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PROVE THAT :COS 2A – COS 2B + COS 2C = -1 -4 COSA COSB COSC
cosC = cos(180−(A+B)) ........ = −cos(A+B) ........ = −(cosA cosB − sinA sinB) ........ = sinA sinB − cosA cosB LHS = 2(cos²A + cos²B + cos²C) -3 ....... =2 [cos²A + cos²B + (sinA sinB − cosA cosB)²] -3 ....... = 2[cos²A + cos²B + sin²A sin²B − 2 sinA sinB cosA cosB + cos²A cos²B ] -3....... = 2 [cos²A + cos²B + (1−cos²A) (1−cos²B) − 2 sinA sinB cosA cosB + cos²A cos²B] -3....... = 2[cos²A + cos²B + 1 − cos²A − cos²B + cos²A cos²B − 2 sinA sinB cosA cosB + cos²A cos²B ] -3....... = 2 − 4 sinA sinB cosA cosB + 4cos²A cos²B -3 ....... = -1 − 4 cosA cosB (sinA sinB − cosA cosB) ....... = -1 − 4 cosA cosB cosC ....... = RHS RegardsArun (askIITians forum expert)
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