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**[SOLVED] another Viete problem**

## Homework Statement

Prove that for every positive integer n,

[tex]\Pi_{k=1}^n \tan \frac{k \pi}{2n+1} = \sqrt{2n+1}[/tex]

## 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

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

? 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. :(