How to Find the Apothem of a Regular Polygon

In summary, the given equation can be proven by using complex numbers and manipulating them to reach the desired form. Alternatively, it can also be proven geometrically by considering the sum of the sines as the vertical projections of segments in a regular polygon.
  • #1
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Prove that
[tex] \sum_{k=0}^{n} \sin\left( \frac{k \pi}{n} \right) = \cot \left( \frac{\pi}{2n} \right) [/tex]
 
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  • #2
This is a fairly straightforward calculation:

[tex]\begin{align*}\sum_{k=0}^{n} \sin\left(\frac{k\pi}{n}\right) &= \Im \left(\sum_{k=0}^{n} \exp \left(\frac{k\pi i}{n} \right) \right) \\
&= \Im \left( \frac{1 - \exp \left(\frac{(n+1)\pi i}{n}\right)}{1 - \exp \left(\frac{\pi i}{n}\right)}\right)\\
&= \Im \left( \frac{1+\exp \left(\frac{\pi i}{n}\right)}{1 - \exp \left(\frac{\pi i}{n}\right)}\right)\\
&= \Im \left(\frac{\exp \left(-\frac{\pi i}{2n}\right) + \exp \left(\frac{\pi i}{2n} \right)}{\exp \left(-\frac{\pi i}{2n}\right) - \exp \left(\frac{\pi i}{2n}\right)}\right) \\
&= \Im \left(\frac{2 \cos \left(-\frac{\pi }{2n}\right)}{2i \sin \left(-\frac{\pi}{2n} \right)}\right) \\
&= \Im \left(-i \cot \left(-\frac{\pi}{2n} \right) \right) \\
&= \Im \left(i \cot \left(\frac{\pi}{2n} \right)\right) \\
&= \cot \left(\frac{\pi}{2n} \right) \end{align*}[/tex]
 
  • #3
Here's another calculation:
$$\begin{align}
\sin\left(\frac{\pi}{2n}\right) \sum_{k=0}^{n} \sin\left(\frac{k\pi}{n}\right)
&= \sum_{k=0}^{n} \sin\left(\frac{k\pi}{n}\right) \sin\left(\frac{\pi}{2n}\right) \\
&= \frac{1}{2} \sum_{k=0}^{n} \left[ \cos\left( \frac{(2k-1)\pi}{2n}\right) - \cos\left(\frac{(2k+1)\pi}{2n}\right) \right] \\
\end{align}$$
The sum telescopes, so the above reduces to
$$\begin{align}
\frac{1}{2} \left[ \cos\left(-\frac{\pi}{2n}\right) - \cos\left(\frac{(2n+1)\pi}{2n}\right) \right]
&= \frac{1}{2}\left[ \cos\left(\frac{\pi}{2n}\right) - \cos\left(\pi + \frac{\pi}{2n}\right) \right] \\
&= \frac{1}{2} \left[\cos\left(\frac{\pi}{2n}\right) + \cos\left(\frac{\pi}{2n}\right)\right] \\
&= \cos\left(\frac{\pi}{2n}\right)
\end{align}$$
Dividing both sides by ##\sin\left(\frac{\pi}{2n}\right)## gives us what we want.
 
  • #4
Those are some nice solutions! Anyone have another way of doing the calculation?
 
  • #5
The sum is also the distance between opposite sides of a regular polygon with 2n sides of length 1, from where the result comes with simple geometry.
 
  • #6
I stumbled upon a really complicated way to d0 it by induction... too tired now to do the arithmetic and see if it works right now. After easily establishing that both sides equal 0 for n=1, it involves turning the right side into the form (e^ipi/n + 1)/(e^ipi/n -1) and multiplying the numerator and the denominator by e to the power of (i*pi/n^2)/(1+1/n) which turns the denominator of the exponent into n+1 without changing the numerator, but it changes the 1 on the numerator and the -1 in the denominator so you must add something to make them turn back into 1 and -1 without changing the e^ipi/n+1's. So add A/B to that and solve for A and B and then add A/B to the other side and see if you can get it in the n+1 form. Is that on the right track?
 
  • #7
CosmicKitten said:
I stumbled upon a really complicated way to d0 it by induction... too tired now to do the arithmetic and see if it works right now. After easily establishing that both sides equal 0 for n=1, it involves turning the right side into the form (e^ipi/n + 1)/(e^ipi/n -1) and multiplying the numerator and the denominator by e to the power of (i*pi/n^2)/(1+1/n) which turns the denominator of the exponent into n+1 without changing the numerator, but it changes the 1 on the numerator and the -1 in the denominator so you must add something to make them turn back into 1 and -1 without changing the e^ipi/n+1's. So add A/B to that and solve for A and B and then add A/B to the other side and see if you can get it in the n+1 form. Is that on the right track?

It might work, why don't you try going through the details to see if you can hammer it out?

chingel, that's a pretty cool way of attacking this problem. Can you explain a bit more how the geometry works out to give it?
 
  • #8
Since on an x-y plane the sine is the vertical projection of a segment of length 1 at a particular angle and we are trying to find the sum of sines, that is the sum of the vertical projections, we can add up all the segments and then find the vertical projection of the sum. Noticing that the angles increase with regular intervals and that the segments make up half of an 2n-gon, then the vertical projection of the sum is just twice the apothem of the 2n-gon and using the well known simple formula for the apothem we get the answer.

http://www.buzzle.com/articles/finding-apothem-of-a-regular-polygon.html
 
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  • #9
chingel said:
Since on an x-y plane the sine is the vertical projection of a segment of length 1 at a particular angle and we are trying to find the sum of sines, that is the sum of the vertical projections, we can add up all the segments and then find the vertical projection of the sum. Noticing that the angles increase with regular intervals and that the segments make up half of an 2n-gon, then the vertical projection of the sum is just twice the apothem of the 2n-gon and using the well known simple formula for the apothem we get the answer.

http://www.buzzle.com/articles/finding-apothem-of-a-regular-polygon.html

This is a very cool proof.
 

What is Challenge 13: Sums of Sines?

Challenge 13: Sums of Sines is a mathematical challenge that involves finding the sum of several sine functions, with each function having a different amplitude, frequency, and phase shift.

What is the purpose of this challenge?

The purpose of this challenge is to test one's understanding of sine functions, as well as their ability to manipulate and combine multiple functions.

What background knowledge is required to solve this challenge?

To solve this challenge, one should have a solid understanding of trigonometric functions, specifically sine functions, as well as knowledge of amplitude, frequency, and phase shift.

What is the most efficient approach to solving this challenge?

The most efficient approach to solving this challenge would be to use mathematical identities and properties of sine functions to simplify the sum before calculating the final answer.

Can this challenge be applied in real-world situations?

While this challenge is primarily a mathematical exercise, the concept of combining multiple sine functions can be applied in fields such as engineering, physics, and signal processing.

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