Question

find the value of cos(18 degree)

Answer 1

Let A = 18°           5A = 90°
2A = 90 - 3A
Sin 2A = Sin (90-3A) = Cos 3A

2 Sin A Cos A = 4Cos³ A - 3 Cos A ,  Cancel Cos A as it is not zero.

2 Sin A = 4 Cos² A - 3  = 1 - 4 Sin² A

4 Sin² A + 2 Sin A - 1 = 0 

Solving quadratic equation :   Sin A = (√5 - 1)/4
Sin² A = (1-2Sin A)/4 = (3-√5 )/ 8
Cos² A = (5+√5)/8 

$Cos\ A = \sqrt{\frac{5+\sqrt5}{8}}=\frac{1}{4}\sqrt{10+2\sqrt5}$

Answer 2

Answer:

Step-by-step explanation:

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 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 A - 4 (1 - sin2 A) + 3 = 0

⇒ 4 sin2 A + 2 sin A - 1 = 0, which is a quadratic in sin A

Therefore, sin A = −2±−4(4)(−1)√2(4)

⇒ sin A = −2±4+16√8

⇒ sin A = −2±25√8

⇒ sin A = −1±5√4

Now sin 18° is positive, as 18° lies in first quadrant.

Therefore, sin 18° = sin A = √5−14

Now cos 18° = √(1 - sin2 18°), [Taking positive value, cos 18° > 0]

⇒ cos 18° = 1−(5√−14)2−−−−−−−−−−√

⇒ cos 18° = 16−(5+1−25√)16−−−−−−−−−−√

⇒ cos 18° = 10+25√16−−−−−−√

Therefore, cos 18° =  10+25√√4