Answers

2015-10-29T19:38:52+05:30
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 = 4cos^3 A - 3 cos A
⇒ 2 sin A cos A - 4cos^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)
⇒ sin θ = −2±4+168
⇒ sin θ = −2±25√8
⇒ sin θ = −1±5√4
Now sin 18° is positive, as 18° lies in first quadrant.
Therefore, sin 18° = sin A = −1±5√4
Hope this helps u frn

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