若sin(α-π)=2cos(2π-α),求(sinα+5cosα)/(sinα-3cosα)的值,

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若sin(α-π)=2cos(2π-α),求(sinα+5cosα)/(sinα-3cosα)的值,

若sin(α-π)=2cos(2π-α),求(sinα+5cosα)/(sinα-3cosα)的值,
若sin(α-π)=2cos(2π-α),求(sinα+5cosα)/(sinα-3cosα)的值,

若sin(α-π)=2cos(2π-α),求(sinα+5cosα)/(sinα-3cosα)的值,
(sinα+5cosα)/(sinα-3cosα)=-3/5
sin(α-π)=sin(-π+α)=-sinα
2cos(2π-α)=2cosα=-sinα
所以 sinα+2cosα=0
(sinα+5cosα)/(sinα-3cosα)
=((sinα+2cosα)+3cosα)/((sinα+2cosα)-5cosα)
=3cosα/(-5cosα)
=-3/5

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