If A + B + C = 0° , then prove that Cos^2 A + Cos^2 B + Cos^2 C = 1 + 2 Cos ACosBCosC
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If A + B + C = 0° , then prove that Cos^2 A + Cos^2 B + Cos^2 C = 1
If A + B + C = 0° , then prove that Cos^2 A + Cos^2 B + Cos^2 C = 1 + 2 Cos ACosBCosC
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Answer:
We write cos2A=1−sin2A
and as in ΔABC A+B+C=180
⟹cosC=cos(180−A−B)=−cos(A+B)
L.H.S.=1−sin2A+cos2B+cos2C
=1+(cos2B−sin2A)+cos2C
=1+cos(B+A)cos(B−A)+cos2C ..... (cos2C−sin2D=cos(C+D)cos(C−D))
=1−cosCcos(B−A)+cos