• # question_answer ${{\sin }^{4}}\frac{\pi }{4}+{{\sin }^{4}}\frac{3\pi }{8}+{{\sin }^{4}}\frac{5\pi }{8}+{{\sin }^{4}}\frac{7\pi }{8}=$ [Roorkee 1980] A) $\frac{1}{2}$ B) $\frac{1}{4}$ C) $\frac{3}{2}$ D) $\frac{3}{4}$

${{\sin }^{4}}\frac{\pi }{8}+{{\sin }^{4}}\frac{3\pi }{8}+{{\sin }^{4}}\frac{5\pi }{8}+{{\sin }^{4}}\frac{7\pi }{8}$ =$\frac{1}{4}\,\left[ {{\left( 2{{\sin }^{2}}\frac{\pi }{8} \right)}^{2}}+{{\left( 2{{\sin }^{2}}\frac{3\pi }{8} \right)}^{2}} \right]$ $+\frac{1}{4}\,\left[ {{\left( 2{{\sin }^{2}}\frac{\pi }{8} \right)}^{2}}+{{\left( 2{{\sin }^{2}}\frac{3\pi }{8} \right)}^{2}} \right]$ = $\frac{1}{4}\,\left[ {{\left( 1-\cos \frac{\pi }{4} \right)}^{2}}+{{\left( 1-\cos \frac{3\pi }{4} \right)}^{2}} \right]$ $+\frac{1}{4}\,\left[ {{\left( 1-\cos \frac{\pi }{4} \right)}^{2}}+{{\left( 1-\cos \frac{3\pi }{4} \right)}^{2}} \right]$ = $\frac{1}{4}\,\left[ {{\left( 1-\frac{1}{\sqrt{2}} \right)}^{2}}+{{\left( 1+\frac{1}{\sqrt{2}} \right)}^{2}} \right]+\frac{1}{4}\,\left[ {{\left( 1-\frac{1}{\sqrt{2}} \right)}^{2}}+{{\left( 1+\frac{1}{\sqrt{2}} \right)}^{2}} \right]$ = $\frac{1}{4}(3)\,+\frac{1}{4}(3)=\frac{3}{2}$.