asosB+bcosA=csinC怎样变形得 sinAcosB+sinBcosA=sin^2C?

sinAcosB+sinBcosA=sin(A+B)=sin(π-C)=sinC

如果A,B,C是三角形内角,
acosB+bcosA=csinC→ac cosB+bc cosA=c^2 sinC
根据余弦定理,2ac cosB=a^2+c^2-b^2;
2bc cosA=b^2+c^2-a^2
∴ac cosB+bc cosA=2c^2/2=c^2;
则c^2=c^2 sinC
sinC=1
则sin^2C=1

∴sinAcosB+sinBcosA=sin(A+B)=sin(π-C)=sinC=sin^2C=1