Question

Prove that cos2A – cos2B + cos2C = 1 – 4 sinA · cosB · sinC

Answer 1

L.H.S. = cos 2A – cos2B + cos2C 

= -2 sin(A + B) . sin(A – B) + 1 – 2sin²C 

= 1 – 2 sinC sin (A – B) – 2 sin2C [∵ sin(A+B)-sinc] 

= 1 – 2 sinC [sin(A – B) + sinC] 

= 1 – 2 sinC [sin(A + B) + sin(A – B)] 

= 1 – 2 sinC [2sinA . cosB] 

= 1 – 4 sinA cosB sinC = R.H.S.

Comments

can u pls tell me how did u get the first step pls?

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sin(A + B).sin(A - B) = (sin A cos B + cos A sin B).(sin A cos B - cos A sin B)
= sin^2 A cos^2 B - cos^2 A sin^2 B
= sin^2 A(1 - sin^2 B) - (1 - sin^2 A)sin^2 B
= sin^2 A - sin^2 A sin^2 B - sin^2 B + sin^2 A sin^2 B
= sin^2 A - sin^2 B
= (1/2)(-1 + 2 sin^2 A + 1 - 2 sin^2 B)
= (1/2)(cos 2B - cos 2A)
⇒ cos 2A - cos 2B = -2 sin(A + B).sin(A - B)