[cos(α-π)*cos(19/2*π-α)]/[sin(π/2-α)* tan[(2k+1)π+α]]

来源：学生作业帮助网 编辑：作业帮 时间：2024/05/06 00:58:07

[cos(α-π)*cos(19/2*π-α)]/[sin(π/2-α)* tan[(2k+1)π+α]]
[cos(α-π)*cos(19/2*π-α)]/[sin(π/2-α)* tan[(2k+1)π+α]]

[cos(α-π)*cos(19/2*π-α)]/[sin(π/2-α)* tan[(2k+1)π+α]]
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 alpha * (-)sin alpha)/(cos alpha * (-1) tan alpha)
　　= - cos alpha

[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

cos^2α+cos^2(α+π/3)-cosαcos(α+π/3)化简 Cos(α-π) Cos(α- π/2 )如何化简 2sinα*cosα*cos(2π-α)+cos(π+2α)*cosα*tanα化简! [sin(5π-α)cos(-π-α)]/[cos(α-π)cos(π/2+α)]化简 化简:cosα+cos(2π/3-α)+cos(2π/3+α) 【紧急求助】计算:cosα+cos(2π/3 +α)+cos（4π/3 +α） 化简:cos(-π-α) cos^3(-π-α) 计算：cos（π/5)+cos（2π/5)+cos(3π/5)+cos(4π/5) cos(π/17)*cos(2π/17)*cos(4π/17)*cos(8π/17) 求值:cos(π/11)cos(2π/11)cos(3π/11)cos(4π/11)cos(5π/11) 求值：cos(π/11)-cos(2π/11)+cos(3π/11)-cos(4π/11)+cos(5π/11) cos(π/11)cos(2π/11)cos(3π/11)cos(4π/11)cos(5π/11) sin(3π+α )=1/4,求cos(π+α)/cosα[cos(π+α)-1]+cos(α-2π)/cos(α+2π)cos(π+α)+coa(-α) cos^2(π/4+α)=(cos(π/4)cosα-sin(π/4)sinα)² cos^2(π/4+α)是怎么等于(cos(π/4)cosα-sin(π/4)sinα)²的? 2cos(α-2π)=2cos(-2π+α)=2cosα2cos(α-2π)=-2cos（2π-α）=-2cosα那个是对的 化简sin(π/2+α)cos(3π-α)tan(π+α)/cos(π/2-α)cos(-α-π) 化简{sin(π+α)^3cos(-α)cos(π+α)}/{tan(π+α)^3cos(-π-α)^2}