在三角形ABC中,cos A cos B+cos Asin B+sin Asin B=2,则三角形ABC是
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sin²A+sin²B+sin²C

=sin²A+sin²B+sin²(A+B)

=sin²A+sin²B+(sinAcosB+cosAsinB)²

=sin²A+sin²B+sin²Acos²B+cos²Asin²B+2sinAcosAsinBcosB

=sin²A+sin²B+(1-cos²A)cos²B+cos²A(1-cos²B)+2sinAcosAsinBcosB

=2-2cos²Acos²B+2sinAcosAsinBcosB

=2-2cosAcosB(cosAcosB-sinAsinB)

=2-2cosAcosBcos(A+B)

=2+2cosAcosBcosC

所以,若sin²A+sin²B+sin²C=2,则cosAcosBcosC=0,其中一个角是直角,所以△ABC是直角三角形