已知tan(α+π)=3,求【2cos(π-α)-3sin(π+α)】/[4cos(-a)+sin(2π-α)]的值

问题描述:

已知tan(α+π)=3,求【2cos(π-α)-3sin(π+α)】/[4cos(-a)+sin(2π-α)]的值

由tan(α+π)=3 得 tanα=3
【2cos(π-α)-3sin(π+α)】/[4cos(-a)+sin(2π-α)]
=(-2cosα+3sinα)/(4cosa-sinα)
=(-2+3tanα)/(4-tanα)
=(-2+3x3)/(4-3)
= 7