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  4. The value of sin4 (π/8) + sin4 (3 π/8) + sin4 (5π/8) + sin4 (7π/8) is

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Q. The value of $\sin^{4} \frac{\pi}{8} + \sin^{4} \frac{3 \pi}{8} + \sin^{4} \frac{5\pi}{8} + \sin^{4} \frac{7\pi}{8} $ is

UPSEEUPSEE 2018

Solution:

We have, $\frac{7\pi}{8}=\pi-\frac{\pi}{8}$
and $\frac{5\pi}{8}=\pi-\frac{3\pi}{8}$
$\Rightarrow \sin \frac{7\pi}{8}=\sin\left(\pi-\frac{\pi}{8}\right)$
and $\sin \frac{5\pi}{8}=\sin\left(\pi-\frac{3\pi}{8}\right)$
$\Rightarrow \sin \frac{7\pi}{8}=\sin \frac{\pi}{8}$ and $\sin \frac{5\pi}{8}=\sin \frac{3\pi}{8}$
$\Rightarrow \sin^{4}\left(\frac{7\pi}{8}\right)=\sin^{4}\left(\frac{\pi}{8}\right)$
and $\sin^{4}\left(\frac{5\pi}{8}\right)=\sin^{4}\left(\frac{3\pi}{8}\right)$
Now, $\sin^{4}\left(\frac{\pi}{8}\right)+\sin^{4}\left(\frac{3\pi}{8}\right)+\sin^{4}\left(\frac{5\pi}{8}\right)+\sin^{4}\left(\frac{7\pi}{8}\right)$
$=\sin^{4}\left(\frac{\pi}{8}\right)+\sin^{4}\left(\frac{3\pi}{8}\right)+\sin^{4}\left(\frac{3\pi}{8}\right)+\sin^{4}\left(\frac{\pi}{8}\right)$
$=2\sin^{4}\left(\frac{\pi}{8}\right)+2\sin^{4}\left(\frac{3\pi}{8}\right)$
$=2\left[\left(\sin^{2} \frac{\pi}{8}\right)^{2}+\left(\sin^{2}\left(\frac{3\pi}{8}\right)\right)^{2}\right]$
$=2\left[\left(\frac{1-\cos\,2\left(\frac{\pi}{8}\right)}{2}\right)^{2}+\left(\frac{1-\cos\,2\left(\frac{3\pi}{8}\right)}{2}\right)^{2}\right]$
$=2\left[\frac{\left(1-\cos \frac{\pi}{4}\right)^{2}}{4}+\frac{\left(1-\cos \frac{3\pi}{4}\right)^{2}}{4}\right]$
$=\frac{1}{2}\left[\left(1-\frac{1}{\sqrt{2}}\right)^{2}+\left(1+\frac{1}{\sqrt{2}}\right)^{2}\right]$
$\left[\because \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}} \,\text{and} \,\cos \frac{3\pi}{4}=\frac{-1}{\sqrt{2}}\right]$
$=\frac{1}{2}\left[1+\frac{1}{2}-\frac{2}{\sqrt{2}}+1+\frac{1}{2}+\frac{2}{\sqrt{2}}\right]$
$=\frac{1}{2}\left[2+\frac{2}{2}\right]=\frac{1}{2}\left[2+1\right]=\frac{3}{2}$