Why does this FFT based signal deliver a sinusoidal result?
Let me see if I understood you correctly. You have one period of a sinusoidal signal:

$$x[n]=\sin\left (\frac{2\pi n}{N}\right ),\quad n=0,1,\ldots N-1$$

Its DFT is given by

$$X[k]=\sum_{n=0}^{N-1}x[n]e^{-j2\pi nk/N}$$

For $k=1$ we have

$$X[1]=\sum_{n=0}^{N-1}x[n]e^{-j2\pi n/N}$$

For the given signal $x[n]$ we get

$$X[1]=\frac{N}{2j}$$

For the left-shifted versions of $x[n]$ we get

$$X_l[1]=\sum_{n=0}^{N-1}x[n+l]e^{-j2\pi n/N}=e^{j2\pi l/N}\sum_{n=0}^{N-1}x[n]e^{-j2\pi n/N}=e^{j2\pi l/N}\frac{N}{2j}$$

So we have

$$\Re\{X_l[1]\}=\Re\left\{e^{j2\pi l/N}\frac{N}{2j}\right\}=\frac{N}{2}\sin\left (\frac{2\pi l}{N}\right )=\frac{N}{2}x[l],\quad l=0,1,\ldots, N-1$$

Q.E.D.