Question

Solve The Following

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

Answer 1

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

$\Rightarrow {\left({\sin }^2\frac{\pi }{8}\right)}^2+{\left({\sin }^2\frac{4\pi }{8}\right)}^2+{\left({\sin }^2\frac{5\pi }{8}\right)}^2+{\left({\sin }^2\frac{7\pi }{8}\right)}^2$

$\because \cos 2\theta =1-2{\sin }^2\theta$

$\Rightarrow {\left(\frac{1-\cos \frac{\pi }{4}}{2}\right)}^2+{\left(\frac{1-\cos \pi }{2}\right)}^2+{\left(\frac{1-\cos \frac{5\pi }{4}}{2}\right)}^2+{\left(\frac{1-\cos \frac{7\pi }{4}}{2}\right)}^2$

Using $\cos \frac{5\pi }{4}=\cos (\pi +\frac{\pi }{4})=-\cos \frac{\pi }{4}$

$\cos \frac{7\pi }{4}=\cos (2\pi -\frac{\pi }{4})=+\cos \frac{\pi }{4}$

$\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}$

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

$\Rightarrow \frac{1}{4}[1+\frac{1}{2}-\frac{2}{\sqrt{2}}+1+1+\frac{1}{2}+\frac{2}{\sqrt{2}}+1-\frac{2}{\sqrt{2}}+\frac{1}{2}]$

$\Rightarrow \frac{1}{4}[4+\left(\frac{1}{2}\right)\times \left(3\right)-\frac{2}{\sqrt{2}}]$

$\Rightarrow \frac{1}{4}[4+\frac{3}{2}-\frac{2\sqrt{2}}{2}]$

$\Rightarrow \frac{1}{4}\left[\frac{8+3-2\sqrt{2}}{2}\right]$

$\Rightarrow \frac{11-2\sqrt{2}}{8}$