sin( α+kπ)=cos(α+π+kπ),则α的值是?答案是kπ+3/4π (k是整数）

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] sin(kπ-α）*cos〔（k-1)π-α〕/sin〔(k+1)π+α〕*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[α+(k+1)π]+sin[α-(k+1)π]}/[sin(α+kπ)*cos(α-kπ)],k∈Z cosαcos[(2k+1)π]-sinαsin[(2k+1)π]为什么等于-cosα 化简：cos[(k+1)π-α]*sin（kπ-α）/cos（kπ+α）*sin[(k+1)π+α]拜托了各位 sin[α+(k+1)л]+sin[α-(k+1)л]/sin(α+kл)*cos(α-kл)= sin( α+kπ)=cos(α+π+kπ),则α的值是?答案是kπ+3/4π (k是整数） [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) 求证（1-cosα）/sinα=sinα/（1+cosα）=tan（α/2）(α≠kπ,k∈z) 快回答! 已知sin^4α+cos^4α=1,求：sin^kα+cos^kα(k∈Z）. sin（kπ－α）•cos（kπ+α）∕sin[（k+1）π+α]•cos[（k+1）π－α]=1. 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 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 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)