Question

在三角形ABC中,求证sin²A+sin²B+sin²C=2+2cosAcosBcosC

Answer 1

三角形ABC，有A+B+C=π
而：
sinC=sin[π-(A+B)]=sin(A+B)=sinAcosB+cosAsinB
cosC=cos[π-(A+B)]=-cos(A+B)=sinAsinB-cosAcosB
所以：
左边=sin²A+sin²B+sin²C
=sin²A+sin²B+(sinAcosB+cosAsinB)²
=sin²A+sin²B+sin²Acos²B+cos²Asin²B+2sinAcosBcosAsinB
=(sin²A+cos²Asin²B)+(sin²B+sin²Acos²B)+2sinAsinBcosAcosB
=[sin²A+cos²A(1-cos²B)]+[sin²B+(1-cos²A)cos²B]+2sinAsinBcosAcosB
=[sin²A+cos²A-cos²Acos²B]+[sin²B+cos²B-cos²Acos²B]+2sinAsinBcosAcosB
=2-2cos²Acos²B+2sinAsinBcosAcosB
=2+2cosAcosB(sinAsinB-cosAcosB)
=2+2cosAcosBcosC
=右边
证毕。