tan(a+b）=3,tan(a-b)=1/2,求sin2a/sin2b
问题描述：
tan(a+b）=3,tan(a-b)=1/2,求sin2a/sin2b
答
tan(a+b）=3,tan(a-b)=1/2,
tan2a=tan(a+b+a-b)=[tan(a+b)+tan(a-b)]/[1-tan(a+b)tan(a-b)]=(7/2)/(1-3/2)=-7
sin2a=-7/5√2,或7/5√2
同理
tan2b=tan(a+b-a+b)=[tan(a+b)-tan(a-b)]/[1+tan(a+b)tan(a-b)]=(5/2)/(1+3/2)=1
sin2b=√2/2,-√2/2
sin2a/sin2b=7/5或-7/5