### Indialbedo's blog

By Indialbedo, history, 4 months ago, $\sum_{i=1}^n\sin(ix)=\dfrac{\cos(x/2)-\cos((2n+1)x/2)}{2\sin(x/2)}$
$\sum_{i=1}^n\sin(ix)=\dfrac{\sin(\frac{n+1}2x)\sin(\frac n 2x)}{\sin(\frac x 2)}$

Now I give my proof.

notice that $\sin(x)\sin(y)=\frac{1}{2}(cos(x-y)-cos(x+y))$

$\sum_{i=1}^n\sin(ix)=\dfrac{\sin(x/2)(\sin(x)+\sin(2x)+\sin(3x)+\ldots)}{\sin(x/2)}$
$=\dfrac{\cos(x/2)-\cos(3x/2)+\cos(3x/2)-\cos(5x/2)\ldots+\cos((2n+1)x/2)}{2\sin(x/2)}$
$=\dfrac{\cos(x/2)-\cos((2n+1)x/2)}{2\sin(x/2)}$

inverse $\sin(x)\sin(y)=\frac{1}{2}(cos(x-y)-cos(x+y))$ to be $\cos(a)-\cos(b)=-2\sin(\frac{a+b}2)\sin(\frac{a-b}2)$

$\text{LHS}=\dfrac{\sin(\frac{n+1}2x)\sin(\frac n 2x)}{\sin(\frac x 2)}$

maybe a nice idea! What do u think? math, Comments (26)
 » Isn't it taught at high school?
•  » » Sometimes I am grateful, that I don't study in such schools.
•  » » I heard chineseman learn something about it in primary, craaaaaaaazy
•  » » » As a Chinese, I can tell you that most of us don't. (a huuuge Orz to any primary school students kicking my butt at math though)
•  » » » » you dont know, Botswana knowsthis is liexiang, do you kno liexiangplz support Botswana
•  » » » » » I do know about 裂项(liexiang), and although in China primary school students do learn about it, it's restricted to fractional kinds, but this one is trig, and we don't even know about the existence of trig until 9th grade...
•  » » » » » » 裂项(lièxiàng) in Chinese is translated to telescoping. See also: Wikipedia and Section 2.6 in Concrete Mathematics. (IDK why in one Chinese version it was translated into 叠缩)
•  » » » That's too exaggerated，We Just Learn this at junior high school..
•  » » LS Dini Pisa orz
•  » » 4 months ago, # ^ | ← Rev. 3 →   Snob
•  » » LS Dini Pisa orz
•  » » not most of the times sir...
 » Random Chinese guy comes and say: 'Well I learnt it when I was six, lol noob' (not me though)btw this is a pretty neat trig conclusion.
 » Yes it's a nice identity though it's fairly standard. You can also use complex numbers to get the same result with $\sum_{k=1}^{n} \sin(kx) = \text{Im} \sum_{k=1}^{n} e^{i k \theta}$
•  » » wow! A even nicer proof, clearer than me!
•  » » » It's a very well-known thing in optics when you have rays of light interfering, just with real component (sum of cosines) rather than imaginary.
 » Another trigonometric sum I found mildly interesting is, $\sum_{k=0}^{n} 2^k \tan{2^k x} = \cot{x} - 2^{n+1}\cot{2^{n+1} x}$Which comes from, $\cot{x} - 2 \cot 2x = \tan x$
 » Don't care + didn't ask
 » If you sum up $\cos kx$ rather than $\sin kx$, you'll end up with a family of functions known as the Dirichlet kernel.Dirichlet kernel has a great importance in the Fourier analysis, as the convolution of any function with $n$-th Dirichlet kernel will provide the $n$-th degree Fourier approximation of the function.I would personally prefer the complex numbers way to compute the sum: $\sum\limits_{k=1}^n \cos kx + i\sum\limits_{k=1}^n \sin kx = \sum\limits_{k=1}^n e^{ikx} = \frac{e^{ix} - e^{i(n+1)x}}{1-e^{ix}}.$Multiplying the numerator and the denominator by $e^{-\frac{ix}{2}}$ and using $\sin x = \frac{e^{ix} - e^{-ix}}{2i}$ formula, we get $\frac{e^{ix} - e^{i(n+1)x}}{1-e^{ix}} = \frac{i(e^{\frac{ix}{2}} - e^{i(n+\frac{1}{2})x})}{2\sin \frac{x}{2}}.$The real part of the nominator is $\sin(n+\frac{1}{2})x-\sin \frac{x}{2}.$The imaginary part of the nominator is $\cos \frac{x}{2} - \cos (n+\frac{1}{2})x.$Therefore, $\sum\limits_{k=1}^n \cos kx = \frac{\sin(n+\frac{1}{2})x}{2\sin \frac{x}{2}} - \frac{1}{2}$and $\sum\limits_{k=1}^n \sin kx = \frac{\cos \frac{x}{2} - \cos (n+\frac{1}{2})x}{2\sin \frac{x}{2}}.$