Find the value of sin 18^° — JEE Mathematics

Let $\theta = 18^\circ \implies 5\theta = 90^\circ$.
We can write this as $2\theta + 3\theta = 90^\circ \implies 2\theta = 90^\circ - 3\theta$.
Take sine on both sides:

$$\sin 2\theta = \sin(90^\circ - 3\theta) = \cos 3\theta$$

Expand using double and triple angle formulas:

$$2\sin\theta\cos\theta = 4\cos^3\theta - 3\cos\theta$$

Since $\theta = 18^\circ$, $\cos\theta \ne 0$. Divide both sides by $\cos\theta$:

$$2\sin\theta = 4\cos^2\theta - 3$$

Replace $\cos^2\theta$ with $1 - \sin^2\theta$:

$$2\sin\theta = 4(1 - \sin^2\theta) - 3$$

$$2\sin\theta = 4 - 4\sin^2\theta - 3 \implies 4\sin^2\theta + 2\sin\theta - 1 = 0$$

Let $x = \sin\theta$:

$$4x^2 + 2x - 1 = 0$$

Using the quadratic formula:

$$x = \frac{-2 \pm \sqrt{4 - 4(4)(-1)}}{8} = \frac{-2 \pm \sqrt{20}}{8} = \frac{-2 \pm 2\sqrt{5}}{8} = \frac{-1 \pm \sqrt{5}}{4}$$

Since $18^\circ$ is in the first quadrant, $\sin 18^\circ > 0$. We choose the positive root:

$$\sin 18^\circ = \frac{\sqrt{5} - 1}{4}$$

Answer: $\frac{\sqrt{5} - 1}{4}$