prove that sin18=√5-1/4
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To 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 we know , sin(90 - A) = cos A
⇒ 2 sin A cos A = 4cos^3 A - 3 cos A using formulas for sin 2A and cos 3A
⇒ 2 sin A cos A - 4cos^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
Using SHRI DHAR ACHRYA formula
x = [-b ± √(b^2 - 4ac)] / 2a

18° lies in 1st quadrant,and sine is positive in 1st quadrant, so we take only positive value.
Therefore, sin 18° = sin A = −1+5–√4−1+54 = 0.30901699437
Thank you .
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 we know , sin(90 - A) = cos A
⇒ 2 sin A cos A = 4cos^3 A - 3 cos A using formulas for sin 2A and cos 3A
⇒ 2 sin A cos A - 4cos^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
Using SHRI DHAR ACHRYA formula
x = [-b ± √(b^2 - 4ac)] / 2a

18° lies in 1st quadrant,and sine is positive in 1st quadrant, so we take only positive value.
Therefore, sin 18° = sin A = −1+5–√4−1+54 = 0.30901699437
Thank you .
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19
Answer:
since a €q1 we take only positive value
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