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प्रश्न
Prove the following:
sin 18° = `(sqrt(5) - 1)/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`
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