What is the Value of Sin 18 Degrees ?

Solution :

The value of sin 18 degrees is $\sqrt{5} – 1\over 4$.

Proof : 

Let $\theta$ = 18 degrees. Then,

$5\theta$ = 90 degrees

$\implies$   $2\theta$ + $3\theta$ = 90

$\implies$   $2\theta$ = 90 – $3\theta$

$\implies$   $sin 2\theta$ = $sin (90 – 3\theta)$

$\implies$  $sin 2\theta$ = $cos 3\theta$

By using the formula of $sin 2\theta$ and $cos 3\theta$

$\implies$  $2 sin \theta cos \theta$ = $4 cos^3 \theta – 3 cos \theta$

$\implies$  $cos \theta$ ($2 sin \theta – 4 cos^2 \theta + 3$ = 0

$\implies$   $2 sin \theta$ – $4 cos^2 \theta$ + 3 = 0

[ $\because$  $cos \theta$ = cos 18 $\ne$ 0 ]

$\implies$  $2 sin \theta$ – $4(1 – sin^2 \theta)$ + 3 = 0

$\implies$   $4 sin^2 \theta$ + $2 sin \theta$ – 1 = 0

Solving this quadratic equation by using quadratic formula,

$\implies$  $sin \theta$ = $-2 \pm \sqrt{4 + 16}\over 8$

$\implies$  $sin \theta$ = $-1 \pm \sqrt{5}\over 4$

Since  $\theta$ lies in 1st quadrant  $\therefore$  $sin \theta$ > 0

$\implies$   $sin \theta$ = $-1 + \sqrt{5}\over 4$ = $\sqrt{5} – 1\over 4$

Hence, sin 18 degrees = $\sqrt{5} – 1\over 4$