Let θ = 18°, so that 5θ = 90°. Use 2θ = 90° - 3θ to show that sin 18° = (√5 - 1)/4, and give it to…

Step 1. An equation. $\sin 2\theta = \cos 3\theta$, so $2\sin\theta\cos\theta = 4\cos^3\theta - 3\cos\theta$.

Step 2. Divide by $\cos\theta$, which is not $0$: $2\sin\theta = 4\cos^2\theta - 3 = 4(1 - \sin^2\theta) - 3 = 1 - 4\sin^2\theta$.

Step 3. The quadratic. With $s = \sin\theta$: $4s^2 + 2s - 1 = 0$, so $s = \dfrac{-2 \pm \sqrt{4 + 16}}{8} = \dfrac{-1 \pm \sqrt{5}}{4}$.

Step 4. The sign. $18^\circ$ is acute, so $s > 0$: $\sin 18^\circ = \dfrac{\sqrt{5} - 1}{4} \approx 0.309$.