Indialbedo's blog

I found a brilliant math conclusion
By Indialbedo, history, 5 years ago, In English
$$$\sum_{i=1}^n\sin(ix)=\dfrac{\cos(x/2)-\cos((2n+1)x/2)}{2\sin(x/2)}$$$
$$$\sum_{i=1}^n\sin(ix)=\dfrac{\sin(\frac{n+1}2x)\sin(\frac n 2x)}{\sin(\frac x 2)}$$$

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?

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5 years ago, hide # |
 
Vote: I like it +63 Vote: I do not like it

Isn't it taught at high school?

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5 years ago, hide # |
 
Vote: I like it -11 Vote: I do not like it

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.

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5 years ago, hide # |
 
Vote: I like it +125 Vote: I do not like it

Yes it's a nice identity though it's fairly standard. You can also use complex numbers to get the same result with $$$\sum_{k=1}^{n} \sin(kx) = \text{Im} \sum_{k=1}^{n} e^{i k \theta}$$$

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5 years ago, hide # |
 
Vote: I like it +21 Vote: I do not like it

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, $$$\cot{x} - 2 \cot 2x = \tan x$$$

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5 years ago, hide # |
 
Vote: I like it -102 Vote: I do not like it

Don't care + didn't ask

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5 years ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Completely unrelated response:

Spoiler
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5 years ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

12th class stuff!!

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5 years ago, hide # |
 
Vote: I like it +26 Vote: I do not like it

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

$$$ \sum\limits_{k=1}^n \sin kx = \frac{\cos \frac{x}{2} - \cos (n+\frac{1}{2})x}{2\sin \frac{x}{2}}. $$$
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