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3.Trigonometrical Ratios, Functions and Identities
easy
यदि $\sin \alpha = \frac{{ - 3}}{5},$ जहाँ $\pi < \alpha < \frac{{3\pi }}{2},$ तो $\cos \frac{1}{2}\alpha = $
A
$\frac{{ - 1}}{{\sqrt {10} }}$
B
$\frac{1}{{\sqrt {10} }}$
C
$\frac{3}{{\sqrt {10} }}$
D
$\frac{{ - 3}}{{\sqrt {10} }}$
Solution
$\cos (\alpha /2) = – \sqrt {\frac{{1 + \cos \alpha }}{2}} $
$\cos \alpha = – \sqrt {1 – {{\sin }^2}\alpha } $ [$\because \alpha$ तृतीय चतुर्थांश में है]
$ = – \sqrt {1 – \frac{9}{{25}}} = – \frac{4}{5}$
$\therefore \,\,\,\cos (\alpha /2) = – \sqrt {\frac{{1 – \frac{4}{5}}}{2}} = – \frac{1}{{\sqrt {10} }}$.
Standard 11
Mathematics
