evaluate sin 18 degree
Answers
Answered by
3
Heya here's the answer,
sin 18° = 4.24
Hope answer helps you....✌✌✌✌
Please mark as brainliest dear....✌✌✌✌
:
:
:
Jiwanshi ❤❤❤
sin 18° = 4.24
Hope answer helps you....✌✌✌✌
Please mark as brainliest dear....✌✌✌✌
:
:
:
Jiwanshi ❤❤❤
Answered by
9
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+168
⇒ sin θ = −2±2√58
⇒ sin θ = −1±√54
Now sin 18° is positive, as 18° lies in first quadrant.
Therefore, sin 18° = sin A = −1±√54 or
it can be written as -0.751
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+168
⇒ sin θ = −2±2√58
⇒ sin θ = −1±√54
Now sin 18° is positive, as 18° lies in first quadrant.
Therefore, sin 18° = sin A = −1±√54 or
it can be written as -0.751
Similar questions