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So I've been working on this problem for a while and I was able to get up to here:

$$\sum_{k=1}^{2004}\cos^{2}\left(\frac{k\pi}{2\cdot2005}\right)$$

With the trigonometric identity that $1+\tan^2\left(\theta\right) = \sec^2\left(\theta\right)$. And that $\cos\left(\theta\right) = \frac{1}{\sec\left(\theta\right)}$.

I'm stuck at this point, does anyone have anything on how to move it forward?

Sidenote: I'm a precalculus student

Angina Seng
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Alsagol
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4 Answers4

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\begin{align} \sum_{k=1}^{2004}\cos^{2}\left(\frac{k\pi}{2\cdot2005}\right) & =\frac12 \sum_{k=1}^{2004}(1+ \cos\frac{k\pi}{2005})\\ & = \frac{2004}2+ \frac12 \sum_{k=1}^{1002}\left[\cos\frac{k\pi}{2005} + \cos\frac{(2005-k)\pi}{2005}\right]\\ & = 1002+ \frac12 \sum_{k=1}^{1002}0=1002\\ \end{align}

Quanto
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  • a couple questions (sorry): 1. Howcome the cos^2 didn't stay squared when you used the half angle formula (I assume that's what you used). 2. Where did that 1002 come from? 3. Where did the \left[\cos\frac{k\pi}{2005}+\cos\frac{\left(2005-k\right)\pi}{2005}\right] come from? I see the summation went from 2004 to 1002 but I don't know how it might've changed to what you got. 4. How does \left[\cos\frac{k\pi}{2005}+\cos\frac{\left(2005-k\right)\pi}{2005}\right] = 0? This isn't me bashing your work I just want to know how you got there. – Alsagol May 14 '20 at 18:50
  • @Alsagol 1 half angle identity $2\cos^2t=1+\cos2t$. 2. 1002 is first half of the sum. 3. Write the sum in two part 0 to 1002 and 1003 to 2004. 4. Use $\cos t + \cos(\pi-t)=0$ – Quanto May 14 '20 at 18:58
  • Yes! I understand now, thank you so much! – Alsagol May 14 '20 at 19:15
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Now use the half-angle formula $$\cos^2 \frac{x}{2} = \frac{1 + \cos x}{2}.$$ This gives $$1002 + \frac{1}{2}\sum_{k=1}^{2004} \cos \frac{k\pi}{2005}.$$ Then recall Euler's formula $$e^{i\theta} = \cos \theta + i \sin \theta,$$ which then gives $$\cos \theta = \frac{e^{i\theta} + e^{-i\theta}}{2}.$$ This will give you a geometric series that you can sum.

heropup
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Using the double angle formula for cosine and then replacing $k$ with $2005-k$ while using that $\cos(\pi-\theta)+\cos\theta =0 $ yields the sum. Another approach can to be use the formula for sum of series of cosine when angles are in arithmetic progression.

$$\begin{aligned}\sum_{k=1}^{2004}\cos^2\left(\frac{k\pi}{2005}\right)&=\frac{1}{2}\sum_{k=1}^{2004}\left[1+\cos\left(\frac{k\pi}{2005}\right)\right]\\ & =\frac{1}{2}\sum_{k=1}^{2004}\left[1+\cos\left(\pi-\frac{k\pi}{2005}\right)\right]\end{aligned}$$

Paras Khosla
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$$S=\sum_{k=1}^{n-1}\cos^2(\frac{k\pi}{2n})=\frac{1}{2}\sum_{k=1}^{n-1}(1+\cos(\frac{k\pi}{n}))$$

But with $\omega=e^{i\pi/n}$:

$$\sum_{k=1}^{n-1}\cos(\frac{k\pi}{n})=\Re\sum_{k=1}^{n-1}e^{i\frac{k\pi}{n}}=\Re\omega\frac{\omega^{n-1}-1}{\omega-1}=\Re\frac{(-1)^{n}-\omega}{\omega-1}=\Re\begin{Bmatrix}-1&,n=2k\\-i\tan(\frac{\pi}{4n}) &,n=2k+1\end{Bmatrix}$$

and thus, since $n=2005$

$$S=1002$$

DinosaurEgg
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