已知tanα=-1/3,求下列各式的值
问题描述：
已知tanα=-1/3,求下列各式的值
求sin²α+2sinαcosα-3cos²α的值
答
已知tanα=-1/3
所以,sin2α=2tanα/(1+tan^2 α)=(-2/3)/[1+(1/9)]=-3/5
cos2α=(1-tan^2 α)/(1+tan^2 α)=4/5
sin^2 α+2sinαcosα-3cos^2 α
=sin^2 α+sin2α-3(1-sin^2 α)
=4sin^2 α+sin2α-3
=4*[(1-cos2α)/2]+sin2α-3
=2-2cos2α+sin2α-3
=sin2α-2cos2α-1
=(-3/5)-(8/5)-1
=-16/5