设k∈Z,化简sin(kπ−α)cos[(k−1)π−α]sin[(k+1)π+α]cos(kπ+α)的结果是( )
设k∈Z,化简sin(kπ−α)cos[(k−1)π−α]sin[(k+1)π+α]cos(kπ+α)的结果是( )
sin(kπ-α)*cos〔(k-1)π-α〕/sin〔(k+1)π+α〕*cos(kπ+α) ,k属于Z
化简[sin(kπ-α)*cos(kπ+α)]/{sin[(k+1)π+α]*cos[(k+1)π-α]}
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)
sin(kπ-α)cos【(k-1)π-α】/sin【(k+1)π+α】cos(kπ+α) (k∈Z) 希望老师能详细解
已知sin^4α+cos^4α=1,求:sin^kα+cos^kα(k∈Z).
【1】求证sin(kπ-a)cos(kπ+a)/sin[(k+1)π+a]cos[(k+1)π+a]=-1,k∈Z
弧度制下的角的表示sin(2kπ+α)=sinα (k∈Z) cos(2kπ+α)=cosα (k∈Z) tan(2
当2kπ-π/4≤α≤2kπ+π/4(k∈Z),化简√(1-2sinα×cosα)+√(1+2sinα×cosα)
设K为整数,化简sin(k∏-α)cos((k-1)∏-α)/sin((k+1)∏+α)cos(k∏+α)
设k为整数,化简sin(kπ-a)cos[(k-1)π-a]/sin[(k+1)π+a]cos(kπ+a)