化简 sin[(k+1)π+θ]*cos[(k+1)π-θ]/sin(kπ-θ)*cos(kπ+θ) k∈z
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/17 02:12:27
化简 sin[(k+1)π+θ]*cos[(k+1)π-θ]/sin(kπ-θ)*cos(kπ+θ) k∈z
化简 sin[(k+1)π+θ]*cos[(k+1)π-θ]/sin(kπ-θ)*cos(kπ+θ) k∈z
化简 sin[(k+1)π+θ]*cos[(k+1)π-θ]/sin(kπ-θ)*cos(kπ+θ) k∈z
当k为偶数,即 k=2n,n∈z时,
sin[(k+1)π+θ]*cos[(k+1)π-θ]/sin(kπ-θ)*cos(kπ+θ)=sinθ*cosθ/sin(-θ)*cosθ=-1
当k为奇数,即 k=2n+1,n∈z时,
sin[(k+1)π+θ]*cos[(k+1)π-θ]/sin(kπ-θ)*cos(kπ+θ)=sinθ*cosθ/sinθ*(-cosθ)=-1
综上,得
sin[(k+1)π+θ]*cos[(k+1)π-θ]/sin(kπ-θ)*cos(kπ+θ)=-1 ,(k∈z)
化简:sin[(k+1)π+θ]×cos[(k+1)π-θ] / sin(kπ-θ)×cos(kπ+θ) (k∈Z)
化简 sin[(k+1)π+θ]*cos[(k+1)π-θ]/sin(kπ-θ)*cos(kπ+θ) 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]
设k为整数,化简sin(kπ-a)cos[(k-1)π-a]/sin[(k+1)π+a]cos(kπ+a)
化简:sin(kπ-a)cos[(k-1)π-a]/sin[(k+1)π+a]cos(kπ+a) k∈Z
化简:sin(kπ-a)cos[(k-1)π-a]/sin[(k+1)π+a]cos(kπ+a) k∈Z
化简:cos[(k+1)π-a]·sin(kπ-a)/cos[(kπ+a)·sin[(k+1)π+a] (k属于整数)
化简:sin(kπ-2)cos[(k-1)π-2]/sin[(k+1)π+2]cos(kπ+2),k属于Z
化简 {sin[α+(k+1)π]+sin[α-(k+1)π]}/[sin(α+kπ)*cos(α-kπ)],k∈Z
化简:cos[(k+1)π-α]*sin(kπ-α)/cos(kπ+α)*sin[(k+1)π+α]拜托了各位
[sin(kπ-a)cos(kπ-π-a)]/[sin(kπ+π+a)cos(kπ+a)] 化[sin(kπ-a)cos(kπ-π-a)]/[sin(kπ+π+a)cos(kπ+a)] 化简
已知sin(θ+kπ)=-2cos(θ+kπ)(k∈Z) 求1/4sin^2θ+2/5cos^2θ
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
化简sin(4k-1/4)π-a+cos(4k+1/4)π-a
cosαcos[(2k+1)π]-sinαsin[(2k+1)π]为什么等于-cosα
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
已知sin(θ+kπ)=2cos[θ+(k+1)π],k∈Z,求4sinθ-2cosθ/5cosθ+3sinθ的值