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# if A+B+C=pi then prove the following parts: 1)cos2A+cos2B+cos2C= -1-4cosAcosBcosC 2)sin2A+sin2B+sin2c=4cosAsinBcosC 3)SinA+sinB-SinC=4sin(A/2)sin(B/2)cos(C/2)

Asked by 29th September 2013, 12:34 AM
Answered by Expert
Answer:
Please ask one question at a time. The solution to your first question is as follows.

cos2A + cos2B + cos2C = 2cos(A+B).cos(A-B) + 2cos^2C -1
=2cos(180-C).cos(A-B) + 2cos^2C -1
= - 2cosC.cos(A-B) +2cos^2C -1
= -2cosC(cos(A-B) - cosC) -1
= -2cosC(cos(A-B) - cos(180-(A+B)) ) -1
= -2cosC(cos(A-B) + cos(A+B)) -1
= -2cosC( 2cosA.cosB) -1
= -4cosAcosBcosC -1
Answered by Expert 29th September 2013, 11:49 AM
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