Please help me to solve this problem

Use mathematical induction to prove that
sum(for j=1 to j<=n)cosjx= cos[(n+1)x/2] sin(nx/2) sin(x/2) whenever n is a positive integer and
sin(x/2) is not equal to 0.
Thanks much!!!

∑(j=1,2,3...n) cos(jx)= cos([n+1]x/2)sin(nx/2)/sin(x/2).

1) n=1
∑(j=1) cos(jx)= cos([1+1]x/2)sin(x/2)/sin(x/2)
cos(x)=cos(x) ok!
2) let true for n=k
∑(j=1,2,3...k) cos(jx)= cos([k+1]x/2)sin(kx/2)/sin(x/2).
3) prove to n=k+1
∑(j=1,2,3...k+1) cos(jx)= ∑(j=1,2,3...k) cos(jx) +cos((k+1)x) =
=cos([k+1]x/2)sin(kx/2)/sin(x/2) +cos((k+1)x) =
=[cos([k+1]x/2)sin(kx/2)+ cos((k+1)x)sin(x/2)]/sin(x/2)

sin(kx/2)cos([k+1]x/2)= ½[sin(kx/2-[k+1]x/2)+sin(kx/2+[k+1]x/2)] =
½[sin(kx+x/2)-sin(x/2)]

sin(x/2)cos((k+1)x)= ½[sin(x/2-(k+1)x)+sin(x/2+(k+1)x)] =
=½[-sin(kx+x/2)+sin(3x/2+kx)]

hence
[cos([k+1]x/2)sin(kx/2)+ cos((k+1)x)sin(x/2)]/sin(x/2) =
=½[sin(kx+x/2)-sin(x/2)-sin(kx+x/2)+ sin(3x/2+kx)]/sin(x/2) =
=½[sin(3x/2+kx)-sin(x/2)]/sin(x/2)

sin(3x/2+kx)-sin(x/2) =2sin[((3x/2+kx)- x/2)/2]cos[((3x/2+kx)+x/2)/2] =
=2sin[(1+k)x/2]cos[(2+k)x/2]

hence
½[sin(3x/2+kx)-sin(x/2)]/sin(x/2) =½[2sin[(1+k)x/2]cos[(2+k)x/2]]/sin(x/2) =
=cos[(2+k)x/2]sin[(1+k)x/2]/sin(x/2)
Ok!