在△ABC中,求证：sin^2A+sin^2B-sin^2C=2sinAsinBcosC
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证明左边=1/2(1-cos2A)+1/2(1-cos2B)-(1-cos²C)
=cos²C-1/2(cos2A+cos2B)
=cos²C-cos(A+B)·cos(A-B)
=cos²C+cosC·cos(A-B)
=cosC[cosC+cos(A-B)]
=cosC2cos1/2(C+A-B)cos1/2(C-A+B)
=2cosCcos1/2(180°-2B)cos(1/2)(180°-2A)
=2cosCcos(90°-B)cos(90°-A)
=2sinAsinBcosC=右边
=cos²C-1/2(cos2A+cos2B)
=cos²C-cos(A+B)·cos(A-B)
=cos²C+cosC·cos(A-B)
=cosC[cosC+cos(A-B)]
=cosC2cos1/2(C+A-B)cos1/2(C-A+B)
=2cosCcos1/2(180°-2B)cos(1/2)(180°-2A)
=2cosCcos(90°-B)cos(90°-A)
=2sinAsinBcosC=右边
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