Trigonometric-Identities Question 399
Question: The value of $ \sin \frac{\pi }{16}\sin \frac{3\pi }{16}\sin \frac{5\pi }{16}\sin \frac{7\pi }{16} $ is
[MP PET 2004]
Options:
A) $ \frac{1}{16} $
B) $ \frac{\sqrt{2}}{16} $
C) $ \frac{1}{8} $
D) $ \frac{\sqrt{2}}{8} $
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Answer:
Correct Answer: B
Solution:
We have $ \sin \frac{\pi }{16}.\sin \frac{3\pi }{16}.\sin \frac{5\pi }{16}.\sin \frac{7\pi }{16} $ $ =\frac{1}{4}[ 2\sin \frac{\pi }{16}\sin \frac{3\pi }{16}.2\sin \frac{5\pi }{16}\sin \frac{7\pi }{16} ] $ $ =\frac{1}{4}[ ( \cos \frac{\pi }{8}-\cos \frac{\pi }{4} )( \cos \frac{\pi }{8}-\cos \frac{3\pi }{4} ) ] $ $ =\frac{1}{4}[ ( \cos \frac{\pi }{8}-\frac{1}{\sqrt{2}} )( \cos \frac{\pi }{8}+\frac{1}{\sqrt{2}} ) ] $ $ =\frac{1}{4}[ ( {{\cos }^{2}}\frac{\pi }{8}-\frac{1}{2} ) ]=\frac{1}{8}[ 2{{\cos }^{2}}\frac{\pi }{8}-1 ] $ $ =\frac{1}{8}[ \cos \frac{\pi }{4} ]=\frac{1}{8}\times \frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{16} $ .
BETA
