Indialbedo's blog

I found a brilliant math conclusion
By Indialbedo, history, 2 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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Comments Comments (20)
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2 years ago, # |
  Vote: I like it +63 Vote: I do not like it

Isn't it taught at high school?

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    2 years ago, # ^ |
      Vote: I like it +148 Vote: I do not like it

    Sometimes I am grateful, that I don't study in such schools.

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    2 years ago, # ^ |
      Vote: I like it +83 Vote: I do not like it

    I heard chineseman learn something about it in primary, craaaaaaaazy

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      2 years ago, # ^ |
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      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)

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        2 years ago, # ^ |
          Vote: I like it +34 Vote: I do not like it

        you dont know, Botswana knows

        this is liexiang, do you kno liexiang

        plz support Botswana

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          2 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          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...

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            22 months ago, # ^ |
              Vote: I like it 0 Vote: I do not like it

            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)

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    2 years ago, # ^ |
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    not most of the times sir...

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2 years ago, # |
  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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2 years ago, # |
  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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    2 years ago, # ^ |
      Vote: I like it +34 Vote: I do not like it

    wow! A even nicer proof, clearer than me!

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      2 years ago, # ^ |
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      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.

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

Don't care + didn't ask

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    2 years ago, # ^ |
      Vote: I like it +14 Vote: I do not like it

    who cares about your opinion?

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      2 years ago, # ^ |
        Vote: I like it -58 Vote: I do not like it

      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.

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    2 years ago, # ^ |
      Vote: I like it +13 Vote: I do not like it

    ratio

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

Completely unrelated response:

Spoiler
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2 years ago, # |
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12th class stuff!!

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2 years ago, # |
  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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