Question

Show that $si{n}^4\frac{\pi }{8}+si{n}^4\frac{3\pi }{8}+si{n}^4\frac{5\pi }{8}+si{n}^4\frac{7\pi }{8}=\frac{3}{2}$

Answer 1

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

$={\left[{\sin }^2\frac{\pi }{8}\right]}^2+{\left[{\sin }^2\frac{3\pi }{8}\right]}^2+{\left[{\sin }^2\frac{5\pi }{8}\right]}^2+{\left[{\sin }^2\frac{7\pi }{8}\right]}^2$

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

$={\left[\frac{1-\frac{1}{\sqrt{2}}}{2}\right]}^2+{\left[\frac{1+\frac{1}{\sqrt{2}}}{2}\right]}^2+{\left[\frac{1+\frac{1}{\sqrt{2}}}{2}\right]}^2+{\left[\frac{1-\frac{1}{\sqrt{2}}}{2}\right]}^2$

$=\frac{1}{8}\left[\right(4-4\sqrt{2}+2)+(4+4\sqrt{2}+2\left)\right]$

$=\frac{12}{8}$

$=\frac{3}{2}$