\cos \frac{9\pi}{13}$ $+2\cos(\frac{3\pi+5\pi}{13*2})\cos(\frac{3\pi-5\pi}{13*2})$
= $2\cos \frac{\pi}{13}\cos \frac{9\pi}{13}$ $+2\cos(\frac{4\pi}{13})\cos(\frac{-\pi}{13})$
= $2\cos(\frac{\pi}{13})\cos(\pi-\frac{4\pi}{13})$ $+2\cos(\frac{4\pi}{13})\cos(\frac{\pi}{13})$
= $-2\cos(\frac{\pi}{13})\cos(\frac{4\pi}{13})$ $+2\cos(\frac{4\pi}{13})\cos(\frac{\pi}{13})$ =0
L.H.S.=R.H.S.
2. $(\sin 3x+\sin x)\sin x$ $+(\cos 3x-\cos x)\cos x$ =0
L.H.S.,
= $[2\sin (\frac{3x+x}{2})\cos (\frac{3x-x}{2})]$ $\sin x$ $+[-2\sin (\frac{3x+x}{2})\sin (\frac{3x-x}{2})]$ $\cos x$
= $2\sin 2x \cos x \sin x$ $-2\sin 2x \cos x \sin x$
=0
L.H.S.=R.H.S.
3. $(\cos x+\cos y)^{2}$ $+(\sin x-\sin y)^{2}$ $=4\cos^{2}(\frac{x+y}{2})$
L.H.S.,
= $[2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})]^{2}$ $+[2\cos(\frac{x+y}{2})\sin (\frac{x-y}{2})]^{2}$
= $4\cos^{2}(\frac{x+y}{2})\cos^{2}(\frac{x-y}{2})$ $+4\cos^{2}(\frac{x+y}{2})\sin^{2}(\frac{x-y}{2})$
= $4\cos^{2}(\frac{x+y}{2})$ $[\cos^{2}(\frac{x-y}{2})+\sin^{2}(\frac{x-y}{2})]$
= $4\cos^{2}(\frac{x+y}{2})$ [cos$\displaystyle \small ^{2}$ x+sin$\displaystyle \small ^{2}$ x=1]
L.H.S.=R.H.S.
4. $(\cos x-\cos y)^{2}$ $+(\sin x-\sin y)^{2}$ $=4\sin^{2}(\frac{x-y}{2})$
L.H.S.,
= $[-2\sin(\frac{x+y}{2})\sin(\frac{x-y}{2})]^{2}$ $+[2\cos(\frac{x+y}{2})\sin (\frac{x-y}{2})]^{2}$
= $4\sin^{2}(\frac{x+y}{2})\sin^{2}(\frac{x-y}{2})$ $+4\cos^{2}(\frac{x+y}{2})\sin^{2}(\frac{x-y}{2})$
= $4\sin^{2}(\frac{x-y}{2})$ $[\sin^{2}(\frac{x+y}{2})+\cos^{2}(\frac{x+y}{2})]$
= $4\sin^{2}(\frac{x-y}{2})$ [cos$\displaystyle \small ^{2}$ x+sin$\displaystyle \small ^{2}$ x=1]
L.H.S.=R.H.S.
5. $\sin x+\sin 3x$ $+\sin 5x+\sin 7x$ $=4\cos x\cos 2x\sin 4x$
L.H.S.,
= $\left (\sin x+\sin 7x \right )$ $+\left (\sin 3x+\sin 5x \right )$
= $2\sin \left ( \frac{x+7x}{2} \right )\cos\left ( \frac{x-7x}{2} \right )$ $+2\sin \left ( \frac{3x+5x}{2} \right )\cos\left ( \frac{3x-5x}{2} \right )$
= $2\sin 4x\cos(-3x)$ $+2\sin 4x\cos(-x)$
= $2\sin 4x$ $[\cos 3x+\cos x]$
= $2\sin 4x$ $\left [ 2\cos(\frac{3x+x}{2})\cos(\frac{3x-x}{2}) \right ]$
= $2\sin 4x$ $[2\cos 2x\cos x]$
= $4\cos x\cos 2x\sin 4x$
L.H.S.=R.H.S.
6. $\frac{(\sin 7x+\sin 5x)+(\sin 9x+\sin 3x)}{(\cos 7x+\cos 5x)+(\cos 9x+\cos 3x)}$ $=\tan 6x$
L.H.S.,
= $\frac{2\sin(\frac{7x+5x}{2})\cos(\frac{7x-5x}{2})+2\sin(\frac{9x+3x}{2})\cos(\frac{9x-3x}{2})}{2\cos (\frac{7x+5x}{2})\cos(\frac{7x-5x}{2})+2\cos (\frac{9x+3x}{2})\cos(\frac{9x-3x}{2})}$
= $\frac{2\sin 6x\cos x+2\sin 6x\cos 3x}{2\cos 6x\cos x+2\cos 6x\cos 3x}$
= $\frac{2\sin 6x}{2\cos 6x}$ $\frac{\cos x+\cos 3x}{\cos x+\cos 3x}$
= $\tan 6x$
L.H.S.=R.H.S.
7. $\sin 3x+\sin 2x-\sin x$ $=4\sin x\cos \frac{x}{2}\cos \frac{3x}{2}$
L.H.S.,
= $\sin 3x$ $+[2\cos(\frac{2x+x}{2})\sin (\frac{2x-x}{2})]$
= $\sin 3x+2\cos \frac{3x}{2}\sin \frac{x}{2}$
= $\sin 2(\frac{3x}{2})+2\cos \frac{3x}{2}\sin \frac{x}{2}$
= $2\sin \frac{3x}{2}\cos \frac{3x}{2}$ $+2\cos \frac{3x}{2}\sin \frac{x}{2}$
= $2\cos \frac{3x}{2}$ $\left [ \sin \frac{3x}{2}+\sin \frac{x}{2} \right ]$
= $2\cos \frac{3x}{2}$ $\left [ 2\sin (\frac{3x+x}{2*2})\cos(\frac{3x-x}{2*2}) \right ]$
= $2\cos \frac{3x}{2}$ $[2\sin x\cos \frac{x}{2}]$
= $4\sin x\cos \frac{x}{2}\cos \frac{3x}{2}$
L.H.S.=R.H.S.
Find $\sin \frac{x}{2}$ , $\cos \frac{x}{2}$ and $\tan \frac{x}{2}$ in each of the following:
8. $\tan x=-\frac{4}{3}$ , x in quadrant II
x is in quadrant II, i.e. $\frac{\pi}{2}< x< \pi$
⇒ $\frac{\pi}{4}< \frac{x}{2}< \frac{\pi}{2}$
⇒ $\sin \frac{x}{2}$ , $\cos \frac{x}{2}$ and $\tan \frac{x}{2}$ lies in quadrant I
Given, $\tan x=-\frac{4}{3}$
$\sec^{2}x=1+\tan^{2}x$
= $1+\left ( -\frac{4}{3} \right )^{2}$ = $1+\frac{16}{9}$
= $\frac{25}{9}$
∴ $\sec x=\pm \frac{5}{3}$
⇒ $\cos x = \pm \frac{3}{5}$
Since x lies in quadrant II, cos x is negative
⇒ $\cos x = -\frac{3}{5}$
$\cos x=2\cos^{2}\frac{x}{2}-1$ [cos 2x=2cos$\displaystyle \small ^{2}$ x-1]
⇒ $-\frac{3}{5}=2\cos^{2}\frac{x}{2}-1$
⇒ $2\cos^{2}\frac{x}{2}=-\frac{3}{5}+1=\frac{2}{5}$
⇒ $\cos^{2}\frac{x}{2}=\frac{1}{5}$
⇒ $\cos \frac{x}{2}=\pm \frac{1}{\sqrt{5}}$
Since $\frac{x}{2}$ lies in quadrant I, $\cos \frac{x}{2}$ is positive
$\cos \frac{x}{2}=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}$ [$\frac{1}{\sqrt{5}}*\frac{\sqrt{5}}{\sqrt{5}}=\frac{\sqrt{5}}{5}$ ]
$\sin^{2}\frac{x}{2}+\cos^{2}\frac{x}{2}=1$
⇒ $\sin^{2}\frac{x}{2}=1-(\frac{1}{\sqrt{5}})^{2}$
= $1-\frac{1}{5}=\frac{4}{5}$
∴ $\sin \frac{x}{2}=\pm \frac{2}{\sqrt{5}}$
Since $\frac{x}{2}$ lies in quadrant I, $\sin \frac{x}{2}$ is positive
$\sin \frac{x}{2}= \frac{2}{\sqrt{5}}= \frac{2\sqrt{5}}{5}$
$\tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}$
= $\frac{2/\sqrt{5}}{1/\sqrt{5}}$
$\tan \frac{x}{2}=2$
9. $\cos x=-\frac{1}{3}$, x in quadrant III
x is in quadrant III, i.e. $\pi< x< \frac{3\pi}{2}$
⇒ $\frac{\pi}{2}< \frac{x}{2}< \frac{3\pi}{4}$
⇒ $\sin \frac{x}{2}$ , $\cos \frac{x}{2}$ and $\tan \frac{x}{2}$ lies in quadrant II
Given, $\cos x = -\frac{1}{3}$
$\cos x=2\cos^{2}\frac{x}{2}-1$ [cos 2x=2cos$\displaystyle \small ^{2}$ x-1]
⇒ $-\frac{1}{3}=2\cos^{2}\frac{x}{2}-1$
⇒ $2\cos^{2}\frac{x}{2}=-\frac{1}{3}+1=\frac{2}{3}$
⇒ $\cos^{2}\frac{x}{2}=\frac{1}{3}$
⇒ $\cos \frac{x}{2}=\pm \frac{1}{\sqrt{3}}$
Since $\frac{x}{2}$ lies in quadrant II, $\cos \frac{x}{2}$ is negative
$\cos \frac{x}{2}=-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3}$
$\sin^{2}\frac{x}{2}+\cos^{2}\frac{x}{2}=1$
⇒ $\sin^{2}\frac{x}{2}=1-(-\frac{1}{\sqrt{3}})^{2}$
= $1-\frac{1}{3}=\frac{2}{3}$
∴ $\sin \frac{x}{2}=\pm \sqrt{\frac{2}{3}}$
Since $\frac{x}{2}$ lies in quadrant II, $\sin \frac{x}{2}$ is positive
$\sin \frac{x}{2}=\frac{\sqrt{2}}{\sqrt{3}}=\frac{\sqrt{6}}{3}$
$\tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}$
= $\frac{\sqrt{2}/\sqrt{3}}{-1/\sqrt{3}}$
$\tan \frac{x}{2}=-\sqrt{2}$
10. $\sin x=\frac{1}{4}$ , x in quadrant II
x is in quadrant II, i.e. $\frac{\pi}{2}< x< \pi$
⇒ $\frac{\pi}{4}< \frac{x}{2}< \frac{\pi}{2}$
⇒ $\sin \frac{x}{2}$ , $\cos \frac{x}{2}$ and $\tan \frac{x}{2}$ lies in quadrant I
Given, $\sin x=\frac{1}{4}$
$\cos^{2}x=1-\sin^{2}x$
= $1-(\frac{1}{4})^{2}=1-\frac{1}{16}$ $=\frac{15}{16}$
$\cos x=\pm \frac{\sqrt{15}}{4}$
Since x lies in quadrant II, cos x is negative
$\cos x=- \frac{\sqrt{15}}{4}$
$\cos x=2\cos^{2}\frac{x}{2}-1$ [cos 2x=2cos$\displaystyle \small ^{2}$ x-1]
⇒ $-\frac{\sqrt{15}}{4}=2\cos^{2}\frac{x}{2}-1$
⇒ $2\cos^{2}\frac{x}{2}=-\frac{\sqrt{15}}{4}+1=\frac{-\sqrt{15}+4}{4}$
⇒ $\cos\frac{x}{2}=\pm \sqrt{\frac{4-\sqrt{15}}{8}*\frac{2}{2}}$
⇒ $\cos\frac{x}{2}=\pm \sqrt{\frac{8-2\sqrt{15}}{16}}$
Since $\frac{x}{2}$ lies in quadrant I, $\cos \frac{x}{2}$ is positive
$\cos\frac{x}{2}= \frac{\sqrt{8-2\sqrt{15}}}{4}$
$\sin^{2}\frac{x}{2}+\cos^{2}\frac{x}{2}=1$
⇒ $\sin^{2}\frac{x}{2}=1-\left ( \sqrt{\frac{4-\sqrt{15}}{8}} \right )^{2}$
= $1- \frac{4-\sqrt{15}}{8}$ $=\frac{4+\sqrt{15}}{8}$
$\sin\frac{x}{2}=\pm \sqrt{\frac{4+\sqrt{15}}{8}}$
Since $\frac{x}{2}$ lies in quadrant I, $\sin \frac{x}{2}$ is positive
$\sin\frac{x}{2}=\sqrt{\frac{4+\sqrt{15}}{8}*\frac{2}{2}}$ $=\sqrt{\frac{8+2\sqrt{15}}{16}}$
$\sin\frac{x}{2}=\frac{\sqrt{8+2\sqrt{15}}}{4}$
$\tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}$= $\frac{(\sqrt{8+2\sqrt{15}})/4}{(\sqrt{8-2\sqrt{15}})/4}$
= $\sqrt{\frac{8+2\sqrt{15}}{8-2\sqrt{15}}*\frac{8+2\sqrt{15}}{8+2\sqrt{15}}}$
= $\sqrt{\frac{(8+2\sqrt{15})^{2}}{8^{2}-(2\sqrt{15})^{2}}}$
= $\sqrt{\frac{(8+2\sqrt{15})^{2}}{64-60}}$
= $\sqrt{\frac{(8+2\sqrt{15})^{2}}{4}}$
= $\frac{8+2\sqrt{15}}{2}$ $=4+\sqrt{15}$
$\tan \frac{x}{2}=4+\sqrt{15}$
1. $2\cos \frac{\pi}{13}\cos \frac{9\pi}{13}$ $+\cos\frac{3\pi}{13}+\cos\frac{5\pi}{13}=0$
L.H.S.,
= $2\cos \frac{\pi}{13}\cos \frac{9\pi}{13}$ $+2\cos(\frac{3\pi+5\pi}{13*2})\cos(\frac{3\pi-5\pi}{13*2})$
= $2\cos \frac{\pi}{13}\cos \frac{9\pi}{13}$ $+2\cos(\frac{4\pi}{13})\cos(\frac{-\pi}{13})$
= $2\cos(\frac{\pi}{13})\cos(\pi-\frac{4\pi}{13})$ $+2\cos(\frac{4\pi}{13})\cos(\frac{\pi}{13})$
= $-2\cos(\frac{\pi}{13})\cos(\frac{4\pi}{13})$ $+2\cos(\frac{4\pi}{13})\cos(\frac{\pi}{13})$ =0
L.H.S.=R.H.S.
2. $(\sin 3x+\sin x)\sin x$ $+(\cos 3x-\cos x)\cos x$ =0
L.H.S.,
= $[2\sin (\frac{3x+x}{2})\cos (\frac{3x-x}{2})]$ $\sin x$ $+[-2\sin (\frac{3x+x}{2})\sin (\frac{3x-x}{2})]$ $\cos x$
= $2\sin 2x \cos x \sin x$ $-2\sin 2x \cos x \sin x$
=0
L.H.S.=R.H.S.
3. $(\cos x+\cos y)^{2}$ $+(\sin x-\sin y)^{2}$ $=4\cos^{2}(\frac{x+y}{2})$
L.H.S.,
= $[2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})]^{2}$ $+[2\cos(\frac{x+y}{2})\sin (\frac{x-y}{2})]^{2}$
= $4\cos^{2}(\frac{x+y}{2})\cos^{2}(\frac{x-y}{2})$ $+4\cos^{2}(\frac{x+y}{2})\sin^{2}(\frac{x-y}{2})$
= $4\cos^{2}(\frac{x+y}{2})$ $[\cos^{2}(\frac{x-y}{2})+\sin^{2}(\frac{x-y}{2})]$
= $4\cos^{2}(\frac{x+y}{2})$ [cos$\displaystyle \small ^{2}$ x+sin$\displaystyle \small ^{2}$ x=1]
L.H.S.=R.H.S.
4. $(\cos x-\cos y)^{2}$ $+(\sin x-\sin y)^{2}$ $=4\sin^{2}(\frac{x-y}{2})$
L.H.S.,
= $[-2\sin(\frac{x+y}{2})\sin(\frac{x-y}{2})]^{2}$ $+[2\cos(\frac{x+y}{2})\sin (\frac{x-y}{2})]^{2}$
= $4\sin^{2}(\frac{x+y}{2})\sin^{2}(\frac{x-y}{2})$ $+4\cos^{2}(\frac{x+y}{2})\sin^{2}(\frac{x-y}{2})$
= $4\sin^{2}(\frac{x-y}{2})$ $[\sin^{2}(\frac{x+y}{2})+\cos^{2}(\frac{x+y}{2})]$
= $4\sin^{2}(\frac{x-y}{2})$ [cos$\displaystyle \small ^{2}$ x+sin$\displaystyle \small ^{2}$ x=1]
L.H.S.=R.H.S.
5. $\sin x+\sin 3x$ $+\sin 5x+\sin 7x$ $=4\cos x\cos 2x\sin 4x$
L.H.S.,
= $\left (\sin x+\sin 7x \right )$ $+\left (\sin 3x+\sin 5x \right )$
= $2\sin \left ( \frac{x+7x}{2} \right )\cos\left ( \frac{x-7x}{2} \right )$ $+2\sin \left ( \frac{3x+5x}{2} \right )\cos\left ( \frac{3x-5x}{2} \right )$
= $2\sin 4x\cos(-3x)$ $+2\sin 4x\cos(-x)$
= $2\sin 4x$ $[\cos 3x+\cos x]$
= $2\sin 4x$ $\left [ 2\cos(\frac{3x+x}{2})\cos(\frac{3x-x}{2}) \right ]$
= $2\sin 4x$ $[2\cos 2x\cos x]$
= $4\cos x\cos 2x\sin 4x$
L.H.S.=R.H.S.
6. $\frac{(\sin 7x+\sin 5x)+(\sin 9x+\sin 3x)}{(\cos 7x+\cos 5x)+(\cos 9x+\cos 3x)}$ $=\tan 6x$
L.H.S.,
= $\frac{2\sin(\frac{7x+5x}{2})\cos(\frac{7x-5x}{2})+2\sin(\frac{9x+3x}{2})\cos(\frac{9x-3x}{2})}{2\cos (\frac{7x+5x}{2})\cos(\frac{7x-5x}{2})+2\cos (\frac{9x+3x}{2})\cos(\frac{9x-3x}{2})}$
= $\frac{2\sin 6x\cos x+2\sin 6x\cos 3x}{2\cos 6x\cos x+2\cos 6x\cos 3x}$
= $\frac{2\sin 6x}{2\cos 6x}$ $\frac{\cos x+\cos 3x}{\cos x+\cos 3x}$
= $\tan 6x$
L.H.S.=R.H.S.
7. $\sin 3x+\sin 2x-\sin x$ $=4\sin x\cos \frac{x}{2}\cos \frac{3x}{2}$
L.H.S.,
= $\sin 3x$ $+[2\cos(\frac{2x+x}{2})\sin (\frac{2x-x}{2})]$
= $\sin 3x+2\cos \frac{3x}{2}\sin \frac{x}{2}$
= $\sin 2(\frac{3x}{2})+2\cos \frac{3x}{2}\sin \frac{x}{2}$
= $2\sin \frac{3x}{2}\cos \frac{3x}{2}$ $+2\cos \frac{3x}{2}\sin \frac{x}{2}$
= $2\cos \frac{3x}{2}$ $\left [ \sin \frac{3x}{2}+\sin \frac{x}{2} \right ]$
= $2\cos \frac{3x}{2}$ $\left [ 2\sin (\frac{3x+x}{2*2})\cos(\frac{3x-x}{2*2}) \right ]$
= $2\cos \frac{3x}{2}$ $[2\sin x\cos \frac{x}{2}]$
= $4\sin x\cos \frac{x}{2}\cos \frac{3x}{2}$
L.H.S.=R.H.S.
Find $\sin \frac{x}{2}$ , $\cos \frac{x}{2}$ and $\tan \frac{x}{2}$ in each of the following:
8. $\tan x=-\frac{4}{3}$ , x in quadrant II
x is in quadrant II, i.e. $\frac{\pi}{2}< x< \pi$
⇒ $\frac{\pi}{4}< \frac{x}{2}< \frac{\pi}{2}$
⇒ $\sin \frac{x}{2}$ , $\cos \frac{x}{2}$ and $\tan \frac{x}{2}$ lies in quadrant I
Given, $\tan x=-\frac{4}{3}$
$\sec^{2}x=1+\tan^{2}x$
= $1+\left ( -\frac{4}{3} \right )^{2}$ = $1+\frac{16}{9}$
= $\frac{25}{9}$
∴ $\sec x=\pm \frac{5}{3}$
⇒ $\cos x = \pm \frac{3}{5}$
Since x lies in quadrant II, cos x is negative
⇒ $\cos x = -\frac{3}{5}$
$\cos x=2\cos^{2}\frac{x}{2}-1$ [cos 2x=2cos$\displaystyle \small ^{2}$ x-1]
⇒ $-\frac{3}{5}=2\cos^{2}\frac{x}{2}-1$
⇒ $2\cos^{2}\frac{x}{2}=-\frac{3}{5}+1=\frac{2}{5}$
⇒ $\cos^{2}\frac{x}{2}=\frac{1}{5}$
⇒ $\cos \frac{x}{2}=\pm \frac{1}{\sqrt{5}}$
Since $\frac{x}{2}$ lies in quadrant I, $\cos \frac{x}{2}$ is positive
$\cos \frac{x}{2}=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}$ [$\frac{1}{\sqrt{5}}*\frac{\sqrt{5}}{\sqrt{5}}=\frac{\sqrt{5}}{5}$ ]
$\sin^{2}\frac{x}{2}+\cos^{2}\frac{x}{2}=1$
⇒ $\sin^{2}\frac{x}{2}=1-(\frac{1}{\sqrt{5}})^{2}$
= $1-\frac{1}{5}=\frac{4}{5}$
∴ $\sin \frac{x}{2}=\pm \frac{2}{\sqrt{5}}$
Since $\frac{x}{2}$ lies in quadrant I, $\sin \frac{x}{2}$ is positive
$\sin \frac{x}{2}= \frac{2}{\sqrt{5}}= \frac{2\sqrt{5}}{5}$
$\tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}$
= $\frac{2/\sqrt{5}}{1/\sqrt{5}}$
$\tan \frac{x}{2}=2$
9. $\cos x=-\frac{1}{3}$, x in quadrant III
x is in quadrant III, i.e. $\pi< x< \frac{3\pi}{2}$
⇒ $\frac{\pi}{2}< \frac{x}{2}< \frac{3\pi}{4}$
⇒ $\sin \frac{x}{2}$ , $\cos \frac{x}{2}$ and $\tan \frac{x}{2}$ lies in quadrant II
Given, $\cos x = -\frac{1}{3}$
$\cos x=2\cos^{2}\frac{x}{2}-1$ [cos 2x=2cos$\displaystyle \small ^{2}$ x-1]
⇒ $-\frac{1}{3}=2\cos^{2}\frac{x}{2}-1$
⇒ $2\cos^{2}\frac{x}{2}=-\frac{1}{3}+1=\frac{2}{3}$
⇒ $\cos^{2}\frac{x}{2}=\frac{1}{3}$
⇒ $\cos \frac{x}{2}=\pm \frac{1}{\sqrt{3}}$
Since $\frac{x}{2}$ lies in quadrant II, $\cos \frac{x}{2}$ is negative
$\cos \frac{x}{2}=-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3}$
$\sin^{2}\frac{x}{2}+\cos^{2}\frac{x}{2}=1$
⇒ $\sin^{2}\frac{x}{2}=1-(-\frac{1}{\sqrt{3}})^{2}$
= $1-\frac{1}{3}=\frac{2}{3}$
∴ $\sin \frac{x}{2}=\pm \sqrt{\frac{2}{3}}$
Since $\frac{x}{2}$ lies in quadrant II, $\sin \frac{x}{2}$ is positive
$\sin \frac{x}{2}=\frac{\sqrt{2}}{\sqrt{3}}=\frac{\sqrt{6}}{3}$
$\tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}$
= $\frac{\sqrt{2}/\sqrt{3}}{-1/\sqrt{3}}$
$\tan \frac{x}{2}=-\sqrt{2}$
10. $\sin x=\frac{1}{4}$ , x in quadrant II
x is in quadrant II, i.e. $\frac{\pi}{2}< x< \pi$
⇒ $\frac{\pi}{4}< \frac{x}{2}< \frac{\pi}{2}$
⇒ $\sin \frac{x}{2}$ , $\cos \frac{x}{2}$ and $\tan \frac{x}{2}$ lies in quadrant I
Given, $\sin x=\frac{1}{4}$
$\cos^{2}x=1-\sin^{2}x$
= $1-(\frac{1}{4})^{2}=1-\frac{1}{16}$ $=\frac{15}{16}$
$\cos x=\pm \frac{\sqrt{15}}{4}$
Since x lies in quadrant II, cos x is negative
$\cos x=- \frac{\sqrt{15}}{4}$
$\cos x=2\cos^{2}\frac{x}{2}-1$ [cos 2x=2cos$\displaystyle \small ^{2}$ x-1]
⇒ $-\frac{\sqrt{15}}{4}=2\cos^{2}\frac{x}{2}-1$
⇒ $2\cos^{2}\frac{x}{2}=-\frac{\sqrt{15}}{4}+1=\frac{-\sqrt{15}+4}{4}$
⇒ $\cos\frac{x}{2}=\pm \sqrt{\frac{4-\sqrt{15}}{8}*\frac{2}{2}}$
⇒ $\cos\frac{x}{2}=\pm \sqrt{\frac{8-2\sqrt{15}}{16}}$
Since $\frac{x}{2}$ lies in quadrant I, $\cos \frac{x}{2}$ is positive
$\cos\frac{x}{2}= \frac{\sqrt{8-2\sqrt{15}}}{4}$
$\sin^{2}\frac{x}{2}+\cos^{2}\frac{x}{2}=1$
⇒ $\sin^{2}\frac{x}{2}=1-\left ( \sqrt{\frac{4-\sqrt{15}}{8}} \right )^{2}$
= $1- \frac{4-\sqrt{15}}{8}$ $=\frac{4+\sqrt{15}}{8}$
$\sin\frac{x}{2}=\pm \sqrt{\frac{4+\sqrt{15}}{8}}$
Since $\frac{x}{2}$ lies in quadrant I, $\sin \frac{x}{2}$ is positive
$\sin\frac{x}{2}=\sqrt{\frac{4+\sqrt{15}}{8}*\frac{2}{2}}$ $=\sqrt{\frac{8+2\sqrt{15}}{16}}$
$\sin\frac{x}{2}=\frac{\sqrt{8+2\sqrt{15}}}{4}$
$\tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}$= $\frac{(\sqrt{8+2\sqrt{15}})/4}{(\sqrt{8-2\sqrt{15}})/4}$
= $\sqrt{\frac{8+2\sqrt{15}}{8-2\sqrt{15}}*\frac{8+2\sqrt{15}}{8+2\sqrt{15}}}$
= $\sqrt{\frac{(8+2\sqrt{15})^{2}}{8^{2}-(2\sqrt{15})^{2}}}$
= $\sqrt{\frac{(8+2\sqrt{15})^{2}}{64-60}}$
= $\sqrt{\frac{(8+2\sqrt{15})^{2}}{4}}$
= $\frac{8+2\sqrt{15}}{2}$ $=4+\sqrt{15}$
$\tan \frac{x}{2}=4+\sqrt{15}$
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