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

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

人气：133 ℃　时间：2020-06-13 10:25:30

解答

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...