I have to show that
$\sum_{n=0}^\infty A_n\cos\left({xn\frac{2\pi}{T}-\theta_n}\right) \equiv \sum_{n=-\infty}^\infty c_n \mathrm{e}^{\left({ixn\frac{2\pi}{T}}\right)}$
I have tried two approaches: firstly I started with the right hand side, used $e^{i\theta} = \cos\theta + i\sin\theta$ to convert the complex exponential into a trigonometric form and then tried to express this in harmonic form to obtain the left hand side. I obtained two inconsistent equations, namely $c_nR\cos\alpha = 1$ and $c_nR\sin\alpha = i$ for some constants $R$ and $\alpha$.
My second approach was to start with the left hand side and use the fact that $\cos\theta = \frac{e^{i\theta}+e^{-i\theta}}{2}$ to obtain
$\sum_{n=0}^\infty \frac{A_n}{2} \left({e^{i\left({xn\frac{2\pi}{T}+\theta_n}\right)}+e^{-i\left({xn\frac{2\pi}{T}+\theta_n}\right)}}\right)$
which I cannot simplify to being equivalent to $\sum_{n=-\infty}^\infty c_n \mathrm{e}^{\left({ixn\frac{2\pi}{T}}\right)}$.
Are my approaches correct, or do I need to rethink the problem? Please can someone give me a hint as to how I can continue?