Prove the following:sin 18° = √5-1/4

1.	Prove the following:sin 18° = √5-1/4
Answer» 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 cos³ θ – 3 cos θ ∴ 2 sin θ = 4 cos² θ – 3 …..[∵ cos θ ≠ 0] ∴ 2 sin θ = 4 (1 – sin² θ) – 3 ∴ 2 sin θ = 1 – 4 sin² θ ∴ 4 sin² θ + 2 sin θ – 1 = 0 ∴ sin θ = \(\frac{-2±\sqrt{4+16}}{8}\) =\(\frac{-2±2\sqrt5}{8}\) =\(\frac{-1±\sqrt5}{4}\) Since, sin 18° > 0 ∴ sin 18°=\(\frac{\sqrt5-1}{4}\)

Answer»

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 cos³ θ – 3 cos θ

∴ 2 sin θ = 4 cos² θ – 3 …..[∵ cos θ ≠ 0]

∴ 2 sin θ = 4 (1 – sin² θ) – 3

∴ 2 sin θ = 1 – 4 sin² θ

∴ 4 sin² θ + 2 sin θ – 1 = 0

∴ sin θ = \(\frac{-2±\sqrt{4+16}}{8}\)

=\(\frac{-2±2\sqrt5}{8}\)

=\(\frac{-1±\sqrt5}{4}\)

Since, sin 18° > 0

∴ sin 18°=\(\frac{\sqrt5-1}{4}\)

Discussion