当k为偶数时,
sin(kpi-a)=sin(-a)=-sina
cos[(k-1)pi-a]=-cosa
sin[(k+1)pi+α]=-sina
cos(k∏+α)=cosa
所以原式=-1
当k为奇数时,
sin(kpi-a)=sina
cos[(k-1)pi-a]=cosa
sin[(k+1)pi+α]=sina
cos(k∏+α)=-cosa
所以原式=-1
综上原式=-1
设K为整数,化简sin(k∏-α)cos((k-1)∏-α)/sin((k+1)∏+α)cos(k∏+α)
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