求和1 +(1/2)*cosx+ (1/4)*cos2x +… +(1/2∧n)cosnx

C(N,1)COSX+C(N,2)COS2X+-----+C(N,N)COSNX COSX+COS2X+COS3X+COS4X+COS5X+COS6X+...+COSNX=1/2|{SIN(N+1/2) 求证:cosx+cos2x+...+cosnx={[cos(n+1)x/2]*[sin(n/2)x]}/[sin(x/2 lim[(1-cosx*cos2x****cosnx)/x^2]在x趋于0时 三角恒等变换证明cosx+cos2x+…+cosnx=[cos（n+1/2）·sinnx/2]/sinx/2怎么证明? 求和Sn=cosx+cos2x+cos3x+……+cosnx 设n∈N*，且sinx+cosx=-1，则sinnx+cosnx=______． 请问怎么证明cosnx*sinx+sinnx*cosx=sin(n+1)*x? (cos2x-sin2x)/[(1-cos2x)(1-tan2x)] =cos2x/(1-cos2x)=[cosx)^2 （1+cos2x)/2cosx=sin2x/(1-cos2x) 已知向量m=(根号3sinx,cos2x),向量n=(cosx,-1/2),…… 2(cosx)^2-1=cos2x