求证：(sin(kπ-α)cos(kπ+α))/(sin((k+1)π+α)cos((k+1)π+α))=-1,k∈Z

求证：(sin(kπ-α)cos(kπ+α))/(sin((k+1)π+α)cos((k+1)π+α))=-1,k∈Z 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] 求证（1-cosα）/sinα=sinα/（1+cosα）=tan（α/2）(α≠kπ,k∈z) 快回答! 已知sinα=4sin(α+β),α+β≠kπ+π/2(k∈Z).求证tan(α+β)=sinβ/(cosβ-4) sin(kπ-α）*cos〔（k-1)π-α〕/sin〔(k+1)π+α〕*cos(kπ+α) ,k属于Z 化简 {sin[α+(k+1)π]+sin[α-(k+1)π]}/[sin(α+kπ)*cos(α-kπ)],k∈Z 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 cosαcos[(2k+1)π]-sinαsin[(2k+1)π]为什么等于-cosα [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 化简：cos[(k+1)π-α]*sin（kπ-α）/cos（kπ+α）*sin[(k+1)π+α]拜托了各位 已知sin(α-3π)+cos(π/2+α)=m,求证cos(α-7π/2)+2sin(2kπ-α)=3m/2 已知θ≠kπ（k∈Z）求证：tan(θ/2)=(1-cosθ)/sinθ 已知α是第四象限角,化简sin(kπ+α)√[1+cos(kπ+α)]/[1-cos(kπ+α)],k属于z cos[(k-1)π-α]=cos[(k+1)π+α]=-cos(kπ+α）sin[（k+1）π+α]=-sin(kπ+α） 上述两个式子为什么相等 刚学这部分 还不太懂. 化简 sin(4k-1/4π- α)+cos(4k+1/4π -α)(k∈Z) 化简sin(4k-1/4π- α)+cos(4k+1/4π -α)(k∈Z) 化简 tan(kπ-a).sin(kπ-a).cos(kπ+a)/cos(kπ-a).sin(kπ+a)