Question

value of 2 cos 18 degree​

Answer 1

Answer:

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

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

let 18° = A

5A = 90°

3A + 2A = 90°

2A = 90° - 3A

applying sin both side

sin2A = sin(90° - 3A)

sin2A = cos3A

2sinAcosA = 4cos^3A - 3cosA

2 sinAcosA = cosA(4cos^2A - 3)

2 sinA = 4cos^2A - 3

2sinA = 4(1-sin^2A) - 3

2sinA = 4-4sin^2A - 3

4sin^2A + 2sinA - 1 = 0

sinA = (-2 +_ √(2^2 - 4*4*(-1)))/2*4

sinA = (-2 +_ √(4+16))/8

sinA = (-2 +_ √20)/8

sinA = (-1 +√5)/4. (positive value only as A is 18°)

now

cosA = √(1-sin^2A

= √1- ((√5-1)/4)^2

= √1-(5+1-2√5)/16

= √16-6+2√5)/16

= √(10+2√5/16)

= (10+2√5)^(1/2)/4

2cosA = √10+2√5/2