Trigonometric Identities Question 101
Question: $ ( 1+\cos \frac{\pi }{8} ),( 1+\cos \frac{3\pi }{8} ),( 1+\cos \frac{5\pi }{8} ),( 1+\cos \frac{7\pi }{8} )= $
[IIT 1984; WB JEE 1992]
Options:
A) $ \frac{1}{2} $
B) $ \frac{1}{4} $
C) $ \frac{1}{8} $
D) $ \frac{1}{16} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ ( 1+\cos \frac{\pi }{8} ),( 1+\cos \frac{3\pi }{8} ),( 1+\cos \frac{5\pi }{8} ),( 1+\cos \frac{7\pi }{8} ) $ $ =( 1+\cos \frac{\pi }{8}+\cos \frac{7\pi }{8}+\cos \frac{\pi }{8}\cos \frac{7\pi }{8} ) $ $ ( 1+\cos \frac{5\pi }{8}+\cos \frac{3\pi }{8}+\cos \frac{3\pi }{8}\cos \frac{5\pi }{8} ) $ $ =( 1+\cos \frac{\pi }{8}-\cos \frac{\pi }{8}+\cos \frac{\pi }{8}\cos \frac{7\pi }{8} ) $ $ ( 1+\cos \frac{5\pi }{8}-\cos \frac{5\pi }{8}+\cos \frac{3\pi }{8}\cos \frac{5\pi }{8} ) $ $ =( 1+\cos \frac{\pi }{8}\cos \frac{7\pi }{8} ),( 1+\cos \frac{3\pi }{8}\cos \frac{5\pi }{8} ) $ $ =\frac{1}{4}( 2+2\cos \frac{\pi }{8}\cos \frac{7\pi }{8} )( 2+2\cos \frac{3\pi }{8}\cos \frac{5\pi }{8} ) $ $ =\frac{1}{4}( 2+\cos \frac{3\pi }{4}+\cos \pi )( 2+\cos \frac{\pi }{4}+\cos \pi ) $ $ =\frac{1}{4},( 1+\cos \frac{3\pi }{4} ),( 1+\cos \frac{\pi }{4} )=\frac{1}{4}( 1-\cos \frac{\pi }{4} ),( 1+\cos \frac{\pi }{4} ) $ $ =\frac{1}{4}( 1-{{\cos }^{2}}\frac{\pi }{4} )=\frac{1}{4}( 1-\frac{1}{2} )=\frac{1}{8} $ . Aliter: $ ( 1+\cos \frac{\pi }{8} ),( 1+\cos \frac{7\pi }{8} ),( 1+\cos \frac{3\pi }{8} ),( 1+\cos \frac{5\pi }{8} ) $ $ =( 1+\cos \frac{\pi }{8} ),( 1-\cos \frac{\pi }{8} ),( 1+\cos \frac{3\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 \frac{\pi }{8}.\sin \frac{3\pi }{8} )}^{2}} $ $ =\frac{1}{4}{{( \cos \frac{\pi }{4}-\cos \frac{\pi }{2} )}^{2}}=\frac{1}{8} $ .
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