Question

Prove: ${\sin }^4\frac{\pi }{8}+{\sin }^{4}3\frac{\pi }{8}+{\sin }^{4}5\frac{\pi }{8}+{\sin }^{4}7\frac{\pi }{8}=\frac{3}{2}$

Answer 1

$\begin{array}{c}{\sin }^4\frac{\pi }{8}+{\sin }^{4}3\frac{\pi }{8}+{\sin }^{4}5\frac{\pi }{8}+{\sin }^{4}7\frac{\pi }{8}\\ \sin 5\frac{\pi }{8}=\cos \frac{\pi }{2}-\frac{5\pi }{8}=\cos \frac{3\pi }{8}\\ similarly,\\ \sin \frac{7\pi }{8}=\cos \frac{\pi }{8}\\ {\sin }^4\frac{\pi }{8}+{\cos }^4\frac{\pi }{8}+{\sin }^{4}3\frac{\pi }{8}+{\cos }^{4}3\frac{\pi }{8}\\ ={\left({\sin }^2\frac{\pi }{8}+{\cos }^2\frac{\pi }{8}\right)}^2+{\left({\sin }^{2}3\frac{\pi }{8}+{\cos }^{2}3\frac{\pi }{8}\right)}^2-2{\sin }^2\frac{\pi }{8}{\cos }^2\frac{\pi }{8}-{\sin }^{2}3\frac{\pi }{8}{\cos }^{2}3\frac{\pi }{8}\\ {\sin }^2\frac{\pi }{8}+{\cos }^2\frac{\pi }{8}=\frac{1}{2}.4{\sin }^2\frac{\pi }{8}co{s}^2\frac{\pi }{8}\\ =\frac{1}{2}{\left(2\sin \frac{\pi }{8}cos\frac{\pi }{8}\right)}^2=\frac{1}{2}{\sin }^2\frac{\pi }{4}\\ samefor2{\sin }^{2}3\frac{\pi }{8}{\cos }^{2}3\frac{\pi }{8}\\ So,\\ {\left({\sin }^2\frac{\pi }{8}+{\cos }^2\frac{\pi }{8}\right)}^2+{\left({\sin }^{2}3\frac{\pi }{8}+{\cos }^{2}3\frac{\pi }{8}\right)}^2-2{\sin }^2\frac{\pi }{8}{\cos }^2\frac{\pi }{8}-{\sin }^{2}3\frac{\pi }{8}{\cos }^{2}3\frac{\pi }{8}\\ =1+1-\frac{1}{2}{\sin }^2\frac{\pi }{4}-\frac{1}{2}{\sin }^{2}3\frac{\pi }{4}\\ =2-\frac{1}{2}\left(\frac{1}{2}+\frac{1}{2}\right)=2-\frac{1}{2}=\frac{3}{2}\end{array}$