tan(a)-tan(b)=2tan^2(a)tan(b),a,b均不等与K派/2,试求sin(2a+b)/sin(b)

题目

tan(a)-tan(b)=2tan^2(a)tan(b),a,b均不等与K派/2,试求sin(2a+b)/sin(b)

答案

将tan(a)-tan(b)=2tan^2(a)tan(b)变形为tan(b)=tan(a)/(1+2tan^2(a))
sin(2a+b)/sin(b)
=[sin(2a)cos(b)+cos(2a)sin(b)]/sin(b)
=sin(2a)/tan(b)+cos(2a)(将tan(b)=tan(a)/(1+2tan^2(a))代入,即)
=sin(2a)*(1+2tan^2(a))/tan(a)+cos(2a)
=2sin(a)cos(a)*[1+2sin^2(a)/cos^2(a)]*[cos(a)/sin(a)]+cos(2a)
=2cos^2(a)+4sin^2(a)+cos^2(a)-sin^2(a)
=3cos^2(a)+3sin^2(a)
=3