How Does Viète's Formula Apply to Trigonometric Products?

[ehrenfest]

[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. :(

 

[tiny-tim]

Hi ehrenfest!

Hint: use de Moivre's theorem.

 

[ehrenfest]

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]

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!

 

[ehrenfest]

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]

$$\cos \theta + i \sin \theta = ?$$

 

[ehrenfest]

$$\cos \theta + i \sin \theta = ?$$

$$e^{i \theta}$$

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?

 

[George Jones]

$$e^{i \theta}$$

But how does that help simplify the RHS?

Substituting this shows that this series is a special type of series.

What type of series?

 

[ehrenfest]

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]

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!

 

[ehrenfest]

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?

 

[tiny-tim]

… 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. …

ah, but if you use de Moivre again, then the n+1 cancels, and you get … ?

 

[ehrenfest]

ah, but if you use de Moivre again, then the n+1 cancels, and you get … ?

What? There is a 2n+1 not an n+1 in the denominator of the cosine and the sine...

 

[ehrenfest]

Well, this was really bothering so I looked at the solution in my book that employs Viete. Using de Moivre was a good idea but I think you really do need Viete also.