2016/089) Evaluate $\sin\,18^\circ$ and $\cos\,18^\circ$

Let, $A = 18^\circ$
Then $2A = 90^\circ - 3A$
Taking sine on both sides, we get
$\sin 2A = \sin (90^\circ - 3A) = \cos 3A$
$=> 2 \sin\, A \cos\, A = 4 \cos^3 A - 3 \cos\, A$
or $2 \sin\, A \cos\, A - 4 \cos^3 A + 3 \cos\, A = 0$
or $\cos\, A (2 \sin\, A - 4 \cos^2 A + 3) = 0$
Dividing both sides by $\cos\, A = \cos 18^\circ$ which is not zero we get
$2 \sin\, A - 4 (1 - \sin^2 A) + 3 = 0$
or $4 \sin^2 A + 2 \sin\ A - 1 = 0$ which is a quadratic in $\sin\ A$
hence $\sin\,A = \frac{-1\pm\sqrt{5}}{4}$
but as $\sin\, 18^\circ$ is positive we have $\sin 18^\circ = \frac{-1+\sqrt{5}}{4}$
now $\cos^2 18^\circ= 1- \sin ^2 18^\circ = 1 - (\frac{-1+\sqrt{5}}{4})^2$
$= 1 - \frac{5+1-2\sqrt{5}}{16} = \frac{10+2\sqrt{5}}{16}$
 $\cos\,18^\circ= \frac{\sqrt{10+2\sqrt{5}}}{4}$