For a function \(\sin(\omega_0 t)\), it has Fourier Series
\[\sin(\omega_0 t) = \frac{1}{2j} e^{j\omega_0 t} - \frac{1}{2j}e^{-j\omega_0 t}\]For such a function with zero order hold with frequency \(\omega_s\), assuming
\[\omega_s = N \omega_0\]and
\[T_s = \frac{2\pi}{\omega_s}\]then we can calculate its Fourier Series \(\sum a_k e^{j k \omega_0 t}\)
\[\begin{align} a_k &= \frac{\omega_0}{2\pi} \sum_{m=0}^{N-1} \int_{m \cdot T_s}^{(m+1)T_s} \sin(\omega_0 \cdot m \cdot T_s) e^{-jk\omega_0 t} dt \end{align}\]we can calculate, if \(k=0\)
\[\begin{align} a_0 &= \frac{\omega_0}{2\pi} \cdot T_s \sum_{m=0}^{N-1} \sin(\omega_0 \cdot m \cdot T_s)\\ &= \frac{1}{N} \cdot \frac{1}{2j} \sum_{m=0}^{N-1} \bigg( e^{j m \omega_0 T_s} - e^{-jm\omega_0 T_s}\bigg)\\ &= 0 \end{align}\]if \(k \ne 0\)
\[\begin{align} &\int_{m \cdot T_s}^{(m+1)T_s} \sin(\omega_0 \cdot m \cdot T_s) e^{-jk\omega_0 t} dt = \sin(\omega_0 \cdot m \cdot T_s) \int_{m\cdot T_s}^{(m+1)T_s} e^{-jk\omega_0 t}\\ =& \frac{1}{2j}\bigg( e^{jm\omega_0 T_s} - e^{-jm\omega_0 T_s}\bigg) \cdot \frac{1}{-jk\omega_0} \bigg( e^{-jk(m+1)\omega_0 T_s} - e^{-jkm\omega_0 T_s} \bigg)\\ =& \frac{1}{2k \omega_0} \bigg( Terms \bigg) \end{align}\]where
\[\begin{align} Terms =& e^{-jk\omega_0 T_s} \cdot e^{j(1-k)m \omega_0 T_s}\\ & - e^{j(1-k)m \omega_0 T_s}\\ & - e^{-jk\omega_0 T_s} e^{-j(1+k)m \omega_0 T_s}\\ & + e^{-j(1+k)m \omega_0 T_s} \end{align}\]using the theorm from DSSP (FFT chapter), summation of terms will be nonzero only when
\[e^{j(1-k)m\omega_0 T_s} = 1 \quad \text{or} \quad e^{-j(1+k)m\omega_0 T_s} = 1\]that is when
\[k = 1 + qN \quad \text{or} -1 + qN\]the results
\[a_k = \frac{1}{2j} \cdot \frac{1}{1+qN} \cdot \frac{\sin(\pi/N)}{\pi/N} \cdot e^{-j\pi/N} \quad \text{when} \quad k = 1+qN\]and
\[a_k = \frac{1}{2j} \cdot \frac{1}{-1+qN} \cdot \frac{\sin(\pi/N)}{\pi/N} \cdot e^{j\pi/N} \quad \text{when} \quad k = -1+qN\]If subtract the sampling signal from the original signal, the remaining parts has fourier series as (approximately)
\[\begin{equation*} a_k = \left\{ \begin{aligned} & -\frac{\pi}{2N}, \quad \quad \quad \quad \quad \quad \quad k=1 \\ &- \frac{\pi}{2N}, \quad \quad \quad \quad \quad \quad k=-1\\ & \frac{1}{2j} \cdot \frac{1}{1+qN}, \quad \quad \quad k = 1 + qN\\ & \frac{1}{2j} \cdot \frac{1}{-1+qN}, \quad \quad k = -1 + qN\\ & 0, \quad \quad \quad \quad \quad \quad \quad \quad \text{otherwise} \end{aligned} \right. \end{equation*}\]