# 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

$$\implies$$  $$sin \theta$$ = $$-2 \pm \sqrt{4 + 16}\over 8$$
$$\implies$$  $$sin \theta$$ = $$-1 \pm \sqrt{5}\over 4$$
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 = $$\sqrt{5} – 1\over 4$$