Various methods to find value of $\sin 18^\circ$ To find value of $\sin 18^\circ$. Now my textbook gives a proof in which it takes $\theta=18^\circ$ and then multiply it out by 5 and write again as sum of $2\theta+3\theta$ and then taking sin on both sides forms a quadratic equation.
Are there any other elegant ways to find its value?
Thanks

$$S=\cos A+\cos3A=2\cos A\cos2A=\dfrac{(2\cos A\sin A)\cos2A}{\sin A}=\cdots=\dfrac{\sin4A}{2\sin A}$$
Now if $S=\dfrac12,\sin4A=\sin A$ with $\sin A\ne0$ 
either $4A=m360^\circ+A\iff A=120^\circ m$ with $3\nmid m$ as $\sin A\ne0$ $\implies A \equiv\pm120^\circ\pmod{360^\circ}$
or $4A=(2m+1)180^\circ-A\iff A=(2m+1)36^\circ$ with $m\not\equiv2\pmod5$ as $\sin A\ne0$
$\implies A \equiv\pm36^\circ,\pm108^\circ\pmod{360^\circ}$
All the four cases reduce to $$\cos36^\circ+\cos108^\circ=\dfrac12$$
Now use $\cos36^\circ=1-2\sin^218^\circ$
and $\cos108^\circ=\cos(90+18)^\circ=-\sin18^\circ$ to form a Quadratic Equation in $\sin18^\circ$ 
Use the fact that $\sin18^\circ>0$
See also : Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$
