Question

derive the value of sin 18°

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Answer 1

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√8−2±√4+16/8

⇒ sin θ = −2±2√5/8−2±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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Answer 2

Answer:

Taking 18° as Ø 90° will be 5Ø. Then

3Ø+2Ø=90°=>3Ø=90°-20 taking cos function of both sides it will be cos3Ø=cos ( 90-20 )

=> cos3Ø=sin2Ø=>4cos3Ø-3cosØ=2sinØcosØ

=>4cos2Ø-3=2sinØ=>4-4sin2Ø-3=2sinØ

=> 4sin2Ø+2sinØ-1=0

Applying their respective identities we can find the value of x which is 5^0.5-1÷4.

Hope it's help you ☑