Answer to Question #260340 in Trigonometry for prettygirl

Question

Prove that cos 3theta - cos theta/sin 3 theta sin theta = 4 cos theta/ 1- 4 cos^2 theta

Answer 1

let theta = k

We need to prove,

(cos 3k - cos k)/(sin 3k*sin k) = 4 cos k/(1- 4 cos² k)

We know that,

cos² k + sin²k = 1

cos 3k = 4 cos³ k - 3 cos k

sin3k = 3 sin k - 4 sin³ k

taking the Left Hand side,

(cos 3k - cos k)/(sin 3k*sin k) = (4 cos ³ k - 3 cos k - cos k) / (3 sin k - 4 sin³ k)(sin k)

= (4cos ³ k - 4 cos k)/(3 - 4 sin² k)(sin ² k)

=4 cos k(cos² k - 1) / (3 - 4 sin²k)(sin ² k)

=4 cos k(-sin² k) / (3 - 4 sin² k)(sin² k)

=4 cos k / (4 sin² k - 3)

= 4 cos k / (4(1 - cos² k) - 3)

= 4 cos k / (4 - 4 cos² k - 3)

= 4 cos k / (1 - 4 cos² k)

= Right Hand Side

Hence proved