三角形ABC中,A:B:C=1:2:4 求证c^2-a^2=bc

证:△ABC中A:B:C=1:2:4,由A+B+C=π解方程组得到
A=π/7,B=π/7,C=π/7.
设a/sinA=b/sinB=c/sinC=R
c^2-a^2=R²[(sinC)²-(sinA)²] 半角公式
=R²[(1-cos2C)/2-(1-cos2A)/2]
=R²[(cos2A-cos2C)/2] 和差化积
=R²*[-sin(A+C)sin(C-A)]
=R²*[-sinBsin(3pi/7)], [sin(3pi/7)=sin(pi-4pi/7)=-sin4pi/7]
=R²*sinBsin(4pi/7)
=R²*sinBsinC
=(RsinB)(RsinC)
=bc.
答题完毕，祝你开心！