在△ABC中,求证:sin^2A+sin^2B+cos^2C+2sinAsinBcos(A+B)=1

问题描述:

在△ABC中,求证:sin^2A+sin^2B+cos^2C+2sinAsinBcos(A+B)=1

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-s...