已知tan(x+y)=3,tan(x-y)=1/2,则sin2x/sin2y=
问题描述：
已知tan(x+y)=3,tan(x-y)=1/2,则sin2x/sin2y=
答
tan(x+y)+tan(x-y)=sin(2x)/(cosxcosy)=3+1/2=7/2tan(x+y)-tan(x-y)=sin(2y)/(cosxcosy)=3-1/2=5/2sin(2x)/sin(2y)=[tan(x+y)+tan(x-y)]/[tan(x+y)-tan(x-y)]=7/2/(5/2)=7/5tan(x+y)+tan(x-y)=sin(2x)/(cosxcosy)为什么？根据和差化积公式：tanA+tanB=sin(A+B)/(cosAcosB)tanA-tanB=sin(A-B)/(cosAcosB)tan(x+y)+tan(x-y)=sin[(x+y)+(x-y)]/(cosxcosy)=sin(2x)/(cosxcosy)tan(x+y)-tan(x-y)=sin[(x+y)-(x-y)]/(cosxcosy)=sin(2y)/(cosxcosy)那我觉得后面应该是cos(x+y)cos(x-y)，但是对答案没影响，所以采纳了。