Trigonometric Identities Question 320
Question: $ {{\sin }^{4}}\frac{\pi }{4}+{{\sin }^{4}}\frac{3\pi }{8}+{{\sin }^{4}}\frac{5\pi }{8}+{{\sin }^{4}}\frac{7\pi }{8}= $
[Roorkee 1980]
Options:
A) $ \frac{1}{2} $
B) $ \frac{1}{4} $
C) $ \frac{3}{2} $
D) $ \frac{3}{4} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ {{\sin }^{4}}\frac{\pi }{8}+{{\sin }^{4}}\frac{3\pi }{8}+{{\sin }^{4}}\frac{5\pi }{8}+{{\sin }^{4}}\frac{7\pi }{8} $ = $ \frac{1}{4},[ {{( 2{{\sin }^{2}}\frac{\pi }{8} )}^{2}}+{{( 2{{\sin }^{2}}\frac{3\pi }{8} )}^{2}} ] $ $ +\frac{1}{4},[ {{( 2{{\sin }^{2}}\frac{\pi }{8} )}^{2}}+{{( 2{{\sin }^{2}}\frac{3\pi }{8} )}^{2}} ] $ = $ \frac{1}{4},[ {{( 1-\cos \frac{\pi }{4} )}^{2}}+{{( 1-\cos \frac{3\pi }{4} )}^{2}} ] $ $ +\frac{1}{4},[ {{( 1-\cos \frac{\pi }{4} )}^{2}}+{{( 1-\cos \frac{3\pi }{4} )}^{2}} ] $ = $ \frac{1}{4},[ {{( 1-\frac{1}{\sqrt{2}} )}^{2}}+{{( 1+\frac{1}{\sqrt{2}} )}^{2}} ]+\frac{1}{4},[ {{( 1-\frac{1}{\sqrt{2}} )}^{2}}+{{( 1+\frac{1}{\sqrt{2}} )}^{2}} ] $ = $ \frac{1}{4}(3),+\frac{1}{4}(3)=\frac{3}{2} $ .
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