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Prove that cos2A + cos2B + cos2C = -1 – 4 cosA cosB cosC

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Prove that cos2A + cos2B + cos2C = -1 – 4 cosA cosB cosC

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L.H.S. = cos2A + cos2B + cos2C 

= 2cos(A + B) · cos(A – B) + 2cos2C – 1 

= -2cosC . cos(A – B) + 2cos2C – 1 

= -2cosC[cos (A – B) – cos C] – 1 

= -2cosC{cos (A – B) + cos(A + B)} – 1 

= -2 cosC (2cosA · cosB) – 1 

= – 1 – 4 cos A cosB cosC = R.H.S.

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