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How can I calculate the sum of $\sin\alpha + \sin3\alpha +\cdots+ \sin(2n-1)\alpha$ using De Moivre's formula and geometric series?

Attempt: I know that $\mathrm{cis}\alpha + \mathrm{cis}3\alpha +\cdots+ \mathrm{cis}(2n-1)\alpha$ is a geometric series. So, $$\sum_{k=1}^{2n-1} r^k(\cos \alpha+i\sin \alpha)^k=\sum_{k=1}^{2n-1} z^k,\qquad\text{with}\ z=r(\cos \alpha+i\sin \alpha)$$ But I don't know how to continue.

Arturo Magidin
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user447085
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  • The formula for the sum of a geometric series in terms of the base, first and last terms, is well known. You can probably look it up... – Arturo Magidin Nov 07 '22 at 20:09
  • You only want odd powers of $z$, i.e. the sum should be $z\sum_{k=0}^{n-1}z^{2k}$. At least $=1$ makes things simpler. – J.G. Nov 07 '22 at 20:13
  • Sorry, $r=1$. Anyway, you should find the answer is of the form $\sin^2x/\sin y$ (I leave you to find what $x,,y$ are). – J.G. Nov 07 '22 at 20:25
  • Also here: https://math.stackexchange.com/a/1967352/42969 – Martin R Nov 08 '22 at 09:11

2 Answers2

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As you have more or less noted, the result of interest can be expressed in the form \begin{align} \sum_{k=0}^{n-1} \sin([2k+1]\alpha) &= \sum_{k=0}^{n-1} \operatorname{Im}[\mathrm{cis}([2k+1]\alpha)] \\ & = \operatorname{Im}\left[\sum_{k=0}^{n-1}[\mathrm{cis}([2k+1]\alpha)]\right] \\ &= \operatorname{Im}\left[\sum_{k=0}^{n-1}z^{2k+1}\right] = \operatorname{Im}\left[z\sum_{k=0}^{n-1}(z^2)^{k}\right], \end{align} where $z = \mathrm{cis}(\alpha)$. With that, we can rewrite $\sum_{k=0}^{n-1}(z^2)^{k} = \frac{1-z^{2n}}{1-z^2}$, and the rest comes down to routine simplification. A helpful step in this simplification is to rewrite $$ z\cdot \frac{1 - z^{2n}}{1-z^2} = \frac{1 - z^{2n}}{z^{-1}-z}, $$ noting that $z^{-1} - z$ is a purely imaginary number.

Ben Grossmann
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Rather than considering a sum of an odd number ($2n\pm1$) of terms, put the odd number in the argument to $\sin$ and take a sum of $n$ terms. For $n\in\Bbb N\cup\{0\}$,

$$\begin{align*} S &= \sin(\alpha) + \sin(3\alpha) + \sin(5\alpha) + \cdots + \sin((2n+1)\alpha) \\[1ex] &= \Im e^{i\alpha} + \Im e^{i3\alpha} + \Im e^{i5\alpha} + \cdots + \Im e^{i(2n+1)\alpha} \tag{1} \\[1ex] &= \Im e^{i\alpha} \sum_{k=0}^n \left(e^{i2\alpha}\right)^k \\[1ex] &= \Im e^{i\alpha} \frac{1 - \left(e^{i2\alpha}\right)^{n+1}}{1 - e^{i2\alpha}} \tag{2} \\[1ex] &= \Im \frac{i - i\cos((2n+2)\alpha) - i^2\sin((2n+2)\alpha)}{2\sin(\alpha)} \tag{3} \\[1ex] &= \frac{1 - \cos((2n+2)\alpha)}{2\sin(\alpha)} \end{align*}$$

  • $(1)$ : $\Im e^{ix} = \sin(x)$
  • $(2)$ : $|r|<1 \implies \sum\limits_{k=0}^n r^k = \frac{1-r^{n+1}}{1-r}$ (geometric series)
  • $(3)$ : $\left(e^{ix}\right)^n = e^{inx}$ (de Moivre)
user170231
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