k=(πcosπ-sinπ)/π2怎么算的
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k=(πcosπ-sinπ)/π2怎么算的
k=(πcosπ-sinπ)/π2怎么算的
k=(πcosπ-sinπ)/π2怎么算的
cosπ=-1 sinπ=0k=-π/(2π)=-1/2
k=(πcosπ-sinπ)/π2怎么算的
sin(kπ+a)和cos(kπ+a)等于什么 怎么算
cosαcos[(2k+1)π]-sinαsin[(2k+1)π]为什么等于-cosα
2(sin a)^2+(2sin a*cos a)/(1+tan a)=k试用k表示sin a-cos aa∈(π/4,π/2)
已知sin(θ+kπ)=-2cos(θ+kπ)(k∈Z) 求1/4sin^2θ+2/5cos^2θ
已知 sin(θ+kπ)=-2cos (θ+kπ) 求 ⑴4sinθ-2cosθ/5cosθ+3sinθ; ⑵(1/4)sin平方θ+(2/5)cos平方θ已知sin(θ+kπ)=-2cos (θ+kπ)求⑴ 4sinθ-2cosθ/5cosθ+3sinθ⑵(1/4)sin平方θ+(2/5)cos平方θ
化简:sin(kπ-2)cos[(k-1)π-2]/sin[(k+1)π+2]cos(kπ+2),k属于Z
求证:(sin(kπ-α)cos(kπ+α))/(sin((k+1)π+α)cos((k+1)π+α))=-1,k∈Z
已知sin(θ+kπ)=-2cos(θ+kπ),k∈Z,求4sinθ-2cosθ/5cosθ+3sinθ①4sinθ-2cosθ/5cosθ+3sinθ ②sin²θ+2/5cos²θ
[sin(kπ-a)cos(kπ-π-a)]/[sin(kπ+π+a)cos(kπ+a)] 化[sin(kπ-a)cos(kπ-π-a)]/[sin(kπ+π+a)cos(kπ+a)] 化简
[sinπ(/2+α)-cos(3π/2-α)]/[tan(2kπ -α)+cot(-kπ+α)]=[sin(4kπ-α)sin(π/2 -α)]/[cos(5π+α)-cos(π/2+α)
化简[sin(kπ-α)*cos(kπ+α)]/{sin[(k+1)π+α]*cos[(k+1)π-α]}
化简sin(kπ-α)cos(kπ+α)/sin[(k+1)π+α]cos[(k+1)π+α]
o``````计算sin[(2k+1)π/4] + cos[(2k+1)π/4]之值.`` ``怎么计算的呢?
已知sin(θ+kπ)=2cos[θ+(k+1)π],k∈Z,求4sinθ-2cosθ/5cosθ+3sinθ的值
已知sin(θ+kπ)=2cos[θ+(k+1)π],k∈Ζ,求4sinθ-2cosθ/5cosθ+3sinθ的值
求证(1-cosα)/sinα=sinα/(1+cosα)=tan(α/2)(α≠kπ,k∈z) 快回答!
求证sin(π/2 +α)-cos(3π/2 -α)/tan(2kπ -α)+cot(-kπ+α)=sin(4kπ -α)sin(π/2 -α)/cos(5π+α)sin(π/2 +α)-cos(3π/2 -α)/tan(2kπ -α)+cot(-kπ+α)=sin(4kπ -α)sin(π/2 -α)/cos(5π+α)-cos(π/2 + α)谁帮我做做