If A+B+C=π, prove that
cos2A+cos2B−cos2C=−2cosA⋅cosB⋅cosC.
ATTEMPT:
Given A+B+C=π,
A+B=π−C
Taking "cos" on both sides
cos(A+B)=−cosC.
Now,
L.H.S=cos2A+cos2B−cos2C=1+cos2A2+1+cos2B2−cos2C=2+cos2A+cos2B2−cos2C.
How should I complete now?
Answer
The identity
1−cos2A−cos2B−cos2C−2cosAcosBcosC=0
can be seen as a re-writing of the relation cosC=−cos(A+B) without sines of A and B:
cosC=−cos(A+B)=sinAsinB−cosAcosB
cosC+cosAcosB=sinAsinB
(cosC+cosAcosB)2=sin2Asin2B=(1−cos2A)(1−cos2B)
cos2C+2cosAcosBcosC+cos2Acos2B=1−cos2A−cos2B+cos2Acos2B
and the result follows.
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