Sum of sinus.

In this post we ask the question: what is \displaystyle \sum_{k=0}^n \sin(kx)? It is clear that

\displaystyle \sum_{k=0}^n \exp(kx) = \frac{\exp([n+1]x)-1}{\exp(x)-1}


\displaystyle \sum_{k=0}^n \sin(kx) = \frac{1}{2i} \left\{ \sum_{k=0}^n \exp(ikx) - \sum_{k=0}^n \exp(-ikx) \right\}


\displaystyle \sum_{k=0}^n \sin(kx) = \frac{1}{2i} \left\{ \frac{\exp(i[n+1]x)-1}{\exp(ix)-1} - \frac{\exp(-i[n+1]x)-1}{\exp(-ix)-1} \right\}


\displaystyle \sum_{k=0}^n \sin(kx) = \frac{\sin(nx) - \sin([n+1]x) + \sin(x)}{2-2\cos(x)}


\displaystyle \sum_{k=0}^n \sin(kx) = \frac{1}{2} \sin(nx) + \frac{1}{2} \sin(x) \frac{1-\cos(nx)}{1-\cos(x)}

A simple test:

\displaystyle \displaystyle \sum_{k=0}^{n+1} \sin(kx) - \displaystyle \sum_{k=0}^n \sin(kx) = \sin([n+1]x).


\displaystyle \frac{\sin([n+1]x) - \sin([n+2]x) + \sin(x)}{2-2\cos(x)} - \frac{\sin(nx) - \sin([n+1]x) + \sin(x)}{2-2\cos(x)}

can be written as

\displaystyle \frac{2\sin([n+1]x) - \sin([n+2]x) - \sin(nx)}{2-2\cos(x)}

thus we obtain

\displaystyle \frac{\sin(nx)\left[2\cos(x)-\cos(2x)-1\right] + \cos(nx)\left[2\sin(x)-\sin(2x)\right]}{2-2\cos(x)}

which can be written as

\sin(nx)\cos(x) + \cos(nx)\sin(x) = \sin([n+1]x)

Therefore it is clear that IF this expression \displaystyle \sum_{k=0}^n \sin(kx) = \frac{1}{2} \sin(nx) + \frac{1}{2} \sin(x) \frac{1-\cos(nx)}{1-\cos(x)} is TRUE for n THEN it is also TRUE for n+1. As the expression is TRUE for n=1 it is TRUE for all positive integers n. So we obtain

\displaystyle \sum_{k=0}^n \sin(kx) = \frac{1}{2} \sin(nx) + \frac{1}{2} \sin(x) \frac{1-\cos(nx)}{1-\cos(x)}

Now we may wonder about \displaystyle \int_0^z dx \sin(x). It is clear that

\displaystyle \int_0^z dx \sin(x) = \lim_{n \rightarrow \infty} z/n \left( \frac{1}{2} \sin(z) + \frac{1}{2} \sin(z/n) \frac{1-\cos(z)}{1-\cos(z/n)} \right)


\displaystyle \int_0^z dx \sin(x) = 1-\cos(z)


\displaystyle \lim_{n \rightarrow \infty} \frac{z/n \sin(z/n)}{1-\cos(z/n)} = \lim_{\xi \downarrow 0} \frac{\xi \sin(\xi)}{1-\cos(\xi)} = 1

Hope you liked the post!

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