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I'm having a great deal of trouble with this proof.

"Prove $\cos θ + \cos 3θ + \cos 5θ + \cdots + \cos [(2n-1)θ] = \dfrac{\sin 2nθ}{2 \sin θ}$. Prove $\sin θ + \sin 3θ + \sin 5θ + \cdots + \sin [(2n-1)θ] = \dfrac{(\sin nθ)^2}{\sin θ}$."

Relevant Equations

Euler's formula is $e^{iθ} = \cos θ + i \sin θ$. The geometric progression formula is $S_n = a\left(\dfrac{1-r^n}{1-r}\right)$, where $a$ is the first term and $r$ is the constant that each term is multiplied by to get the next term.

My Attempts to Complete the Proof

Obviously, I need to use the geometric progression formula to prove this. With Euler's formula, the initial term $a$ is $e^{iθ}$. The $r$ term is $e^{2iθ}$. I believe these two values for $a$ and $r$ are correct, as the first term will simply be $\cos θ + i \sin θ$, the second will be $e^{iθ + 2iθ} = \cos 3θ + i \sin 3θ$, and so on.

When I plug this into the formula, I get $S_{2n-1} = e^{iθ}\left(\dfrac{1-e^{2iθ (2n-1)}}{1-e^{2iθ}}\right)$.

This is fine, but I can't for the life of me simplify this to anything meaningful. Did I make a mistake somewhere? Are these the correct values for $a$ and $r$? Is there some special trick that I'm missing?

ANY help would be appreciated. I've been stuck on this for days.

Thanks in advance, Leo

EDIT: how do I get the last step? I end up with $\dfrac{1}{2} \dfrac{\sin (4n-2)θ}{2 \sin θ}$. It seems that the $2n-1$ is causing problems. If it was just $n$, then this would all work out perfectly. How can I fix this?

159753x
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1 Answers1

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You have $\cos(k\theta) = \dfrac{e^{ik\theta}+e^{-ik\theta}}{2}$, so \begin{align} & \cos\theta+\cos(3\theta)+\cdots+\cos((2n-1)\theta) \\[8pt] = {} & \frac 1 2 \left(e^{i\theta}+e^{3i\theta}+ \cdots+e^{i\theta(2n-1)}\right) + \frac12\left(e^{-i\theta}+ e^{-3i\theta}+\cdots+e^{-i\theta(2n-1)}\right) \\[8pt] = {} & \frac 1 2 e^{i\theta} \frac{1-e^{2i\theta(2n-1)}}{1-e^{2i\theta}} + e^{-i\theta} \frac 1 2 \frac{1-e^{-2i\theta(2n-1)}}{1-e^{2i\theta}} \\[8pt] = {} & \frac 1 2 \frac{1-e^{2i\theta(2n-1)}}{e^{-i\theta} - e^{i\theta}} + \frac 1 2 \frac{1-e^{-2i\theta(2n-1)}}{e^{i\theta} - e^{-i\theta}} \\[8pt] = {} & \frac 1 2 \frac{e^{2i\theta(2n-1)}- e^{-2i\theta(2n-1)}}{e^{i\theta} - e^{-i\theta}} \end{align}

Michael Hardy
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  • Thanks for the help, Michael. I can understand each step that you wrote. How do I go about turning these into sines? I know that I can do it through Euler's formula, but I run into some trouble when I try it. If I multiply the (2n - 1) in the exponent in the numerator, I get exp(4nθ)/exp(2θ). These don't seem to divide because of the n term. Is there a way around this, or perhaps another way to convert the above to sines? – 159753x Sep 22 '14 at 11:59
  • Use the fact that $\sin\theta=\dfrac{e^{i\theta}-e^{-i\theta}}{2i}$, so that the fraction above is $\dfrac{2i\sin(2(2n-1)\theta)}{2i\sin\theta}$. ${}\qquad{}$ – Michael Hardy Sep 22 '14 at 14:35
  • Right. I was able to do this by converting the expression to cos(θ) + i sin(θ)s, but using the identity you provided was much more efficient. Now I am stuck with that fraction, and I completely understand how we got there, but it seems that the $sin(2(2n-1)θ)$ is a problem. is it possible to get this to $sin(2nθ)$? The double angle identity comes to mind but I don't think it would work here. Hmm... I'll let you know if I get anywhere. – 159753x Sep 22 '14 at 16:32