若-p/8<y<0<x<p/4,且tan(2x-y)=3tan(x-2y),则x+y的最大值= (p就是π)

问题描述:

若-p/8<y<0<x<p/4,且tan(2x-y)=3tan(x-2y),则x+y的最大值= (p就是π)

令t1=tan(2x-y);t2=tan(x-2y) ; 则t1=3t2
tan(x+y)=tan[(t1-t2]=(-2t2)/[1+t2^2]=(-2)/(3t2+1/t2)
-π/8