若tan(π/4-x)=-1/2.求(2sinx-cosx)/(sinx+3cosx)
问题描述：
若tan(π/4-x)=-1/2.求(2sinx-cosx)/(sinx+3cosx)
答
tan(π/4-x)=-1/2.
(tanπ/4-tanx)/(1+tanπ/4tanx)=-1/2
(1-tanx)/(1+tanx)=-1/2
2(1-tanx)=-1-tanx
tanx=3
(2sinx-cosx)/(sinx+3cosx) 分子分母同除以cosx
=(2tanx-1)/(tanx+3)
=(2*3-1)/(3+3)
=5/6
答
tan(π/4-x)
=(tanπ/4-tanx)/(1+tanπ/4tanx)
=(1-tanx)/(1+tanx)
=-1/2.
tanx= 自己算.
(2sinx-cosx)/(sinx+3cosx)上下同除以cosx得(2tanx-1)/(tanx+3)=
把tan带进去