伊然1124
幼苗
共回答了19个问题采纳率:78.9% 举报
1-2sinBsinC=cos2B+cos2C-cos2A
1-2sinBsinC=cos2B-2sin(A+C)sin(C-A)
1-2sinBsinC=1-2sin^2B-2sinBsin(C-A)
sinBsinC=sin^2B+sinBsin(C-A)
sinC=sinB+sin(C-A)
sinC=sin(C+A)+sin(C-A)
sinC=2sinCcosA
cosA=1/2
A=π/3
B+C=2π/3,B=2π/3-C
sinB+sinC=sin(2π/3-C)+sinC=√3/2cosC-1/2sinC+sinC
=√3/2cosC+1/2sinC
=sin(C+π/3)
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1年前
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