tanα=-1／3 则sinα^2+2sinαcosα-3cosα^2=?
问题描述：
tanα=-1／3 则sinα^2+2sinαcosα-3cosα^2=?
答
tanα=-1／3 则tanα^2=1/9(sinα^2+2sinαcosα-3cosα^2)/cosα^2=1/tanα^2-3+2xtanα=9-3+(-2/3)=10/3 sinα/cosα=-1/3sinα^2+cosα^2=1cosα^2=9/10sinα^2+2sinαcosα-3cosα^2=3