问2道三角函数的题目1.已知α、β为锐角,且3(sin(α))^2+2(sin(β))^2=1,3sin(2α)-2si

1.
cos(α+2β)
=cosαcos2β-sinαsin2β
=cosα(1-2(sinβ)^2)-sinαsin2β（注：已知3(sin(α))^2+2(sin(β))^2=1,3sin(2α)-2sin(2β)=0）
=cosα(3(sinα)^2)-sinα(3*sin(2α)/2)
=3sin(2α)sinα/2-3sin(2α)*sinα/2
=0
即cos(α+2β)=0
又因为α,β为锐角,所以0