If A+B+C=π, prove that:
cos(B+2C)+cos(C+2A)+cos(A+2B)=1−4cosB−C2cosC−A2cosA−B2
My Attempt:
Here, A+B+C=π
Now,
LHS=cos(B+2C)+cos(C+2A)+cos(A+2B)=cos(B+C+C)+cos(C+A+A)+cos(A+B+B)=cos(π−(A−C))+cos(π−(B−A))+cos(π−(C−B))=−cos(A−C)−cos(B−A)−cos(C−B)
Please help to continue from here.
Answer
Let C−A=2x,B−C=2y,A−B=2z⟹2(x+y+z)=0
F=cos2x+cos2y+cos2z=2cos(x+y)cos(x−y)+2cos2z−1
Now as cos(x+y)=cos(−z)=cosz,
F=2coszcos(x−y)+2cosz⋅cos(x+y)−1
=2cosz{cos(x+y)+cos(x−y)}−1=2cosz{2cosxcosy}−1=?
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