在△ABC中,求证：sin^2A+sin^2B+cos^2C+2sinAsinBcos(A+B)=1
3个回答
由题意：1-sin^2A=cos^2A
sin^2B+cos^2C+2sinAsinBcos(A+B)=
=sin^2B+cos^2C-2sinAsinBcosC
=sin^2B +cosC(cosC-2sinAsinB)
=sin^2B -cosC[cos(A+B)+2sinAsinB]
=sin^2B-cosC[cosAcosB-sinAsinB+2sinAsinB]
=sin^2B-cosC[cosAcosB+sinAsinB]
=sin^2B +cos(A+B)(cosAcosB+sinAsinB)
=sin^2B+(cosAcosB-sinAsinB)(cosAcosB+sinAsinB)
=sin^2B +cos^2Acos^2B -sin^2Asin^2B
=sin^2B(1-sin^2A)
=sin^2Bcos^2A +cos^2Bcos^2A
=cos^2A
所以sin^2A+sin^2B+cos^2C+2sinAsinBcos(A+B)=1
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