If A + B + C = π, prove that

cos 2A – cos 2B – cos 2C = -1 + 4 cos A sin B sin C


= cos2A – (cos2B + cos2C)


Using formula




= cos2A - {2cos(B+C)cos(B-C)}


since A + B + C = π



= cos2A – {2cos(π – A)cos(B-C)}


And cos(π – A) = -cosA


= cos2A – {-2cosAcos(B-C)}


= cos2A + 2cosAcos(B-C)


Using cos2A = 2cos2A -1


= 2cos2A – 1 + 2cosAcos(B-C)


= 2cosA{cosA + cos(B-C)} – 1







= 2cosA{2sinCsinB} – 1


= 4cosAsinBsinC – 1


= R.H.S


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