[cos(α-π)*cos(19/2*π-α)]/[sin(π/2-α)* tan[(2k+1)π+α]]
问题描述：
[cos(α-π)*cos(19/2*π-α)]/[sin(π/2-α)* tan[(2k+1)π+α]]
答
[cos(α-π)*cos(19/2*π-α)]/[sin(π/2-α)* tan[(2k+1)π+α]]
=[-cosαcos(10π-π/2-α)]/[cosa tan[(2kπ+π+α]]
=[-cosαcos(π/2+α)]/[cosa tan(2kπ+π+α)]
=[-cosα(-sina)]/[cosa tan(π+α)]
=[cosαsina]/[-cosa tanα]
=-sina/ tanα
=-cosa
答
=(-cos alpha * (-)sin alpha)/(cos alpha * (-1) tan alpha)
= - cos alpha
答
cos(α - π) = cos(π - α) = -cosα
cos(19π/2 - α) = cos[(20 - 1)π/2 - α] = cos(10π - π/2 - α) = cos(π/2 + α) = -sinα
sin(π/2-α) = cosα
tan[(2k+1)π+α] = tanα
原式 = -cosα(-sinα)/(cosαtanα) = sinαcosα/(cosα*sinα/cosα) = sinαcosα/sinα = cosα