sin(π/4+α)=k,则cos(π/4-α)=

sin(π/4+α)=k,则cos(π/4-α)= 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 [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) 函数y=sinα+cosα-4sinαcosα+1,且2sin^2α+sin2α/1+tanα=k,π/4 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 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] 已知sinα=4sin(α+β),α+β≠kπ+π/2(k∈Z).求证tan(α+β)=sinβ/(cosβ-4) 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^4α+cos^4α=1,求：sin^kα+cos^kα(k∈Z）. 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 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. 已知sin(θ+kπ)=-2cos(θ+kπ),k∈Z,求4sinθ-2cosθ/5cosθ+3sinθ①4sinθ-2cosθ/5cosθ+3sinθ ②sin²θ+2/5cos²θ sin(π/2+α)·cos(π/2-α)/cos(π+α)+sin（π-α）·cos（π/2+α）/sin（π+α）=?（2）sin²40°+sin²50°=?（3）已知sin10°=k,则cos620°=?（4）已知f（cosx）=cos3x,则f（cos30°）=? 已知A=sin(kπ+α）/sinα +cos(kπ+α）/cosα (k∈Z）,则A值所构成的集合是?