数学归纳法的证明题用数学归纳法证明：1 sin x+2 sin 2x+…+n sin nx=sin[(n+1)x]/4s

数学归纳法的证明题
用数学归纳法证明：1 sin x+2 sin 2x+…+n sin nx=sin[(n+1)x]/4sin^2(x/2)-(n+1)cos{[(2n+1)/2]x}/2sin(x/2)
其中sin(x/2)≠0

前面步骤省略
设:1sin(x)+2sin(2x)+…+nsin(nx)=sin[(n+1)x]/[4sin^2(x/2)]-(n+1)cos[(2n+1)x/2]/[2sin(x/2)]
则需要sin[(n+2)x]/[4sin^2(x/2)]-(n+2)cos[(2n+3)x/2]/[2sin(x/2)]
=sin[(n+1)x]/[4sin^2(x/2)]-(n+1)cos[(2n+1)x/2]/[2sin(x/2)]+(n+1)sin[(n+1)x]
则需要sin[(n+2)x]-(n+2)cos[(2n+3)x/2][2sin(x/2)]
=sin[(n+1)x]-(n+1)cos[(2n+1)x/2][2sin(x/2)]+(n+1)sin[(n+1)x][4sin^2(x/2)]
则需要sin[(n+2)x]-(n+2)cos[(2n+3)x/2][2sin(x/2)]
=sin[(n+1)x]-(n+2)cos[(2n+1)x/2][2sin(x/2)]+cos[(2n+1)x/2][2sin(x/2)]+(n+1)sin[(n+1)x][4sin^2(x/2)]
则需要sin[(n+2)x]-sin[(n+1)x]+(n+2)cos[(2n+1)x/2][2sin(x/2)]-(n+2)cos[(2n+3)x/2][2sin(x/2)]
=cos[(2n+1)x/2][2sin(x/2)]+(n+1)sin[(n+1)x][4sin^2(x/2)]
则需要sin[(n+2)x]-sin[(n+1)x]+(n+2)[2sin(x/2)]{cos[(2n+1)x/2]-cos[(2n+3)x/2]}
=cos[(2n+1)x/2][2sin(x/2)]+(n+1)sin[(n+1)x][4sin^2(x/2)]
则需要2cos[(2n+3)x/2]sin(x/2)+(n+2)[2sin(x/2)][2sin[(n+1)x]sin(x/2)]
=cos[(2n+1)x/2][2sin(x/2)]+(n+1)sin[(n+1)x][4sin^2(x/2)]
则需要(n+2)[sin[(n+1)x][4sin^2(x/2)]-(n+1)sin[(n+1)x][4sin^2(x/2)]
={cos[(2n+1)x/2]-cos[(2n+3)x/2]}[2sin(x/2)]
则需要(n+2)[sin[(n+1)x][4sin^2(x/2)]-(n+1)sin[(n+1)x][4sin^2(x/2)]
=2sin[(n+1)x]sin(x/2)[2sin(x/2)]
则需要(n+2)[sin[(n+1)x][4sin^2(x/2)]-(n+1)sin[(n+1)x][4sin^2(x/2)]
=sin[(n+1)x][4sin^2(x/2)]
很明显,上式是左右相等的
所以问题得证.
需要用到的三角函数：
sinθ-sinφ=2cos[(θ+φ)/2]sin[(θ-φ)/2],则sin[(n+2)x]-sin[(n+1)x]=2cos[(2n+3)x/2]sin(x/2)
cosθ-cosφ=-2sin[(θ+φ)/2]sin[(θ-φ)/2],则cos[(2n+1)x/2]-cos[(2n+3)x/2]=2sin[(n+1)x]sin(x/2)