Question

value of cos 18 degree and sin 18 degree

Answer 1

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

Answer 2

Answer:

Here is your 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 cos^3 A - 3 cos A

⇒ 2 sin A cos A - 4 cos^3 A + 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 A - 4 (1 - sin^2 A) + 3 = 0

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

Therefore, sin A = −2± root(-4(4)(−1)) / 2(4)

⇒ sin A = −2± root(4+16) / 8

⇒ sin A = −2± root(2 rt5 /8)

⇒ sin A = −1± rt 5 / 4

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

Therefore, sin 18° = sin A = (√5−1) /4

Similarly, cos 18° = root(10+2 rt5) / 4

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