# Another Viete problem

[SOLVED] another Viete problem

## Homework Statement

Prove that for every positive integer n,

$$\Pi_{k=1}^n \tan \frac{k \pi}{2n+1} = \sqrt{2n+1}$$

## Homework Equations

http://en.wikipedia.org/wiki/Viète's_formulas

## The Attempt at a Solution

I cannot figure out what polynomial I need to apply Viete to.

Should I let

$$f(x) = \Pi_{k=1}^n \left(x- \tan \frac{k \pi}{2n+1}\right)$$

? That doesn't seem to help at all. I looked for relevant trig identities and couldn't find any. I tried induction (the case n=1 is trivial) but that failed also so alas I am stuck again. :(

Related Calculus and Beyond Homework Help News on Phys.org
tiny-tim
Homework Helper
Hi ehrenfest! Hint: use de Moivre's theorem. Hint: use de Moivre's theorem. Hmm.

$$\cos n \theta (1 + i \tan n \theta) = (\cos \theta + i \sin \theta)^n$$

We can let \theta = pi/(2n+1). We could then sum both sides over n. But somehow we need to get a polynomial and then apply Viete to it. I don't see what polynomial to use. Sorry tiny-tim. :(

tiny-tim
Homework Helper
Hmm.

$$\cos n \theta (1 + i \tan n \theta) = (\cos \theta + i \sin \theta)^n$$

We can let \theta = pi/(2n+1). We could then sum both sides over n. But somehow we need to get a polynomial and then apply Viete to it. I don't see what polynomial to use. Sorry tiny-tim. :( (tears hair out) But the whole point of de Moivre's theorem is that it avoids using Viete! Just do what you said … sum kπ/(2n+1) from 1 to n. ehrenfest … you keep doing this … you get it fixed in your head that you have to do something, even when you yourself have found a way that avoids it! Well I really didn't see how anything would simplify after I summed it:

$$\sum_{k=1}^n \cos \frac{k \pi}{2n+1} (1 + i \tan \frac{k \pi}{2n+1} ) = \sum_{k=1}^n (\cos \frac{\pi}{2n+1}+ i \sin \frac{\pi}{2n+1})^k$$

But I'm sure you have some clever trick up your sleeve tiny-tim.

George Jones
Staff Emeritus
Gold Member
$$\cos \theta + i \sin \theta = ?$$

$$\cos \theta + i \sin \theta = ?$$
$$e^{i \theta}$$

But how does that help simplify the RHS. I know there must be something obvious I am missing? It is just the first n of the 4n+2 roots of unity but does that equal something nice?

Last edited:
George Jones
Staff Emeritus
Gold Member
$$e^{i \theta}$$

But how does that help simplify the RHS?
Substituting this shows that this series is a special type of series.

What type of series?

I'm an idiot.

The RHS is

$$\frac{1-\exp \frac{i (n+1) \pi}{2n+1}}{1-\exp \frac{i \pi}{2n+1}}$$

So

$$\sum_{k=1}^n \cos \frac{k \pi}{2n+1} (1 + i \tan \frac{k \pi}{2n+1} ) = \frac{1-\exp \frac{i (n+1) \pi}{2n+1}}{1-\exp \frac{i \pi}{2n+1}}$$

Sorry to keep prolonging this but I still don't see where this is going. I mean how are we ever going a product of tangents from this!

tiny-tim
Homework Helper
Hi ehrenfest! You're getting confused.

You've used ∑ instead of ∏. Your:
$$\sum_{k=1}^n \cos \frac{k \pi}{2n+1} (1 + i \tan \frac{k \pi}{2n+1} ) = \sum_{k=1}^n (\cos \frac{\pi}{2n+1}+ i \sin \frac{\pi}{2n+1})^k$$
should be: $$\Pi_{k=1}^n \cos \frac{k \pi}{2n+1} (1 + i \tan \frac{k \pi}{2n+1} ) = \Pi_{k=1}^n (\cos \frac{\pi}{2n+1}+ i \sin \frac{\pi}{2n+1})^k = (\cos \frac{\pi}{2n+1}+ i \sin \frac{\pi}{2n+1})^{\sum{k=1}^{n} k}\\,=\,....$$

hmm … the LaTeX hasn't come out right … but I hope you see what I mean! Hi ehrenfest! You're getting confused.

You've used ∑ instead of ∏. Your:

should be: $$\Pi_{k=1}^n \cos \frac{k \pi}{2n+1} (1 + i \tan \frac{k \pi}{2n+1} ) = \Pi_{k=1}^n (\cos \frac{\pi}{2n+1}+ i \sin \frac{\pi}{2n+1})^k = (\cos \frac{\pi}{2n+1}+ i \sin \frac{\pi}{2n+1})^{\sum{k=1}^{n} k}\\,=\,....$$

hmm … the LaTeX hasn't come out right … but I hope you see what I mean! Well you said sum in post #4. Sorry I still don't see where we are going with this. I agree with what you posted above and the sum of the first n integers is n(n+1)/2 but I cannot even see what will happen in the case of n=2.

Then we have

$$\exp i \frac{3 \pi}{5} = \cos \frac{\pi}{5} \cos\frac{2 \pi}{5}(1+i\tan \frac{\pi}{5})(1+i\tan\frac{2\pi}{5})$$

We can take the real part of both sides but then what?

Last edited:
tiny-tim
ah, but if you use de Moivre again, then the n+1 cancels, and you get … ? ah, but if you use de Moivre again, then the n+1 cancels, and you get … ? 