Question

Prove that: sin18°=
$\frac{\sqrt{5} - 1}{4}$

Answer 1

Let A = 18°                          

Therefore, 5A = 90° 

⇒ 2A + 3A = 90˚

⇒ 2θ = 90˚ - 3A

Taking sine on both sides, we get 

sin 2A = sin (90˚ - 3A) = cos 3A 

⇒ 2 sin A cos A = 4 cos^3 A - 3 cos A

⇒ 2 sin A cos A - 4 cos^3A + 3 cos A = 0 

⇒ cos A (2 sin A - 4 cos^2 A + 3) = 0 

Dividing both sides by cos A = cos 18˚ ≠ 0, we get

⇒ 2 sin θ - 4 (1 - sin^2 A) + 3 = 0

⇒ 4 sin^2 A + 2 sin A - 1 = 0, which is a quadratic in sin A

Therefore, sin θ = −2±−4(4)(−1)√2(4)−2±−4(4)(−1)2(4)

⇒ sin θ = −2±4+16√8−2±4+168

⇒ sin θ = −2±25/√8−2±25/8

⇒ sin θ = −1±5/√4−1±5/4

Now sin 18° is positive, as 18° lies in first quadrant.

Therefore, sin 18° = sin A = −1±5/√4

Answer 2

TO PROVE -: sin18°=
$\frac{\sqrt{5} - 1}{4}$

PROOF-:

Let
$\theta = 18° \\ 5\theta = 90° \\ 2\theta + 3\theta = 90° \\ 2\theta = 90 - 3\theta \\ sin2\theta = sin(90° - 3\theta) \\ sin2\theta = cos3\theta \\ 2sin\theta \: cos\theta \: = 4cos {}^{3} \theta - 3cos\theta \\ cos\theta(2sin\theta - 4cos {}^{2} \theta + 3 )= 0 \\ 2sin\theta - 4cos {}^{2} \theta + 3 = 0.............[cos\theta = cos18° = 0] \\2 sin\theta - 4(1 - sin {}^{2} \theta) + 3 = 0 \\ 4sin {}^{2} \theta + 2sin\theta - 1 = 0 \\ \\ sin\theta = \frac{ - 2 ± \sqrt{4 + 16} }{8} \\ sin\theta \: = \frac{ - 1 ± \sqrt{5} }{4} \\ \\ sin\theta = \frac{ - 1 + \sqrt{5} }{4} = \frac{ \sqrt{5 } - 1}{4} ................[\theta \: lies \: in \: 1st \: quadrant sin\theta > 0] \\ \\ \\ Hence, \: sin 18° = \frac{ \sqrt{5} - 1 }{4}$