Question

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

Answer 1

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
=cotα

f(-23π/6)=cot(-23π/6)
=cot(4π-23π/6)
=cot(π/6)
=√3