A) \[\frac{1}{2}\]
B) \[\frac{1}{4}\]
C) \[\frac{3}{2}\]
D) \[\frac{3}{4}\]
Correct Answer: C
Solution :
\[{{\sin }^{4}}\frac{\pi }{8}+{{\sin }^{4}}\frac{3\pi }{8}+{{\sin }^{4}}\frac{5\pi }{8}+{{\sin }^{4}}\frac{7\pi }{8}\] =\[\frac{1}{4}\,\left[ {{\left( 2{{\sin }^{2}}\frac{\pi }{8} \right)}^{2}}+{{\left( 2{{\sin }^{2}}\frac{3\pi }{8} \right)}^{2}} \right]\] \[+\frac{1}{4}\,\left[ {{\left( 2{{\sin }^{2}}\frac{\pi }{8} \right)}^{2}}+{{\left( 2{{\sin }^{2}}\frac{3\pi }{8} \right)}^{2}} \right]\] = \[\frac{1}{4}\,\left[ {{\left( 1-\cos \frac{\pi }{4} \right)}^{2}}+{{\left( 1-\cos \frac{3\pi }{4} \right)}^{2}} \right]\] \[+\frac{1}{4}\,\left[ {{\left( 1-\cos \frac{\pi }{4} \right)}^{2}}+{{\left( 1-\cos \frac{3\pi }{4} \right)}^{2}} \right]\] = \[\frac{1}{4}\,\left[ {{\left( 1-\frac{1}{\sqrt{2}} \right)}^{2}}+{{\left( 1+\frac{1}{\sqrt{2}} \right)}^{2}} \right]+\frac{1}{4}\,\left[ {{\left( 1-\frac{1}{\sqrt{2}} \right)}^{2}}+{{\left( 1+\frac{1}{\sqrt{2}} \right)}^{2}} \right]\] = \[\frac{1}{4}(3)\,+\frac{1}{4}(3)=\frac{3}{2}\].
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