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Question

If A+B+C=π, Prove that
cos4A+cos4B+cos4C=1+4cos2Acos2Bcos2C.

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Solution

We have,

A+B+C=π......(1)

L.H.S.

cos4A+cos4B+cos4C

2cos(4A+4B)2cos(4A4B2)+2cos22C1

=2cos(2A+2B)cos(2A2B)+2cos22C1

=2cos2(A+B)cos(2A2B)+2cos22C1

=2cos2(A+B)cos(2A2B)+2cos22C1

=2cos2(πC)cos(2A2B)+2cos22C1

=2cos2Ccos(2A2B)+2cos22C1

=2cos2C[cos(2A2B)+cos2C]1

=2cos2C[cos(2A2B)+cos2(π(A+B))]1

=2cos2C[cos(2A2B)+cos(2A+2B)]1

=2cos2C[cos(2A+2B)+cos(2A2B)]1

=2cos2C[2cos2Acos2B]1

=4cos2Acos2Bcos2C1

R.H.S.
Hence proved.

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