### 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?

• +61

 » 4 months ago, # |   +63 Isn't it taught at high school?
•  » » 4 months ago, # ^ |   +148 Sometimes I am grateful, that I don't study in such schools.
•  » » 4 months ago, # ^ |   +83 I heard chineseman learn something about it in primary, craaaaaaaazy
•  » » » 4 months ago, # ^ |   0 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)
•  » » » » 4 months ago, # ^ |   +34 you dont know, Botswana knowsthis is liexiang, do you kno liexiangplz support Botswana
•  » » » » » 4 months ago, # ^ |   0 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...
•  » » » » » » 7 weeks ago, # ^ |   0 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)
•  » » » 7 weeks ago, # ^ |   -31 That's too exaggerated，We Just Learn this at junior high school..
•  » » 4 months ago, # ^ |   0 LS Dini Pisa orz
•  » » 4 months ago, # ^ | ← Rev. 3 →   0 Snob
•  » » 4 months ago, # ^ |   0 LS Dini Pisa orz
•  » » 4 months ago, # ^ |   0 not most of the times sir...
 » 4 months ago, # |   -11 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.
 » 4 months ago, # |   +125 Yes it's a nice identity though it's fairly standard. You can also use complex numbers to get the same result with $\sum_{k=1}^{n} \sin(kx) = \text{Im} \sum_{k=1}^{n} e^{i k \theta}$
•  » » 4 months ago, # ^ |   +34 wow! A even nicer proof, clearer than me!
•  » » » 4 months ago, # ^ |   0 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.
 » 4 months ago, # |   +21 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, $\cot{x} - 2 \cot 2x = \tan x$
 » 4 months ago, # |   -102 Don't care + didn't ask
•  » » 4 months ago, # ^ |   +14 who cares about your opinion?
•  » » » 4 months ago, # ^ |   -58 Well, the author asked "What do u think?", so maybe he does care. My point was that trig is awful and kind of not related to cp. This is, of course, very subjective.
•  » » » » 4 months ago, # ^ |   0 yeah but the author didn't ask if you care or no, and he also didn't ask if u asked.do u even understand what does "What do u think?" even mean lol?
•  » » 4 months ago, # ^ |   +13 ratio
 » 4 months ago, # |   0 Completely unrelated response: SpoilerAlbedo is love
 » 4 months ago, # |   0 12th class stuff!!
•  » » 4 months ago, # ^ |   0 For us Asians.
 » 4 months ago, # |   +26 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 $\sum\limits_{k=1}^n \sin kx = \frac{\cos \frac{x}{2} - \cos (n+\frac{1}{2})x}{2\sin \frac{x}{2}}.$