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[SOLVED] summation of cosine

dwsmith

Well-known member
Feb 1, 2012
1,673
Prove that the $\sum\limits_{k = 0}^n\cos k\theta = \text{Re}\left(\frac{1 - e^{i(n + 1)\theta}}{1 - e^{i\theta}}\right)$ simplifies to
$$
\sum\limits_{k = 0}^n\cos k\theta = \frac{\sin\left(\frac{n + 1}{2}\theta\right)}{\sin\frac{\theta}{2}}\cos\frac{n}{2}\theta
$$

So I have that the real part is
$$
\frac{1-\sin^2\theta + \cos\theta\cos(n+1)\theta-\cos\theta-\cos(n+1)\theta}{4\sin^2\frac{\theta}{2}}
$$

However, I don't see which trig identities will help here.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Prove that the $\sum\limits_{k = 0}^n\cos k\theta = \text{Re}\left(\frac{1 - e^{i(n + 1)\theta}}{1 - e^{i\theta}}\right)$ simplifies to
$$
\sum\limits_{k = 0}^n\cos k\theta = \frac{\sin\left(\frac{n + 1}{2}\theta\right)}{\sin\frac{\theta}{2}}\cos\frac{n}{2}\theta
$$

So I have that the real part is
$$
\frac{1-\sin^2\theta + \cos\theta\cos(n+1)\theta-\cos\theta-\cos(n+1)\theta}{4\sin^2\frac{\theta}{2}}
$$

However, I don't see which trig identities will help here.
Hi dwsmith, :)

With reference to your thread Basics of Fourier Series we have,

\[e^{i\theta} - 1 = 2ie^{i\frac{\theta}{2}}\sin\frac{\theta}{2}\]

\[\Rightarrow e^{i(n+1)\theta} - 1 = 2ie^{i\frac{(n+1)\theta}{2}}\sin\frac{(n+1)\theta}{2}\]

Therefore,

\begin{eqnarray}

\text{Re}\left(\frac{1 - e^{i(n + 1)\theta}}{1 - e^{i\theta}}\right)&=&\text{Re}\left(\frac{2ie^{i \frac{(n+1)\theta}{2}}\sin\frac{(n+1)\theta}{2}}{2ie^{i \frac{\theta}{2}}\sin\frac{\theta}{2}}\right)\\

&=&\text{Re}\left(\frac{e^{\frac{i n\theta}{2}}\sin\frac{(n+1)\theta}{2}}{\sin\frac{ \theta}{2}}\right)\\

&=&\frac{\sin\frac{(n+1) \theta}{2}}{\sin\frac{ \theta}{2}}\text{Re}\left(e^{\frac{i n\theta}{2}}\right)\\

&=&\frac{\cos\frac{n\theta}{2}\sin\frac{(n+1) \theta}{2}}{\sin\frac{ \theta}{2}}

\end{eqnarray}

Kind Regards,
Sudharaka.