Question

$ {{\cos }^{4}}\frac{\pi }{8}+{{\cos }^{4}}\frac{3\pi }{8}+{{\cos }^{4}}\frac{5\pi }{8}+{{\cos }^{4}}\frac{7\pi }{8} $ is equal to

• A. $ \frac{3}{2} $
• B. $ -\frac{2}{3} $
• C. $ -1 $
• D. $ 1 $

Answer 1

Answer: (A)

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