已知tanx=-1/2,则sin^2x+3sinxcosx-1=
写出答案和过程.谢谢
已知tanx=-1/2,则sin^2x+3sinxcosx-1=
写出答案和过程.谢谢
tanx=-1/2 -> sinx=-1/2cosx sin^2x=1/4cos^2x sin^2x+cos^2x=1 ->sin^2x=1/5
sin^2x+3sinxcosx-1=sin^2x-6sin^2x-1 =-5sin^2x-1=-1-1=-2
加个分母,1=sin^x+cos^x,然后上下同除以cos^x即可
sin^2x+3sinxcosx-1=sin^2x+3sinxcosx-(sin^2x+cos^2x)=3sinxcosx-cos^2x=(3sinxcosx-cos^2x)/(sin^2x+cos^2x)=(3tanx-1)/(tan^2x+1)
因为tanx=-1/2,所以sin^2x+3sinxcosx-1=(-3/2-1)/(1/4+1)=-2