The value of \(\sin 18^{\circ}\) is

\(
\begin{aligned}
& \sin 90^{\circ}=\sin 5\left(81^{\circ}\right) \text { and let } 18^{\circ}=\mathrm{A} \\
& \sin 90^{\circ}=\sin 5 \mathrm{~A}=\sin (3 \mathrm{~A}+2 \mathrm{~A}) \\
& \therefore 90^{\circ}=3 \mathrm{~A}+2 \mathrm{~A} \Rightarrow \sin \left(90^{\circ}-3 \mathrm{~A}\right)=\sin 2 \mathrm{~A} \\
& \therefore \sin 2 \mathrm{~A}-\cos 3 \mathrm{~A} \Rightarrow 2 \sin \mathrm{A} \cos \mathrm{A}=4 \cos ^3 \mathrm{~A}-3 \cos \mathrm{A} \\
& \therefore \cos \mathrm{A}\left(2 \sin \mathrm{A}-4 \cos ^2 \mathrm{~A}+3\right)=0 \\
& \therefore \cos \mathrm{A}=0 \text { or }\left[2 \sin \mathrm{A}-4\left(1-\sin ^2 \mathrm{~A}\right)+3\right]=0 \\
& \therefore \mathrm{A}=\frac{\pi}{2} \text { or } 4 \sin ^2 \mathrm{~A}+2 \sin \mathrm{A}-1=0
\end{aligned}
\)
Since, \(A=18^{\circ}, A \neq \frac{\pi}{2}\)
\( \therefore 4 \sin ^2 \mathrm{~A}+2 \sin \mathrm{A}-1=0 \)
\( \therefore \sin \mathrm{A}=\frac{-2 \pm \sqrt{4+16}}{2(4)}=\frac{-2 \pm \sqrt{20}}{8}=\frac{2- \pm 2 \sqrt{5}}{8}=\) \(\frac{-1 \pm \sqrt{5}}{4} \)
\( \therefore \sin \mathrm{A}=\frac{-1+\sqrt{5}}{4} \text { or } \sin \mathrm{A}=\frac{-1+\sqrt{5}}{4}\)
Since, \(\sin \mathrm{A}>0, \sin \mathrm{A}=\frac{\sqrt{5}-1}{4}\)