Question

Prove the following:

sin 18° = `(sqrt(5) - 1)/4`

Answer 1

Let θ = 18°

∴ 5θ = 90°

∴ 2θ + 3θ = 90°

∴ 2θ = 90° – 3θ

∴ sin 2θ = sin (90° – 3θ)

∴ sin 2θ = cos 3θ

∴ 2 sin θ cos θ = 4 cos³θ – 3 cos θ

∴ 2 sin θ = 4 cos²θ – 3  ...[∵ cos θ ≠ 0]

∴ 2 sin θ = 4 (1 – sin²θ) – 3

∴ 2 sin θ = 1 – 4 sin²θ

∴ 4 sin²θ + 2 sin θ – 1 = 0

∴ sin θ = `(-2 ± sqrt(4 + 16))/8`

= `(-2 ± 2sqrt(5))/8`

= `(-1 ± sqrt(5))/4`

Since, sin 18° > 0

∴ sin 18° = `(sqrt(5) - 1)/4`