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
[cos(α-π)*cos(19/2*π-α)]/[sin(π/2-α)* tan[(2k+1)π+α]]
3个回答
相关问题
-
..sin^2（π+α）*cos（π+α）*cos（-α-2π）/tan（π+α）*sin^3（π/2+α）*sin（-
-
sin(2kπ-α)=cos(2kπ-α)=tan(2kπ-α)=
-
tan(π-α)*sin^2(α+π/2)*cos(2π-α)/cos^3(-α-π)*tan(2π-α) sin^2就
-
化简：sin²（α+π)cos(-α+π)/tan(α+π)tan(α+2π)cos²(-α-π)
-
化简（1）tan(π−α)cos(2π−α)sin(−α+3π2)cos(−α−π)sin(−π−α)．
-
化简：sin2(α+π)•cos(π+α)tan(π+α)•cos3(−α−π)•tan(−α−2π)．
-
设tan（π+α）=2，则sin(α−π)+cos(π−α)sin(π+α)−cos(π+α)=（ ）
-
(1+tanα)/(1-tanα)=3+2√2,求cos^2(π-α)+sin(π+α)*cos(π-α)+2sin(α
-
sin(π/2+α)·cos(π/2-α)/cos(π+α)+sin（π-α）·cos（π/2+α）/sin（π+α）=
-
化简（1）sin(α+π)cos(-α)sin(-α-π) （2）sin³（-α）cos（2π+α）tan（-