Question

If A + B + C = 180°, Prove that, sin²A + sin²B + sin²C = 2 + 2cosA cosB cosC.

Answer 1

Let S = sin²A + sin²B + sin²C 

so that 2S = 2sin²A + 1 – cos²B +1 – cos²C 

= 2 sin2A + 2 – 2cos(B + C) cos(B – C) 

= 2 – 2 cos2A + 2 – 2cos(B + C) cos(B – C) 

∴ S = 2 + cosA [cos(B – C) + cos(B+ C)] 

since cosA = – cos(B+C) 

∴ S = 2 + 2 cos A cos B cos C