若tan(π/4-x)=-1/2.求(2sinx-cosx)/(sinx+3cosx)
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若tan(π/4-x)=-1/2.求(2sinx-cosx)/(sinx+3cosx)
若tan(π/4-x)=-1/2.求(2sinx-cosx)/(sinx+3cosx)
若tan(π/4-x)=-1/2.求(2sinx-cosx)/(sinx+3cosx)
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
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
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这道题为什么第二步的+到了第三步变成了-1.tan(x/2+π/4)+tan(x/2-π/4)=[tan(x/2)+tan(π/4)]/[1-tan(x/2)tan(π/4)]+[tan(x/2)-tan(π/4)]/[1+tan(x/2)tan(π/4)]=[tan(x/2)+1]/[1-tan(x/2)]+[tan(x/2)-1]/[1+tan(x/2)]=[(tan(x/2)+1)^2-(tan(x/2)-
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