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प्रश्न
Prove the following:
sin 36° = `(sqrt(10 - 2sqrt(5)))/4`
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उत्तर
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`
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