Solve this


If $\cos x=\frac{-3}{5}$ and $\frac{\pi}{2}

(i) $\sin \frac{x}{2}$

(ii) $\cos \frac{x}{2}$

(iii) $\tan \frac{x}{2}$



Given: $\cos x=-\frac{3}{5}$ and $\frac{\pi}{2}

To Find: i) $\sin \frac{x}{2}$ ii) $\cos \frac{x}{2}$ iii) $\tan \frac{x}{2}$

i) $\sin \frac{x}{2}$

Formula used:

$\sin \frac{x}{2}=\pm \sqrt{\frac{1-\cos x}{2}}$

Now, $\sin \frac{x}{2}=\pm \sqrt{\frac{1-\left(\frac{-3}{5}\right)}{2}}=\pm \sqrt{\frac{\frac{8}{5}}{2}}=\pm \frac{2}{\sqrt{5}}$

Since $\sin x$ is positive in Il quadrant, hence $\sin \frac{x}{2}=\frac{2}{\sqrt{5}}$

ii) $\cos \frac{x}{2}$

Formula used:

$\cos \frac{x}{2}=\pm \sqrt{\frac{1+\cos x}{2}}$

now, $\cos \frac{x}{2}=\pm \sqrt{\frac{1+\left(\frac{-3}{5}\right)}{2}}=\pm \sqrt{\frac{\frac{2}{5}}{2}}=\pm \sqrt{\frac{1}{5}}$

since $\cos x$ is negative in $\|$ quadrant, hence $\cos \frac{x}{2}=-\frac{1}{\sqrt{5}}$

iii) $\tan \frac{x}{2}$

Formula used:

$\tan x=\frac{\sin x}{\cos x}$

hence, $\tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{8}{2}}=\frac{\frac{2}{\sqrt{5}}}{-\frac{1}{\sqrt{5}}}=\frac{2}{\sqrt{5}} \times \frac{\sqrt{5}}{-1}=-2$

Here, tanx is negative in II quadrant.


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