高一数学《二倍角》的一道证明题!

3sin2x-2sin2y=0 => 3sinxcosx=2sinycosy 两边平方,并令u=(sinx)^2 v=(siny)^2 9u(1-u)=4v(1-v) .(1) 3u+2v=1 .(2) => 2v=1-3u 代入到 (1) 9u-9u^2=2v(2-2v)=(1-3u)(1+3u)=1-9u^2 => u=1/9 v=(1-3u)/2=1/3 x,y为锐角,所以 sinx=1/3 ,siny=1/√3 cosx=√(1-1/9)=2√2/3 cosy=√(1-1/3)=√6/3 sin2y=2siny*cosy=2*1/√3 *√6/3=2√2/3 cos2y=2(cosy)^2-1=2*6/9-1=1/3 sin(x+2y)=sinxcos2y+cosxsin2y=1/3 * 1/3 +2√2/3 *2√2/3=1/9+8/9=1 x+2y=90°