Question

Proof of sin18°..................

Answer 1

hope it's help u dear........

Answer 2

Hi praneeth,

Here is your answer!!!

WE CAN INITIALLY TAKE "A"

Let A = 18°

Therefore, 5A = 90°

⇒ 2A + 3A = 90˚

⇒ 2A= 90˚ - 3A

Taking sine on both sides, we get

sin 2A = sin (90˚ - 3A) = cos 3A[ we know , sin(90 - A) = cos A]

⇒ 2 sin A cos A = 4cos^3 A - 3 cos A

 [using formulas for sin 2A and cos 3A]

⇒ 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

BY USING QUADRATIC FORMULA,

x = [-b ± √(b^2 - 4ac)] / 2a

18° lies in 1st quadrant,and sine is positive in 1st quadrant, so we take only positive value.

Therefore, sin 18° = sin A = −1+5√4−1+54 = 0.30901699437.

calculation is on above!!

Hope I helped you a little!!!!