化简cos(3k+13
化简sin(kπ + 2/3π )× cos(kπ -π/6 )
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)
设K为整数,化简sin(k∏-α)cos((k-1)∏-α)/sin((k+1)∏+α)cos(k∏+α)
化简:cos[(k+1)π-a]·sin(kπ-a)/cos[(kπ+a)·sin[(k+1)π+a] (k属于整数)
求值tan(kπ+π/6)cos(kπ-4π/3),k属于z
化简f(x)=cos((6k+1)/3*π+2x)+cos((6k-1)/3*π-2x)(x∈R,k∈Z),并求函数f(
化简sin(4k-1/4)π-a+cos(4k+1/4)π-a
已知 sin(θ+kπ)=-2cos (θ+kπ) 求 ⑴4sinθ-2cosθ/5cosθ+3sinθ; ⑵(1/4)
化简f(x)=cos【(6k+1/3)π+2x】+cos【(6k-1/ 3)π-2x】+2sin(π/6-2x)(x∈R
函数f(x)=cos(-x/2)+根号3cos(4k+1/2派-x/2),k属于Zx属于R(1)化简f(x)求最小正周(
化简sin(-2/3pai+kpai)cos(pai/6+kpai)tan(pai/4+kpai),k属于z