The value of $\sin ^{2}\left(\frac{\pi}{8}\right)=$
The value of $\sin ^{2}\left(\frac{\pi}{8}\right)=$
- $\frac{\sqrt{2}+1}{2 \sqrt{2}}$
- $\frac{\sqrt{5}+1}{2 \sqrt{2}}$
- $\frac{\sqrt{5}-1}{2 \sqrt{2}}$
- $\frac{\sqrt{2}-1}{2 \sqrt{2}}$
Solution
$\begin{aligned}
\sin ^{2}\left(\frac{\pi}{8}\right) &=\frac{1-\cos 2\left(\frac{\pi}{8}\right)}{2} \\
&=\frac{1-\cos \frac{\pi}{4}}{2}=\frac{1-\frac{1}{\sqrt{2}}}{2}=\frac{\sqrt{2}-1}{2 \sqrt{2}}
\end{aligned}$
Asked in: MHT CET 2020 (20 Oct Shift 2)
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