Question

Prove the following:

sin 36° = `(sqrt(10 - 2sqrt(5)))/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`

∴ cos²18° = 1 – sin²18°

= `1 - ((sqrt(5) - 1)/4)^2`

= `1 - ((5 - 2sqrt(5) + 1)/16)`

= `(16 - 5 + 2sqrt(5) - 1)/16`

= `(10 + 2sqrt(5))/16`

∴ cos 18° = `sqrt(10 + 2sqrt(5))/4` ...[∵ 18° is an acute angle]

∴ sin 36° = 2 sin 18° · cos 18°

= `2 xx (sqrt(5) - 1)/4 xx sqrt(10 + 2sqrt(5))/4`

= `(sqrt((sqrt(5) - 1)^2) xx sqrt(10 + 2sqrt(5)))/8`

= `sqrt((6 - 2sqrt(5))(10 + 2sqrt(5)))/8`

= `(sqrt(60 + 12sqrt(5) - 20sqrt(5) - 20))/8`

= `sqrt(40 - 8sqrt(5))/8`

= `(2sqrt(10 - 2sqrt(5)))/8`

∴ sin 36° = `(sqrt(10 - 2sqrt(5)))/4`