Question

Find the value of sin 18°​

Answer 1

Step-by-step explanation:

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)−2±−4(4)(−1)2(4)

⇒ sin θ = −2±4+16√8−2±4+168

⇒ sin θ = −2±25√8−2±258

⇒ sin θ = −1±5√4−1±54

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

Therefore, sin 18° = sin A = −1±5√4

Answer 2

Answer:

hey mate here is your 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)−2±−4(4)(−1)2(4)

⇒ sin θ = −2±4+16√8−2±4+168

⇒ sin θ = −2±25√8−2±258

⇒ sin θ = −1±5√4−1±54

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

Therefore, sin 18° = sin A = −1±5√4