Question

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

Answer 1

${\sin }^4\frac{\pi }{8}+{\sin }^4\frac{3\pi }{8}+{\sin }^4\frac{5\pi }{8}+{\sin }^4\frac{7\pi }{8}$

$={\sin }^4\frac{\pi }{8}+{\sin }^4(\pi -\frac{\pi }{8})+{\sin }^4\frac{3\pi }{8}+{\sin }^4(\pi -\frac{3\pi }{8})$

$=2{\sin }^4\frac{\pi }{8}+2{\sin }^4\frac{3\pi }{8}......\left(\sin \right(\pi -\theta )=\sin \theta )$

$=2{\sin }^4\frac{\pi }{8}+2{\cos }^4\frac{\pi }{8}.....(\sin (\frac{\pi }{2}-\theta )=\cos \theta )$

$=2[{({\sin }^2\frac{\pi }{8}+{\cos }^2\frac{\pi }{8})}^2-2{\sin }^2\frac{\pi }{8}{\cos }^2\frac{\pi }{8}]$

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

$=2[1-\left(\frac{{\sin }^2\frac{\pi }{4}}{2}\right)]$

$=2[1-\frac{1}{2\times 2}]=2=\frac{1}{2}$

$=\frac{3}{2}=RHS$

Hence proved