Question

(iv)
sin 162°+ cos 153°
cos 72°- cos 27°​

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

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

Therefore, sin 18° = sin A = √5−14

Now, cos 72° = cos (90° - 18°) = sin 18° = √5−14

I hope it will help you