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