^c 8 =1 2 2-2 - Vidyadip

Question

$\sin \frac{\pi^c}{8}=\frac{1}{2} \sqrt{2-\sqrt{2}}$

✓

Answer

We know that $\cos \frac{\pi}{4}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}$
Also $\frac{\pi}{8}$ lies in the first quadrant, hence $\sin \frac{\pi}{8}$ is positive.
Now, $\cos 2 \theta=1-2 \sin ^2 \theta$
By putting $\theta=\frac{\pi}{8}$, we get,
$\begin{aligned}
& \cos \frac{\pi}{4}=1-2 \sin ^2 \frac{\pi}{8} \\
& \therefore 2 \sin ^2 \frac{\pi}{8}=1-\cos \frac{\pi}{4} \\
& =1-\frac{1}{\sqrt{2}} \\
& =\frac{\sqrt{2}-1}{\sqrt{2}} \\
& \therefore \sin ^2 \frac{\pi}{8}=\frac{\sqrt{2}-1}{2 \sqrt{2}} \\
& =\frac{\sqrt{2}(\sqrt{2}-1)}{4} \\
& \therefore \frac{2-\sqrt{2}}{4} \\
& \therefore \sin \frac{\pi}{8}=\frac{1}{\sqrt{2-\sqrt{2}}}
\end{aligned}$

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