# [SOLVED]Basics of Fourier Series

#### dwsmith

##### Well-known member
If you write
$$e^{ik\theta} = \cos k\theta + i\sin k\theta,$$
then $\sum\limits_{k = 0}^ne^{ik\theta} = \frac{1 - e^{i(n + 1)\theta}}{1 - e^{i\theta}}$ yields two real sums
$$\sum\limits_{k = 0}^n\cos k\theta = \text{Re}\left(\frac{1 - e^{i(n + 1)\theta}}{1 - e^{i\theta}}\right)$$
and
$$\sum\limits_{k = 0}^n\sin k\theta = \text{Im}\left(\frac{1 - e^{i(n + 1)\theta}}{1 - e^{i\theta}}\right).$$
Prove that
$$e^{i\theta} - 1 = 2ie^{i\frac{\theta}{2}}\sin\frac{\theta}{2}.$$

Not to sure on what to do with this one.

#### Sudharaka

##### Well-known member
MHB Math Helper
If you write
$$e^{ik\theta} = \cos k\theta + i\sin k\theta,$$
then $\sum\limits_{k = 0}^ne^{ik\theta} = \frac{1 - e^{i(n + 1)\theta}}{1 - e^{i\theta}}$ yields two real sums
$$\sum\limits_{k = 0}^n\cos k\theta = \text{Re}\left(\frac{1 - e^{i(n + 1)\theta}}{1 - e^{i\theta}}\right)$$
and
$$\sum\limits_{k = 0}^n\sin k\theta = \text{Im}\left(\frac{1 - e^{i(n + 1)\theta}}{1 - e^{i\theta}}\right).$$
Prove that
$$e^{i\theta} - 1 = 2ie^{i\frac{\theta}{2}}\sin\frac{\theta}{2}.$$

Not to sure on what to do with this one.
Hi dwsmith,

\begin{eqnarray}

2ie^{i\frac{\theta}{2}}\sin\frac{\theta}{2}&=&2i \sin\frac{\theta}{2}\left(\cos \frac{\theta}{2}+i\sin \frac{\theta}{2}\right)\\

&=&i\sin \theta-2\sin^{2} \frac{\theta}{2}\\

&=&i\sin \theta+\left(1-2\sin^{2} \frac{\theta}{2}\right)-1\\

&=&\left(i\sin \theta+\cos \theta\right)-1\\

&=&e^{i\theta} - 1

\end{eqnarray}

Kind Regards,
Sudharaka.