..n*sin^n-1 x*cosx*cosnx+nsin^n x*(-sinnx)化简是怎么得到nsin^n-1*cos(n+1)x

sinx+cosx=1 求sin n次方x+cos n次方x 求证:cosx+cos2x+...+cosnx={[cos(n+1)x/2]*[sin(n/2)x]}/[sin(x/2)] 求证:cosx+cos2x+...+cosnx={[cos(n+1)x/2]*[sin(n/2)x]}/[sin(x/2)] 请求sinx cos nx + cosx sin nx变成sin(n+1)x的步骤 ..n*sin^n-1 x*cosx*cosnx+nsin^n x*(-sinnx)化简是怎么得到nsin^n-1*cos(n+1)x 对于任意实数x和整数n,已知f(sinx)=sin(4n+1)x,求f(cosx)f(cosx)=f(sin(90°-x))=sin(4n+1)(90°-x)=sin[360°n+90°-(4n+1)x] =sin[90°-(4n+1)x]=cos(4n+1)x 以下这几步看不懂sin[360°n+90°-(4n+1)x] =sin[90°-(4n+1)x] 已知x∈R ,n∈Z,且f(sinx)=sin(4n+1)x,则f(cosx)= 对任意实数x和整数n,已知f(sinx)=sin(4n+1)x,求f(cosx) 已知x∈R,n∈Z,且f(sinx)=sin(4n+1)x,则f(cosx)=? 已知x属于实数,n属于整数,且f（sinx）=sin(4n+1)x,求f(cosx). 请问怎么证明cosnx*sinx+sinnx*cosx=sin(n+1)*x? cosnx*sinx+sinnx*cosx=sin（n+1）x.是怎么算的? cos(n+1)x=cos(nx)cosx-sin(nx)sinx这个式子是怎推出来的? 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)X-SIN(2/X)}/SIN(2/X)|怎么证明 若函数f(x)=[(2sin(x+π/6)+x^4+x)/(x^4+cosx)]+1在[-π/2,π/2]上的最大值与最小值分别为M和N,则有（ ）A.M-N=2B.M+N=2C.M-N=4D.M+N=4 (-1)^n*cosx=cos(x+nπ) 已知当X趋近于0时,x^2ln(1+x^2)是sin^n(x)的高阶无穷小,sin^n(x)又是1-cosx的高阶无穷小,求正整数n