Another complicated Trigo Sum

I used the same logic as Humberto Bento but perhaps easier to understand.

From $\quad \cos {2\theta} = 1 - 2\sin^2 {\theta}\quad \Rightarrow \sin^2 {\theta} = \frac {1} {2} (1 - \cos {2\theta})$ $\Rightarrow sin^4 {\theta} = \frac {1} {4} (1 - \cos {2\theta})^2$ .

Therefore,

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

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

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

$=\frac {1}{2} \left[ (1- \frac {1}{\sqrt{2}})^2 + (1+ \frac {1}{\sqrt{2}})^2 \right] =\frac {1}{2} \left[ 1- \sqrt{2} + \frac {1}{2} + 1 + \sqrt{2} + \frac{1}{2} \right]$ $= \frac {1}{2} (2+1) = \frac{3}{2} = \boxed{1.5}$

A simple method.

Knowing that: ${{\sin }^{2}}\alpha =\frac{1-\cos (2\alpha )}{2}$

We get: ${{\sin }^{4}}\alpha ={{\left( \frac{1-\cos (2\alpha )}{2} \right)}^{2}}=\frac{1-2\cos (2\alpha )+{{\cos }^{2}}(2a)}{4}=\frac{3-4\cos (2\alpha )+\cos (4a)}{8}$ with:

$\alpha =\frac{(2r-1)\pi }{8}$ , we get: $2\alpha =\frac{(2r-1)\pi }{4}$ $4\alpha =\frac{(2r-1)\pi }{2}$

As r takes the values 1, 2, 3 and 4, $cos(4\alpha) = 0$ and $cos(2\alpha)$ for r=1 and r=2 are simetric (as for r=3 and 4). Therefore the sum equals $4 \times \frac{3 }{8} = 1.5$