已知-pi/2

问题描述:

已知-pi/2求sinx-cosx的值
求(sin2x+2sinx^2)/(1-tanx)的值

-pi/20
sinx+cosx=1/5
平方
sin²x+2sinxcosx+cos²x=1/25
sinxcosx=-12/25
(sinx-cosx)²=sin²x-2sinxcosx+cos²x=1+24/25=49/25
sinx-cosxsinx-cosx=-7/5
(sin2x+2sinx^2)/(1-tanx)
=2sinx(cosx+sinx)/(1-sinx/cosx)
=2sinxcosx(cosx+sinx)/(cosx-sinx)
=24/25* 1/5÷(7/5)
=24/175