已知tan(π+α)=3,则2cos(π-α)-3sin(π+α)/4sin(α+π/2)+sin(2π-α)=
3个回答
tan(π+a)=tana=3
原式=(-2cosa+3sina)/(4cosa-sina)
上下除以cosa
sina/cosa=tana
所以原式=(-2+3tana)/(4-tana)=-7
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