how to find the value of cos36?
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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)244124
⇒ sin θ = −2±4+16√824168
⇒ sin θ = −2±25√82258
⇒ sin θ = −1±5√4154
Now sin 18° is positive, as 18° lies in first quadrant.
Therefore, sin 18° = sin A = −1±5√4154
Now, cos 36° = cos 2 ∙ 18°
⇒ cos 36° = 1 - 2 sin22 18°
⇒ cos 36° = 1 - 2(5√−14)25142
⇒ cos 36° = 16−2(5+1−25√)16162512516
⇒ cos 36° = 1+45√1614516
⇒ cos 36° = 5√+14514
Therefore, cos 36° = 5√+14514
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)244124
⇒ sin θ = −2±4+16√824168
⇒ sin θ = −2±25√82258
⇒ sin θ = −1±5√4154
Now sin 18° is positive, as 18° lies in first quadrant.
Therefore, sin 18° = sin A = −1±5√4154
Now, cos 36° = cos 2 ∙ 18°
⇒ cos 36° = 1 - 2 sin22 18°
⇒ cos 36° = 1 - 2(5√−14)25142
⇒ cos 36° = 16−2(5+1−25√)16162512516
⇒ cos 36° = 1+45√1614516
⇒ cos 36° = 5√+14514
Therefore, cos 36° = 5√+14514
Jaichawla:
no the value of cos 36 is underoot 5+1/4
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i think you have to learn the value of cos18 cos36 sin18 sin36.....the value of cos 36 is underroot 5+1/4
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