Question

If A + B + C = 2π, then prove that cos² B + cos² C – sin² A = 2 cos A cos B cos C.

Answer 1

We know that:

cos² x = \(\frac{1+cos2x}{2}, \) sin²x = \(\frac{1-cos2x}{2}\)

L.H.S.: cos² B + cos² C - sin² A

= \(\frac{1+cos^2B}{2} + \frac{1+cos^2C}{2} - \frac{1+cos^2A}{2}\)

= \(\frac{1}{2}\) [1 + cos 2B + cos 2C + cos 2A]

= \(\frac{1}{2}\) [2cos²B + 2cos(A + B) cos (A - C)]

= [cos²B + cos (2π + B) cos (A – C)]

= cos²B + cos B cos (A – C)

= cos B (cos B + cos (A – C)

= cos B (cos (2π – (A – C)) + cos (A – C) – cos B) [cos (A – B) + cos (A + C)]

= cos B [cos (A + B) + cos (A – C)]

= cos B [2 cos A cos C]

= 2 cos A cos B cos C