Fourier series coefficient for sin(wt+theta)

Question

Am trying to find the Fourier series coefficient ck for the following function \begin{equation} x\left(t\right)=\sin\left(wt+\theta \right) \end{equation} Here is my work \begin{equation} c_k=\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}\:x\left(t\right)e^{-jkwt}dt \end{equation} Now x(t) can be written as \begin{equation} x\left(t\right)=\frac{\left(e^{j\left(wt+\theta \right)}-e^{-j\left(wt+\theta \:\right)}\right)}{2j} \end{equation} However: \begin{equation} c_k=\frac{1}{T}\int _{-\frac{T}{2}}^{\frac{T}{2}}\:\frac{\left(e^{j\left(wt+\theta \:\right)}-e^{-j\left(wt+\theta \:\:\right)}\right)}{2j}e^{-jkwt}dt \end{equation} would look very messy and am ending up with 2 sinc functions for my answer that I don't even know what do with. I can't put all the steps here but if somebody would steer me in the right direction I would really appreciate it.

Answer 1

Just pointers:

1. pull the \$\frac1{2j}\$ out of your integral, it's just a constant.
2. The integral is a linear operation – use that to split your sum, so you have two integrals; one over \$e^{j\omega t + j\theta}e^{-jk\omega t}\$ and one over \$e^{-j\omega t - j\theta}e^{-jk\omega t}\$.
3. \$e^{j\omega t + \theta}=e^{j\omega t}\cdot e^{j\theta}\$; this applies to both integral. \$e^{\pm j\theta}\$ is a constant and can be pulled out of your integral.
4. You end up with an integrated product: \$e^{\pm j\omega t}\cdot e^{-jk\omega t}\$. Combine that into a single exponential.
5. Notice how that integral is over a length of \$T\$, which is one period. It can only be non-zero for a single value.

Answer 2

From the double angle formulas, multiplying any harmonically related sine and/or cosine terms will result in zero when integrated over a period of the fundamental, except for \$sin^2 k\theta\$ and \$cos^2 k\theta\$.