在三角形ABC中2sin^2A=3sin^2B+3sin^2C,cos2A+3cosA+3cos(B-C)=1求三角形ABC三边之比

题目

在三角形ABC中2sin^2A=3sin^2B+3sin^2C,cos2A+3cosA+3cos(B-C)=1求三角形ABC三边之比

答案

解cos2A+3cosA+3cos(B-C)=1 =>3cosA+3cos(B-C)=1-cos2A =2sin^2A =3sin^2B+3sin^2C =>-3cos(B+C)+3cos(B-C)=3sin^2B+3sin^2C =>-(cosBcosC-sinBsinC)+(cosBcosC+sinBsinC)=sin^2B+sin^2C =>sin^2B+sin^2C-2sinBsinC=...