I have a question on a proof regarding this: The question is "Show this is true," although I am not sure if the last minus sign should be an equal sign, or else it doesn't make sense.
I know that this is related to De Moivre's theorem and Euler's theorem. I tried just solving for the real part of
$$1-\frac{1}{2}e^{i\theta}+\frac{1}{4}e^{2i\theta}-\frac{1}{8}e^{3i\theta}+\cdots $$
And as this is a infinite geometric series, I was thinking this might be finding the real part of the sum of infinity. As $e^{i\theta}$ is equivalent to $cos (\theta) + i sin(\theta)$, it could lead me to proving the answer, but I only got $\frac{2}{2+cos \theta}$.
Could anyone advise me on what I did wrong, and what I could do?
Thank you in advance!
Sorry for any wrong tags or title labelling. Working on those!
Answer
As commented, the $\frac14\cos(\theta)$ must be a typo for $\frac14\cos(2\theta)$.
Yes, the sum of the series with the complex exponential is $2/(2+e^{i\theta})$. But the real part of that is not what you think. The real part of $2/z$ is not $2/\Re z$. You need to rationalize the denominator to find the real part: $$\frac{2}{2+e^{i\theta}}
=\frac{2}{2+e^{i\theta}}\frac{2+e^{-i\theta}}{2+e^{-i\theta}}
=\dots=\frac{4+2e^{-i\theta}}{5+4\cos(\theta)}.$$
You can take the real part of that by taking the real part of the numerator, and you get something very close to what the problem says. Now consider we've already spotted two typos in the problem... (Edit Come to think of it, we can easily check the answer by considering $\theta=0$, and sure enough the sum given in the problem is off by a factor of $2$.)
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