已知tan(π+α)=3 求2cos(π-α)-3sin(π+α)/4cos(-α)+sin(2π-α)的值

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已知tan(π+α)=3 求2cos(π-α)-3sin(π+α)/4cos(-α)+sin(2π-α)的值

已知tan(π+α)=3 求2cos(π-α)-3sin(π+α)/4cos(-α)+sin(2π-α)的值
已知tan(π+α)=3 求2cos(π-α)-3sin(π+α)/4cos(-α)+sin(2π-α)的值

已知tan(π+α)=3 求2cos(π-α)-3sin(π+α)/4cos(-α)+sin(2π-α)的值
tan(π+α)=3
tana=3
2cos(π-α)-3sin(π+α)/4cos(-α)+sin(2π-α)
=(-2cosa+3sinα)/(4cosα-sinα)
=(-2+3tanα)/(4-tanα)
=7

首先tan(π+α)=tanα=3
2cos(π-α)-3sin(π+α)/4cos(-α)+sin(2π-α)
=-2cosα+3sinα/4cosα+sinα
=-2+3tanα/4+tanα
=1