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
网友回答
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(2sinα+1)]
=cosα/sinα
=cotαf(-23π/6)=cot(-23π/6)
=cot(4π-23π/6)
=cot(π/6)
=√3======以下答案可供参考======
供参考答案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+