MHB Proving the Basic Identity of Fourier Series

[Dustinsfl]

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]

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.