show that cos 54 / cos 18 = 2 sin 18
Answers
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