Hi Abhinav,
Excellent question
What is value of sin 18
Let θ = 18°
5θ = 5 × 18° = 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 cos θ – 4 cos3 θ + 3 cos θ = 0
cos θ (2 sin θ – 4 cos2 θ + 3) = 0
2 sin θ – 4 cos2 θ + 3 = 0
2 sin θ – 4 (1 – sin2 θ) + 3 = 0
2 sin θ – 4 + 4sin2 θ + 3 = 0
4sin2 θ + 2 sin θ – 1 = 0
Let sin θ = x
4x2 + 2x – 1 = 0
x = (-b±√(b^2 - 4ac))/2a
x = (-2 ± √(2^2 - 4(4)(-1) ))/(2(4))
x = (-2 ± √(4 + 16 ))/8 = (-2 ± √20)/8 = (-2 ± √(4 × 5))/8 = (-2 ±2√5)/8
x = (-1 ± √5)/4
∴ sin θ = sin 18° = (-1 ± √5)/4
Written on March 18, 2017, 7:55 p.m.