化简 sin[(k+1)π+θ]*cos[(k+1)π-θ]/sin(kπ-θ)*cos(kπ+θ) k∈z
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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)
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)
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