In the theory of series in mathematical analysis, there is a common type of problem involving the summation of sine or cosine series such as: \cos\theta+\cos 2\theta+\dots+\cos n\theta \tag{1} and \sin\theta+\sin 2\theta+\dots+\sin n\theta \tag{2} The main goal is usually to prove that these sums are bounded. To prove this, one typically finds the closed-form expression for the general term of the sum. Of course, these series also play an important role in physics, where they represent the resultant amplitude of n identical oscillators. In our mathematical analysis textbooks, the derivation is usually completed by multiplying the series by a term \sin\frac{\theta}{2} and then applying product-to-sum formulas. Admittedly, if one is restricted to the realm of real numbers, this might be the only derivation technique available. However, the calculation process itself is not simple and is difficult to remember; I found this quite distressing during my sophomore year. A few days ago, while reading Feynman’s book, I thought of a derivation technique using complex numbers. It is strange how simple this technique is—writing it down makes this article seem almost trivial—yet I had never noticed it before! It seems my skills were still shallow, and I need to practice more.
Sometimes, calculating more things at once is actually faster than calculating a single part. Let us consider (1) and (2) simultaneously by calculating: \cos\theta+\cos 2\theta+\dots+\cos n\theta+i\left(\sin\theta+\sin 2\theta+\dots+\sin n\theta\right) Using Euler’s formula, the above expression is simply: e^{i\theta}+e^{2i\theta}+\dots+e^{ni\theta} This is just the sum of a geometric series! It is easy to obtain: e^{i\theta}\left(\frac{e^{ni\theta}-1}{e^{i\theta}-1}\right)=e^{(n+1)i\theta /2}\left( \frac{e^{ni\theta /2} - e^{ - n i\theta /2}}{e^{i\theta /2} - e^{ - i\theta /2}} \right) The result inside the parentheses is simply: \frac{\sin\frac{n\theta}{2}}{\sin\frac{\theta}{2}} Therefore, the overall result is: \left(\cos\frac{(n+1)\theta}{2}+i\sin\frac{(n+1)\theta}{2}\right)\frac{\sin\frac{n\theta}{2}}{\sin\frac{\theta}{2}} According to the principle of equating real and imaginary parts, we have: \cos\theta+\cos 2\theta+\dots+\cos n\theta=\cos\frac{(n+1)\theta}{2}\frac{\sin\frac{n\theta}{2}}{\sin\frac{\theta}{2}} and \sin\theta+\sin 2\theta+\dots+\sin n\theta=\sin\frac{(n+1)\theta}{2}\frac{\sin\frac{n\theta}{2}}{\sin\frac{\theta}{2}}
The entire process is so concise, almost seamless. There are no profound or mysterious techniques involved, so it is particularly strange that I only noticed it now.
Similarly, one can also derive the summation formula for: \cos(\omega+\theta)+\cos(\omega+2\theta)+\dots+\cos (\omega+n\theta) (it just involves multiplying by an additional phase factor...)
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