Trigonometric Identities Question 335
Question: $ {{\cos }^{4}}\frac{\pi }{8}+{{\cos }^{4}}\frac{3\pi }{8}+{{\cos }^{4}}\frac{5\pi }{8}+{{\cos }^{4}}\frac{7\pi }{8}= $
Options:
A) $ \frac{1}{2} $
B) $ \frac{1}{4} $
C) $ \frac{3}{2} $
D) $ \frac{3}{4} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ {{\cos }^{4}}\frac{\pi }{8}+{{\cos }^{4}}\frac{3\pi }{8}+{{\cos }^{4}}\frac{5\pi }{8}+{{\cos }^{4}}\frac{7\pi }{8} $ $ ={{\cos }^{4}}\frac{\pi }{8}+{{\cos }^{4}}\frac{3\pi }{8}+{{\cos }^{4}}\frac{3\pi }{8}+{{\cos }^{4}}\frac{\pi }{8} $ $ =2( {{\cos }^{4}}\frac{\pi }{8}+{{\cos }^{4}}\frac{3\pi }{8} ) $ $ =2[ {{( {{\cos }^{2}}\frac{\pi }{8}+{{\cos }^{2}}\frac{3\pi }{8} )}^{2}}-2{{\cos }^{2}}\frac{\pi }{8}{{\cos }^{2}}\frac{3\pi }{8} ] $ $ =2[ 1-\frac{1}{2}( 2{{\cos }^{2}}\frac{\pi }{8} ),( 2{{\cos }^{2}}\frac{3\pi }{8} ) ] $ $ =2-( 1+\cos \frac{\pi }{4} ),( 1+\cos \frac{3\pi }{4} ) $ $ =2-( 1+\cos \frac{\pi }{4} ),( 1-\cos \frac{\pi }{4} ) $ $ =2-( 1-{{\cos }^{2}}\frac{\pi }{4} )=2-( 1-\frac{1}{2} )=2-\frac{1}{2}=\frac{3}{2} $ .
BETA
