Question

Find value of sin 18degree​

Answer 1

Answer:

$sin18 \degree = \frac{ \sqrt{5} - 1 }{4}$

PROOF:-)

Let x = 18° .

then 5 x = 5 x 18°=90° .

⟹ 2x+3x =90°.

⟹2x=90-3x.

Taking Sin of both side

Sin2x=Sin(90-3x).

applying the formula of Sin 2x and sin (90-x) we get:

$2sinx.cosx= 4 {cos}^{3} x - 3cosx \\ \implies2sinx = 4 {cos}^{2} x - 3 \\ \implies \: 2sinx = 4(1 - {sin}^{2} x) - 3$

$\implies 4 { \sin}^{2} x + 2sinx - 1 = 0$

The above equation is equation in sin x

So,

$sinx = \frac{- 2 \pm \sqrt{ {2 - 4 \times 4 \times ( - 1)}^{2} } }{8} \\ = \frac{- 2 \pm \sqrt{20} }{8} \\ = \frac{ - 1 \pm \sqrt{5} }{4}$

As we have taken x = 18° which implies that x lies in first quadrant sin x must be positive.

So,

$sin18 \degree = \frac{ \sqrt{5} - 1}{4} \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \boxed{ Hence Proved }}$

Answer 2

Answer:

Hy mere

here is your answer

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)

⇒ sin θ = −2±4+16√8

⇒ sin θ = −2±25√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