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By considering $\sum_{r=1}^n z^{2r-1}$ where z= $\cos\theta + i\sin\theta$, show that if $\sin\theta$ $\neq$ 0, $$\sum_{r=1}^n \sin(2r-1)\theta=\frac{\sin^2n\theta}{\sin\theta}$$

I couldn't solve this at first but with some hints some of you gave, I was able to come up with my own solution. Here it is:

First, we want to consider what is given in the question, $$\begin{align} \sum_{r=1}^n z^{2r-1} & =z+z^3+z^5...\\ & = z + z(z^2)+z^3(z^2)+...\\ \end{align}$$ In Geometric Progression, sum is given by $$\sum_{r=1}^n z^{2r-1}=S_n = \frac{a (1-r^n)}{1-r}=\frac{z(1-(z^2)^n)}{1-z^2} = \frac {z-z^{2n+1}}{1-z^2}=\frac {1-z^{2n}}{z^{-1}-z} $$ Now, substitute $z = \cos \theta +i \sin \theta$

$$\begin{align} \sum_{r=1}^n z^{2r-1} & =\frac{1-(\cos (2n\theta) + i \sin(2n\theta))}{\cos\theta - i\sin\theta-(\cos\theta+i\sin\theta)} \\ & = \frac{1-\cos (2n\theta) - i \sin(2n\theta)}{-2i\sin\theta} \cdot \frac{(i\sin\theta)}{(i\sin\theta)}\\ & = \frac{i\sin\theta-i\sin\theta \cos (2n\theta) + \sin\theta \sin(2n\theta)}{2\sin^2\theta} \\ \end{align}$$ Equating imaginary parts, $$\sum_{r=1}^n \sin(2r-1)\theta = \frac{\sin\theta (1 - \cos(2n\theta))}{2\sin^2\theta} = \frac{\sin^2n\theta}{2\sin\theta} $$

Martin Sleziak
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Lily L
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    Hint: There are two ways to write $\sum_{r=1}^nz^{2r-1}$. One way is to remember that $z^a=\cos(a\theta)+i\sin(a\theta)$ and the other way is to look at this as a geometric series and find the sum of the geometric series. The summation that you're considering will show up as the imaginary part of the expression. – Michael Burr Oct 04 '15 at 14:20
  • Use $\sin(x)=(e^{ix}-e^{-ix})/2i$ afterwards apply the geometric series to both resulting parts – tired Oct 04 '15 at 14:20
  • Thank you @MichaelBurr . Your hint had helped me solve my problem. (It took me some time to understand completely) – Lily L Oct 09 '15 at 10:21

2 Answers2

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An alternative (maybe easier) approach. Since: $$ 2\sin(\theta)\sin((2r-1)\theta) = \cos((2r-2)\theta)-\cos(2r\theta), $$ the original sum, multiplied by $2\sin\theta$, becomes a telescopic sum, whose value equals $1-\cos(2n\theta) = 2\sin^2(n\theta).$

Jack D'Aurizio
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Using $$\cos \phi+i\sin \phi = e^{i\phi}$$ Then $$\cos (2r-1)\phi+i\sin (2r-1)\phi = e^{i(2r-1)\phi}$$

We have to calculate $$\displaystyle S = \sum_{r=1}^n \sin(2r-1)\theta=\Im \left[\sum^{n}_{r=1}e^{(2r-1)\phi}\right]$$

So $$\displaystyle S = e^{i\phi}+e^{i3\phi}+e^{i5\phi}+.............+e^{i(2n-1)\phi}.............(1)$$

Now Multiply both side by $e^{i2\phi}\;,$ Then

$$\displaystyle S\cdot e^{i2\phi} = e^{i3\phi}+e^{i5\phi}+........+e^{i(2n+1)\phi}...................(2)$$

So $$\displaystyle (1-e^{i2\phi})S=e^{i\phi}-e^{i(2n+1)\phi}\Rightarrow S = \frac{e^{i(2n+1)\phi}-e^{i\phi}}{e^{i2\phi}-1} = \frac{e^{i2n\phi}-1}{e^{i\phi}-e^{-i\phi}}$$

Now Using $$e^{i\phi} = \cos \phi+i\sin \phi$$ and $$e^{-i\phi} = \cos \phi-i\sin \phi$$.

So we get $$\displaystyle e^{i\phi}-e^{-i\phi} = 2i\sin \phi$$ and $$e^{i2n\phi} = \cos 2n\phi+i\sin 2n\phi$$

So we get $$\displaystyle S=\frac{\cos 2n\phi-1+i\sin 2n\phi}{2i\sin \phi} = \frac{i\sin 2n\phi-2\sin^2 n\phi}{2i\sin \phi} = \frac{\sin 2n\phi}{2\sin \phi}+i\frac{\sin^2 n\phi}{\sin \phi}$$

So we get $$\displaystyle S = \sum_{r=1}^n \sin(2r-1)\theta=\Im \left[\sum^{n}_{r=1}e^{(2r-1)\phi}\right] = \frac{\sin^2 n\phi}{\sin \phi}$$

juantheron
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