已知tanα=-1/2,那么sin^2α+2sinαcosα-3cos^2α的值是?

问题描述:

已知tanα=-1/2,那么sin^2α+2sinαcosα-3cos^2α的值是?

原式=(2sin^2a+sinacosa-3cos^2a)/(sin^2a+cos^2a)
=(2tan^2a+tana-3)/(tan^2a+1)
=(2*1/4-1/2-3)/(1/4+1)
=-12/5

答案应该是-3

sin^2α+2sinαcosα-3cos^2α=(sin^2α+2sinαcosα-3cos^2a)/(sin^2α+cos^2α) 【化成分式分母1=(sin^2α+cos^2α】=(tan^2α+2tanα-3)/(tan^2α+1) 【分子分母同时除以cos^2α化切】=(1/4-1-3)/(1/4+1)=-15/5=...