Question

设f(α)=[2sin(π+α)cos(π-α)-cos(π+α)]/[(1+sin^2α+sin(π-α)-cos^2(π-α)

1.若α=-17/6π,求f(α)的值
2.若α是锐角,则sin(α-3/2π)=3/5,求f(α)的值
过程

Answer 1

1) f(α)=[2sin(π+α)cos(π-α)-cos(π+α)]/[(1+sin^2α+sin(π-α)-cos^2(π-α) = [2(-sinα)(-cosα)-(-cosα)]/[(1+sin^2α+sinα-cos^2α] = [2sinαcosα+cosα]/[2sin^2α+sinα] = cosα(2sinα+1)/sinα(2sinα+1) =cosα/sinα = cos(-17/6π)/sin(-17/6π) = cos(-5/6π)/sin(-5/6π) = cos(5/6π)/-sin(5/6π) = √3

2)α是锐角,sin(α-3/2π)=3/5,则sin(3/2π-α)=-3/5，则sinα=3/5, cosα= 4/5,f(α)=cosα/sinα = 4/3