Trigonometric Identities Question 4
Question: What is the value of $ ( 1+\cos \frac{\pi }{8} )( 1+\cos \frac{3\pi }{8} )( 1+\cos \frac{5\pi }{8} )( 1+\cos \frac{7\pi }{8} )? $
Options:
A) $ \frac{1}{2} $
B) $ \frac{1}{2}+\frac{1}{2\sqrt{2}} $
C) $ \frac{1}{2}-\frac{1}{2\sqrt{2}} $
D) $ \frac{1}{8} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ [ 1+\cos \frac{\pi }{8} ][ 1+\cos \frac{3\pi }{8} ][ 1+\cos \frac{5\pi }{8} ][ 1+\frac{\cos 7\pi }{8} ] $ We have, $ \cos \frac{7\pi }{8}=\cos [ \pi -\frac{\pi }{8} ]=-\cos \frac{\pi }{8} $ and $ \cos \frac{5\pi }{8}=\cos [ \pi -\frac{3\pi }{8} ]=-\cos \frac{3\pi }{8} $
$ \therefore =[ 1+\cos \frac{\pi }{8} ][ 1+\cos \frac{3\pi }{8} ][ 1-\cos \frac{\pi }{8} ][ 1-\cos \frac{3\pi }{8} ] $ $ =[ 1-{{\cos }^{2}}\frac{\pi }{8} ][ 1-{{\cos }^{2}}\frac{3\pi }{8} ]={{\sin }^{2}}\frac{\pi }{8}.{{\sin }^{2}}\frac{3\pi }{8} $ $ =\frac{1}{4}[ 2{{\sin }^{2}}\frac{\pi }{8}.2{{\sin }^{2}}\frac{3\pi }{8} ] $ $ =\frac{1}{4}[ ( 1-\cos \frac{\pi }{4} )( 1-\cos \frac{3\pi }{4} ) ] $ $ ( \because 1-\cos \theta =2{{\sin }^{2}}\frac{\theta }{2} ) $ $ =\frac{1}{4}[ ( 1-\frac{1}{\sqrt{2}} )( 1+\frac{1}{\sqrt{2}} ) ]=\frac{1}{8} $
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