Math, asked by raeess, 11 months ago

the value of sing 18or 70 to is

Answers

Answered by shivam10sep
2


Originally Answered: How do I find the value of sin 18?

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 )
⇒ 2 sin A cos A = 4 (cos A)^3 - 3 cos A
⇒ 2 sin A cos A - 4 (cos A)^3 + 3 cos A = 0
⇒ cos A (2 sin A - 4( cos A)^2 + 3) = 0 
Dividing both sides by cos A = cos 18˚ ≠ 0, we get
⇒ 2 sin A - 4 (1 - (sin A)^2) + 3 = 0
⇒ 4 (sin A)^2 + 2 sin A - 1 = 0, which is a quadratic in sin A
Solving the above quadratic equation roots of quadratic equation is [-b +- sqrt(b^2 - 4*a*c)]/(2*a),
Therefore, sin A = (-1 +- sqrt(5))/4
Ignoring negative solution,
Sin A = [sqrt(5) - 1]/4 = 0.30901699437 (approximately).



 Originally Answered: How do I find the value of sin 18?

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 )
⇒ 2 sin A cos A = 4 (cos A)^3 - 3 cos A
⇒ 2 sin A cos A - 4 (cos A)^3 + 3 cos A = 0
⇒ cos A (2 sin A - 4( cos A)^2 + 3) = 0 
Dividing both sides by cos A = cos 18˚ ≠ 0, we get
⇒ 2 sin A - 4 (1 - (sin A)^2) + 3 = 0
⇒ 4 (sin A)^2 + 2 sin A - 1 = 0, which is a quadratic in sin A
Solving the above quadratic equation roots of quadratic equation is [-b +- sqrt(b^2 - 4*a*c)]/(2*a),
Therefore, sin A = (-1 +- sqrt(5))/4
Ignoring negative solution,
Sin A = [sqrt(5) - 1]/4 = 0.30901699437 (approximately).

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