To prove that

$\cos(0) + \cos\left(\frac{2\pi}{n}\right) + \cos\left(\frac{4\pi}{n}\right) + \ldots + \cos\left(\frac{2\pi(n-1)}{n}\right) = 0$

for $n \geq 1$, let's proceed step by step:

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Step 1: Represent the sum using exponential notation

Using Euler's formula, we know:

$\cos(x) = \frac{e^{ix} + e^{-ix}}{2}.$

So, the given sum can be rewritten as:

$\sum_{k=0}^{n-1} \cos\left(\frac{2\pi k}{n}\right) = \frac{1}{2} \sum_{k=0}^{n-1} \left( e^{i\frac{2\pi k}{n}} + e^{-i\frac{2\pi k}{n}} \right).$

Split this into two separate sums:

$\sum_{k=0}^{n-1} \cos\left(\frac{2\pi k}{n}\right) = \frac{1}{2} \left( \sum_{k=0}^{n-1} e^{i\frac{2\pi k}{n}} + \sum_{k=0}^{n-1} e^{-i\frac{2\pi k}{n}} \right).$

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Step 2: Analyze the exponential sum

Let $S = \sum_{k=0}^{n-1} e^{i\frac{2\pi k}{n}}$. This is a geometric progression with the first term $1$ (when $k=0$) and the common ratio $e^{i\frac{2\pi}{n}}$.

The sum of a geometric series is given by:

$S = \frac{1 - r^n}{1 - r},$

where $r$ is the common ratio. Here, $r = e^{i\frac{2\pi}{n}}$, and $r^n = \left(e^{i\frac{2\pi}{n}}\right)^n = e^{i2\pi} = 1$. Thus:

$S = \frac{1 - 1}{1 - e^{i\frac{2\pi}{n}}} = 0.$

This shows:

$\sum_{k=0}^{n-1} e^{i\frac{2\pi k}{n}} = 0.$

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Step 3: The symmetry of the terms

Using the property of complex conjugates:

$\sum_{k=0}^{n-1} e^{-i\frac{2\pi k}{n}} = \overline{\sum_{k=0}^{n-1} e^{i\frac{2\pi k}{n}}}.$

Since $\sum_{k=0}^{n-1} e^{i\frac{2\pi k}{n}} = 0$, we also have:

$\sum_{k=0}^{n-1} e^{-i\frac{2\pi k}{n}} = 0.$

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Step 4: Combine the results

Substitute back into the original equation:

$\sum_{k=0}^{n-1} \cos\left(\frac{2\pi k}{n}\right) = \frac{1}{2} \left( \sum_{k=0}^{n-1} e^{i\frac{2\pi k}{n}} + \sum_{k=0}^{n-1} e^{-i\frac{2\pi k}{n}} \right).$

Since both exponential sums are $0$, we have:

$\sum_{k=0}^{n-1} \cos\left(\frac{2\pi k}{n}\right) = \frac{1}{2} (0 + 0) = 0.$

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Conclusion

Thus, we have shown:

$\cos(0) + \cos\left(\frac{2\pi}{n}\right) + \cos\left(\frac{4\pi}{n}\right) + \ldots + \cos\left(\frac{2\pi(n-1)}{n}\right) = 0,$

for $n \geq 1$.

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Do you have any questions, or would you like further clarification?

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Related Questions:

1. How can we derive the geometric series formula used here?
2. What happens if we use sine terms instead of cosine?
3. How does the symmetry of the unit circle affect this sum?
4. Can we generalize this proof to other roots of unity?
5. What is the significance of this result in Fourier analysis?

Tip:

Understanding the symmetry of the unit circle and the properties of roots of unity is crucial for solving problems involving trigonometric sums.