Sum of sinusoids

Question

Let $f_k : = \sin(\omega t+ k \alpha)$ for $k \in \mathbb{N}_0$ and $\alpha, \omega \in \mathbb{R}$ fixed. The following formula is convenient in some engineering applications: $$\sum_{j=0}^N f_k = \frac{\sin(\frac{N}{2}\alpha)}{\sin(\frac{1}{2}\alpha)}\sin \left(\omega t + \frac{N-1}{2}\alpha\right).$$

As a fun exercise, I have been attempting to prove this useful formula, but cannot seem to get it for $N > 1$. For $N=1$, we have the following elementary argument:

Indeed, \begin{eqnarray*} \sin(\omega t) + \sin(\omega t + \alpha) &=& \text{Im}( (1 + e^{i \alpha})e^{i \omega t}), \end{eqnarray*}

and \begin{eqnarray*} 1+e^{i\alpha} &=& \frac{\sin(\alpha)}{\sin(\frac{\alpha}{2})}e^{i \alpha/2}. \end{eqnarray*}

Can someone help me with the general case?

Answer 1

HINT:

You have already proven the base case. Assume that it is true for some $K$, so that $$\sin\omega t+\sin(\omega t+\alpha)+\cdots+\sin(\omega t+K\alpha)= \frac{\sin(\frac{K}{2}\alpha)}{\sin(\frac{1}{2}\alpha)}\sin \left(\omega t + \frac{K-1}{2}\alpha\right).$$ Then for $K+1$, we have that $$\sum_{k=0}^{K+1}f_k=\sin(\omega t+(K+1)\alpha)+\frac{\sin(\frac{K}{2}\alpha)}{\sin(\frac{1}{2}\alpha)}\sin \left(\omega t + \frac{K-1}{2}\alpha\right)$$ and we wish to show that $$\sum_{k=0}^{K+1}f_k=\frac{\sin(\frac{K+1}{2}\alpha)}{\sin(\frac{1}{2}\alpha)}\sin \left(\omega t + \frac{K}{2}\alpha\right).$$ This is equivalent to proving that $$\small \sin\frac\alpha2\sin(\omega t+(K+1)\alpha)+\sin\frac{K\alpha}2\sin \left(\omega t + \frac{K-1}{2}\alpha\right)=\sin\frac{(K+1)\alpha}2\sin \left(\omega t + \frac{K}{2}\alpha\right)$$ and use the trigonometric identity $\sin A\sin B=\frac12[\cos(A-B)-\cos(A+B)]$.

Answer 2

Hint: $f_k$ is the imaginary part of $e^{i\omega t} e^{ i k \alpha}$ and $\sum _k e^{ i k \alpha}$ can be computed using the formula $\sum\limits_{k=0}^{N}r^{k}=\frac {1-r^{N+1}} {1-r}$

Answer 3

For convenience, we set $\alpha=2\beta$.

$$S:=\sum_{k=0}^{n-1}e^{i(\omega t+2k\beta)}=\sum_{k=0}^{n-1}z^ke^{i\omega t}=\frac{z^n-1}{z-1}e^{i\omega t},$$ where $z:=e^{2i\beta}$.

Now $$z-1=\cos(2\beta)-1+i\sin(2\beta)=-2\sin^2\beta +2i\sin\beta \cos\beta =2i\sin\beta\,e^{i\beta }$$

and

$$S=\frac{\sin(n\beta)\,e^{in\beta }}{\sin\beta\,e^{i\beta }}e^{i\omega t}=\frac{\sin(n\beta)}{\sin\beta }e^{i(\omega t+(n-1)\beta)}.$$