જો sin alpha = frac{{ - 3}}{5}, કે જ્યાં pi & | vedclass

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Question

(a) $\cos (\alpha /2) = - \sqrt {\frac{{1 + \cos \alpha }}{2}} $

$\cos \alpha = - \sqrt {1 - {{\sin }^2}\alpha } $     [$\because&nb

Vedclass · Accepted Answer

$\frac{{ - 1}}{{\sqrt {10} }}$

Vedclass · Answer

(a) $\cos (\alpha /2) = - \sqrt {\frac{{1 + \cos \alpha }}{2}} $

$\cos \alpha = - \sqrt {1 - {{\sin }^2}\alpha } $     [$\because  \alpha$ lies in $III^{rd}$ Quadrant]

$ = - \sqrt {1 - \frac{9}{{25}}} = - \frac{4}{5}$

$	herefore \,\,\,\cos (\alpha /2) = - \sqrt {\frac{{1 - \frac{4}{5}}}{2}} = - \frac{1}{{\sqrt {10} }}$.

જો $\sin \alpha = \frac{{ - 3}}{5},$ કે જ્યાં $\pi < \alpha < \frac{{3\pi }}{2},$ તો $\cos \frac{1}{2}\alpha = $

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