化简:cos[(k+1)π-α]*sin(kπ-α)/cos(kπ+α)*sin[(k+1)π+α]拜托了各位
来源:学生作业帮 编辑:神马作文网作业帮 分类:数学作业 时间:2024/09/30 02:28:24
化简:cos[(k+1)π-α]*sin(kπ-α)/cos(kπ+α)*sin[(k+1)π+α]拜托了各位
k为奇数 sin(kπ-θ)×cos(kπ θ)=-sinθcosθ sin[(k 1)πθ]×cos[(k 1)π-θ] =sinθcosθ k为偶数 sin(kπ-θ)×cos(kπ θ)=-sinθcosθ sin[(k 1)πθ]×cos[(k 1)π-θ] =sinθcosθ 因此 sin[(k 1)πθ]×cos[(k 1)π-θ] / sin(kπ-θ)×cos(kπ θ)=-1
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
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( )
化简:cos[(k+1)π-a]·sin(kπ-a)/cos[(kπ+a)·sin[(k+1)π+a] (k属于整数)
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
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)
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
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)