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