Question
${\sin ^4}\frac{\pi }{4} + {\sin ^4}\frac{{3\pi }}{8} + {\sin ^4}\frac{{5\pi }}{8} + {\sin ^4}\frac{{7\pi }}{8} = $
$= \frac{1}{4}\,\left[ {{{\left( {2{{\sin }^2}\frac{\pi }{8}} \right)}^2} + {{\left( {2{{\sin }^2}\frac{{3\pi }}{8}} \right)}^2}} \right]$
$ + \frac{1}{4}\,\left[ {{{\left( {2{{\sin }^2}\frac{\pi }{8}} \right)}^2} + {{\left( {2{{\sin }^2}\frac{{3\pi }}{8}} \right)}^2}} \right]$
= $\frac{1}{4}\,\left[ {{{\left( {1 - \cos \frac{\pi }{4}} \right)}^2} + {{\left( {1 - \cos \frac{{3\pi }}{4}} \right)}^2}} \right]$
$ + \frac{1}{4}\,\left[ {{{\left( {1 - \cos \frac{\pi }{4}} \right)}^2} + {{\left( {1 - \cos \frac{{3\pi }}{4}} \right)}^2}} \right]$
$= \frac{1}{4}\,\left[ {{{\left( {1 - \frac{1}{{\sqrt 2 }}} \right)}^2} + {{\left( {1 + \frac{1}{{\sqrt 2 }}} \right)}^2}} \right] + \frac{1}{4}\,\left[ {{{\left( {1 - \frac{1}{{\sqrt 2 }}} \right)}^2} + {{\left( {1 + \frac{1}{{\sqrt 2 }}} \right)}^2}} \right]$
$= \frac{1}{4}(3)\, + \frac{1}{4}(3) = \frac{3}{2}$.
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