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I have seen Lagrange's formula for the sum of $\sin(n)$ from $1$ to $n$ during one of my classes last week, but I never saw how it came to be. I tried googling it to find a proof but couldn't seem to find any as it kept bringing up his other work instead and just statements of the formula rather than derivations/proofs.

I'm really interested to see where it comes from so if anyone has any nice proofs that would be appreciated.

Thank you!

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    $\begingroup$ This might help. $\endgroup$
    – Mufasa
    Jan 25, 2015 at 16:21
  • $\begingroup$ @Mufasa +1 for the rec ... I hate how that website is formatted though :/ $\endgroup$ Jan 25, 2015 at 16:32

4 Answers 4

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we will use the fact that $$2 \sin 1 \sin k = \cos(k-1) -\cos(k + 1)$$let $S = \sin 1 + \sin 2 \cdots + \sin n,$ then

$\begin{align} 2S \sin 1 &= 2\sin 1 \sin 1+ 2 \sin 1 \sin 2 + 2 \sin 1 \sin 3\cdots +2 \sin 1 \sin n \\ &=(1 - \cos 2) +(\cos 2 - \cos3) +(\cos 3 - \cos 4)+\cdots +(\cos (n- 1) - \cos ( n + 1)\\ &=1 - \cos (n+1)=2\sin^2\left(\frac{n+1}{2}\right). \end{align}$

that is $$ \sin 1 + \sin 2 + \cdots + \sin n = \dfrac{\sin^2\left(\frac{n+1}{2}\right)}{\sin 1}$$

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  • $\begingroup$ The result $\frac{\sin^2(\frac{n+1}{2})}{\sin 1}$ is wrong for $n=0$ and other values of $n$. Your telescoping was flawed. $\endgroup$
    – Somos
    May 13 at 18:04
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Easiest way I know is to use complex arithemtic. We have to just know that $e^{ik}=\cos k+i\sin k.$ From that $\sin k=\frac{e^{ik}-e^{-ik}}{2i}.$ Hence $$\sum_{k=1}^n\sin k=\sum_{k=1}^n\frac{e^{ik}-e^{-ik}}{2i}=\frac{\sum_{k=1}^n (e^{i})^k-\sum_{k=1}^n (e^{-i})^k}{2i}.$$ Now just use formula for geometric series, i.e $$\sum_{k=1}^n a^k=a\frac{1-a^n}{1-a}.$$ From that $$\frac{\sum_{k=1}^n (e^{i})^k-\sum_{k=1}^n (e^{-i})^k}{2i}=\frac{e^{i}\frac{1-e^{in}}{1-e^i}-e^{-i}\frac{1-e^{-in}}{1-e^{-i}}}{2i}=\frac{1}{2i}\frac{(e^i-1)(1-e^{in})-(e^{-i}-1)(1-e^{-in})}{(1-e^i)(1-e^{-i})}=\frac{1}{2i}\frac{e^i-e^{i(n+1)}+e^{in}-1-e^{-i}+e^{-i(n+1)}+1-e^{-in}}{1-e^i-e^{-i}+1}=\frac{\sin 1 +\sin n-\sin (n+1)}{2- 2\cos 1}.$$ Is it the result you wanted? If not I could simplify this otherwise to fit your formula.

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  • $\begingroup$ It was slightly different overall but that was mainly just notation. Thanks for this it helped a lot. $\endgroup$ Jan 25, 2015 at 17:27
  • $\begingroup$ In my opinion @abel 's answer is great. My is lame. $\endgroup$ Jan 25, 2015 at 17:28
  • $\begingroup$ shouldn't there be n instead of k when the sigmas disappeared? $\endgroup$ Dec 30, 2020 at 1:36
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    $\begingroup$ At least as things stand right at this moment, this answer is correct whereas the top-voted one -- whose answer is $$ \dfrac{\sin^2\left(\frac{n+1}{2}\right)}{\sin 1} $$ -- is incorrect. $\endgroup$
    – David K
    May 13 at 23:33
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Here is how you advance.

$$ \sum_{k=1}^{n} \sin(n) = \frac{1}{2i} \sum_{k=1}^{n} (e^{in} - e^{-in} ) =\dots\,.$$

To finish the problem you need the following identity

$$ \sum_{k=1}^{n} x^k = {\frac {{x}^{n+1}-x}{x-1}} $$

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A comment on a follow-up question suggests using the following identity:

$$ 2 \sin \left(\tfrac12\right) \sin (k) = \cos\left(k - \tfrac12\right) - \cos\left(k + \tfrac12\right). $$

Then \begin{align} 2 \sin \left(\tfrac12\right)&\left(\sin(1) + \sin(2) + \sin(3) + \cdots + \sin(n-1) + \sin(n)\right) \\ &= \left(\cos\left(\tfrac12\right) - \cos\left(\tfrac32\right)\right) +\left(\cos\left(\tfrac32\right) - \cos\left(\tfrac52\right)\right) +\left(\cos\left(\tfrac52\right) - \cos\left(\tfrac72\right)\right)\\ & \qquad + \cdots +\left(\cos\left(n - \tfrac32\right) - \cos\left(n - \tfrac12\right)\right) +\left(\cos\left(n - \tfrac12\right) - \cos\left(n + \tfrac12\right)\right) \\ &= \cos\left(\tfrac12\right) - \cos\left(n + \tfrac12\right) \\ &= \sin\left(\tfrac{n+1}2\right) \sin\left(\tfrac n2\right). \end{align}

Therefore $$ \sin(1) + \sin(2) + \cdots + \sin(n) = \frac{\sin\left(\frac{n+1}2\right) \sin\left(\frac n2\right)} {\sin\left(\frac12\right)}. $$

There is another derivation of this formula (using $2 \sin(1) \sin(k) = \cos(k-1) -\cos(k + 1)$) in another answer.

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