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UPSEE 2018 Solved Paper

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Question : 101 of 150

Marks: +1, -0

Mathematics

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

is

$\frac{\sqrt{3}}{4}$
$\frac{3}{4}$
$\frac{\sqrt{3}}{2}$
$\frac{3}{2}$

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}\right]

+\left(\frac{1-\cos 2\left(\frac{3\pi}{8}\right)}{2}\right)^{2}

=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}}\right]

and

\cos\frac{3\pi}{4}=\frac{-1}{\sqrt{2}}

=\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}[2+1]=\frac{3}{2}

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