Basics of leakage phenomena in DFT and its approximation with sinc function
No, the two peaks don't need to be the same. The discrete-time Fourier transform (DTFT) (samples of which are computed by the DFT) is also not symmetric with respect to the sinusoid's frequency (as explained in Robert's comment). If you compute the DTFT of a truncated sinusoid

$$x[n]=\sin(\omega_0n)(u[n]-u[n-N])$$

you get

$$X(e^{j\omega})=\frac{1}{2j}\left(e^{-j(\omega-\omega_0)\frac{N-1}{2}}\frac{\sin\left(\frac{(\omega-\omega_0)N}{2}\right)}{\sin\left(\frac{\omega-\omega_0}{2}\right)}-e^{-j(\omega+\omega_0)\frac{N-1}{2}}\frac{\sin\left(\frac{(\omega+\omega_0)N}{2}\right)}{\sin\left(\frac{\omega+\omega_0}{2}\right)}\right)$$

The first term in the parentheses has its main contribution around $\omega=\omega_0$, whereas the second term has it around $\omega=-\omega_0$. Note that the two functions are no sinc functions; only for $\omega$ very close to $\omega_0$ (or $-\omega_0$) can they be approximated by a sinc function. And that still wouldn't take into account their overlap.

So your results are correct, and the plot below shows your DFT on top of the DTFT as computed above:



Here's the Matlab/Octave code for the DTFT plot:


N = 32;
k = 8.5;
fs = 32000;
w = 2 * pi * linspace(0,1,500);
w0 = 2*pi*k/N;
X = exp(-1i*(w-w0)*(N-1)/2) .* sin(N/2*(w-w0))./sin((w-w0)/2);
X = X - exp(-1i*(w+w0)*(N-1)/2) .* sin(N/2*(w+w0))./sin((w+w0)/2);
X = X/(2*1i);
plot(w*fs/(2*pi),abs(X))