Question

sin 18 ka man kitna hai​

Answer 1

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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

Using SHRI DHAR ACHRYA 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 = 0.30901699437

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Answer 2

Answer:

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)

⇒ sin θ = −2+√4+16/8

⇒ sin θ = −2+2√5/8

⇒ sin θ = −1+√5/4

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

Therefore, sin 18° = sin A = −1+√5/4

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