solve it find the value of sin 36
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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 cos33 A - 3 cos A
⇒ 2 sin A cos A - 4 cos33 A + 3 cos A = 0
⇒ cos A (2 sin A - 4 cos22 A + 3) = 0
Dividing both sides by cos A = cos 18˚ ≠ 0, we get
⇒ 2 sin θ - 4 (1 - sin22 A) + 3 = 0
⇒ 4 sin22 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√4−1±54
Now, cos 36° = cos 2 ∙ 18°
⇒ cos 36° = 1 - 2 sin22 18°
⇒ cos 36° = 1 - 2(5√−14)2(5−14)2
⇒ cos 36° = 16−2(5+1−25√)1616−2(5+1−25)16
⇒ cos 36° = 1+45√161+4516
⇒ cos 36° = 5√+145+14
Therefore, sin 36° = 1−cos236°−−−−−−−−−√1−cos236°,[Taking sin 36° is positive, as 36° lies in first quadrant, sin 36° > 0]
⇒ sin 36° = 1−(5√+14)2−−−−−−−−−−√1−(5+14)2
⇒ sin 36° = 16−(5+1+25√)16−−−−−−−−−−√16−(5+1+25)16
⇒ sin 36° = 10−25√16−−−−−−√10−2516
⇒ sin 36° = 10−25√√410−254
Therefore, sin 36° = 10−25√√4
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 cos33 A - 3 cos A
⇒ 2 sin A cos A - 4 cos33 A + 3 cos A = 0
⇒ cos A (2 sin A - 4 cos22 A + 3) = 0
Dividing both sides by cos A = cos 18˚ ≠ 0, we get
⇒ 2 sin θ - 4 (1 - sin22 A) + 3 = 0
⇒ 4 sin22 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√4−1±54
Now, cos 36° = cos 2 ∙ 18°
⇒ cos 36° = 1 - 2 sin22 18°
⇒ cos 36° = 1 - 2(5√−14)2(5−14)2
⇒ cos 36° = 16−2(5+1−25√)1616−2(5+1−25)16
⇒ cos 36° = 1+45√161+4516
⇒ cos 36° = 5√+145+14
Therefore, sin 36° = 1−cos236°−−−−−−−−−√1−cos236°,[Taking sin 36° is positive, as 36° lies in first quadrant, sin 36° > 0]
⇒ sin 36° = 1−(5√+14)2−−−−−−−−−−√1−(5+14)2
⇒ sin 36° = 16−(5+1+25√)16−−−−−−−−−−√16−(5+1+25)16
⇒ sin 36° = 10−25√16−−−−−−√10−2516
⇒ sin 36° = 10−25√√410−254
Therefore, sin 36° = 10−25√√4
damini24:
thanku
Answered by
5
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 cos3 A - 3 cos A
⇒ 2 sin A cos A - 4 cos3 A + 3 cos A = 0
⇒ cos A (2 sin A - 4 cos2 A + 3) = 0
Dividing both sides by cos A = cos 18˚ ≠ 0, we get
⇒ 2 sin θ - 4 (1 - sin2 A) + 3 = 0
⇒ 4 sin2 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
Now, cos 36° = cos 2 ∙ 18°
⇒ cos 36° = 1 - 2 sin2 18°
⇒ cos 36° = 1 - 2(5√−14)2
⇒ cos 36° = 16−2(5+1−25√)16
⇒ cos 36° = 1+45√16
⇒ cos 36° = 5√+14
Therefore, cos 36° = 5√+14
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