Trigonometric Identities Question 200
Question: $ \sin 20^{o}\sin 40^{o}\sin 60^{o}\sin 80^{o}= $
[MNR 1976, 81]
Options:
A) $ -3/16 $
B) $ 5/16 $
C) $ 3/16 $
D) $ -5/16 $
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Answer:
Correct Answer: C
Solution:
$ \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 ( 1-2{{\sin }^{2}}20{}^\circ +\frac{1}{2} ) $
$ =\frac{\sqrt{3}}{4}\sin 20{}^\circ ( \frac{3}{2}-2{{\sin }^{2}}20{}^\circ ) $
$ =\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} $ .
BETA
