[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:2024/05/13 14:18:51
[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}