设f(α)=[2sin(π+α)cos(π-α)-cos(π+α)]/[(1+sin^2α+sin(π-α)-cos^2(π-α),求f(-23π/6)的值
问题描述：
设f(α)=[2sin(π+α)cos(π-α)-cos(π+α)]/[(1+sin^2α+sin(π-α)-cos^2(π-α),求f(-23π/6)的值
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+
答
f(α)=[2sin(π+α)cos(π-α)-cos(π+α)]/[(1+sin^2α+sin(π-α)-cos^2(π-α)=[2sinαcosα+cosα]/[1+sin²α+sinα-cos²α]=[cosα(2sinα+1)]/[2sin²α+sinα)=[cosα(2sinα+1)]/[sinα(2sin...