Explanation

(b)
$\sin^{4}{\frac{\pi}{8} + \sin^{4}{\frac{3\pi}{8} + \sin^{4}{\frac{5\pi}{8} + \sin^{4}\frac{7\pi}{8}}}}$
$= \frac{1}{4}\left\lbrack \left( 2\sin^{2}\frac{\pi}{8} \right)^{2} + \left( 2\sin^{2}\frac{3\pi}{8} \right)^{2} \right\rbrack + \frac{1}{4}\left\lbrack \left( 2\sin^{2}\frac{\pi}{8} \right)^{2} + \left( 2\sin^{2}\frac{3\pi}{8} \right)^{2} \right\rbrack$
$= \frac{1}{4}\left\lbrack \left( 1 - \cos\frac{\pi}{4} \right)^{2} + \left( 1 - {\cos\ }\frac{3\pi}{4} \right)^{2} \right\rbrack + \frac{1}{4}\left\lbrack \left( 1 - {\cos\ }\frac{\pi}{4} \right)^{2} + \left( 1 - {\cos\ }\frac{3\pi}{4} \right)^{2} \right\rbrack$
$= \frac{1}{4}\left\lbrack \left( 1 - \frac{1}{\sqrt{2}} \right)^{2} + \left( 1 + \frac{1}{\sqrt{2}} \right)^{2} \right\rbrack + \frac{1}{4}\left\lbrack \left( 1 - \frac{1}{\sqrt{2}} \right)^{2} + \left( 1 + \frac{1}{\sqrt{2}} \right)^{2} \right\rbrack$
$= \frac{1}{4}(3) + \frac{1}{4}(3) = \frac{3}{2}$