Question
$\sin 36^{\circ}=\frac{\sqrt{10-2 \sqrt{5}}}{4}$

Answer

We know that, $\sin ^2 \theta=1-\cos ^2 \theta$
$\begin{aligned}
& \sin ^2 36^{\circ}=1-\cos ^2 36^{\circ} \\
& =1-\left(\frac{\sqrt{5}+1}{4}\right)^2 \\
& =\frac{16-(5+1+2 \sqrt{5})}{16} \\
& =\frac{10-2 \sqrt{5}}{16} \\
& \therefore \sin 36^{\circ}=\frac{\sqrt{10-2 \sqrt{5}}}{4} \ldots \ldots\left[\because \sin 36^{\circ} \text { is positive }\right]
\end{aligned}$

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