## Sum of sinus.

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

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

Then

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

So

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

Thus

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

Therefore

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

A simple test:

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

Now

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

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

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

thus

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

As

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

Hope you liked the post!