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sin(4k-1/4π- α)+cos(4k+1/4π -α)(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] 化简 {sin[α+(k+1)π]+sin[α-(k+1)π]}/[sin(α+kπ)*cos(α-kπ)],k∈Z sin(kπ+α)化简 化简sin(4k-1/4)π-a+cos(4k+1/4)π-a 1.化简:sin(4k-1/4π -a)+cos(4k+1/4π -a)(k∈Z) 已知sin^4α+cos^4α=1,求：sin^kα+cos^kα(k∈Z）. 当2kπ-π/4≤α≤2kπ+π/4(k∈Z）,化简√（1-2sinα×cosα)+√(1+2sinα×cosα) 化简：cos[(k+1)π-α]*sin（kπ-α）/cos（kπ+α）*sin[(k+1)π+α]拜托了各位 （2（sinα）²-sin 2α）/（1+tanα）=k （π/4＜k＜π/2) 用k表示sinα-cosα的值 函数y=sinα+cosα-4sinαcosα+1,且2sin^2α+sin2α/1+tanα=k,π/4 当a=5π/4时,｛sin[a+（2k+1)π]-sin[-a-(2k+1)π]｝/sin(a+2kπ）cos(a-2kπ）（k属于z）的值是 已知sinα=4sin(α+β),α+β≠kπ+π/2(k∈Z).求证tan(α+β)=sinβ/(cosβ-4) 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(2sin^2α+sin2α)/(1+tanα)=k,(π/4 已知2sin^α+sin2α/1+tanα=k(π/4