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3.Trigonometrical Ratios, Functions and Identities
medium
$\sin {20^o}\,\sin {40^o}\,\sin {60^o}\,\sin {80^o} = $
A
$ - 3/16$
B
$5/16$
C
$3/16$
D
$ - 5/16$
Solution
(c) $\sin 20^\circ \sin {40^o}\sin 60^\circ \sin 80^\circ $
$ = \frac{1}{2}\sin 20^\circ \sin 60^\circ \,(2\sin {40^o}\sin 80^\circ )$
$ = \frac{1}{2}\sin 20^\circ \sin 60^\circ (\cos 40^\circ – \cos 120^\circ )$
$ = \frac{1}{2}.\frac{{\sqrt 3 }}{2}\sin 20^\circ \left( {1 – 2{{\sin }^2}20^\circ + \frac{1}{2}} \right)$
$ = \frac{{\sqrt 3 }}{4}\sin 20^\circ \left( {\frac{3}{2} – 2{{\sin }^2}20^\circ } \right)$
$ = \frac{{\sqrt 3 }}{8}(3\sin 20^\circ – 4{\sin ^3}20^\circ )$
$ = \frac{{\sqrt 3 }}{8}\sin 60^\circ = \frac{{\sqrt 3 }}{8}.\frac{{\sqrt 3 }}{2} = \frac{3}{{16}}$.
Standard 11
Mathematics
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