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

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