# Prove the $(2n+1)$-tuple angle identity

It's common to know the triple angle formula, $\tan(3x) = \tan(x) \cdot \tan\left( x - \frac\pi3 \right) \cdot \tan\left( x + \frac\pi3 \right) .$ David Vreken pointed out in his unworldy solution here that quintuple angle formula below also holds! $\tan(5x) = \tan\left(\frac{2\pi}5 - x\right) \tan\left(\frac{\pi}5 - x\right) \tan(x) \tan\left(\frac{\pi}5 + x\right) \tan\left(\frac{2\pi}5 + x\right) .$ He even suggested that the following generalization must be true as well for any positive integer $n.$ Can anyone prove it? $\tan [(2n + 1)x] = (-1)^n \prod_{k=1}^{2n + 1} \tan\left(x + \frac{k - n - 1}{2n + 1}\pi \right)$

Note by Pi Han Goh
4 weeks, 1 day ago

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We have the identities $\sin nx = 2^{n - 1} \prod_{k = 0}^{n - 1} \sin \left( x + \frac{k \pi}{n} \right)$ and $\cos nx = 2^{n - 1} \prod_{k = 1}^n \sin \left( x + \frac{(2k - 1) \pi}{2n} \right).$ If you divide these, you should be able to manipulate the quotient to get what you want.

References:

https://mathworld.wolfram.com/Multiple-AngleFormulas.html

- 4 weeks ago

Summoning @Mark Hennings

- 4 weeks, 1 day ago

I found the $\tan 5x$ equation here. There seems to be a start of a proof for it but I don't understand it. Perhaps someone can figure that out and generalize it.

- 4 weeks, 1 day ago

Here's a start to a proof:

Since $\tan x = \cfrac{e^{ix} - e^{-ix}}{i(e^{ix} + e^{-ix})} = \cfrac{e^{2ix} - 1}{i(e^{2ix} + 1)}$, that means that $\tan(x + a) = \cfrac{e^{2ix}e^{2ia} - 1}{i(e^{2ix}e^{2ia} + 1)}$. Therefore,

$\tan(\cfrac{\pi}{3} - x) \cdot \tan x \cdot \tan(\cfrac{\pi}{3} + x)$

$= -\tan(x - \cfrac{\pi}{3}) \cdot \tan x \cdot \tan(x + \cfrac{\pi}{3})$

$= -\cfrac{e^{2ix}e^{-\frac{2\pi}{3}i} - 1}{i(e^{2ix}e^{-\frac{2\pi}{3}i} + 1)} \cdot \cfrac{e^{2ix} - 1}{i(e^{2ix} + 1)} \cdot \cfrac{e^{2ix}e^{\frac{2\pi}{3}i} - 1}{i(e^{2ix}e^{\frac{2\pi}{3}i} + 1)}$

$= -\cfrac{e^{6ix} - e^{4ix}(e^{\frac{2\pi}{3}i} + e^{-\frac{2\pi}{3}i} + 1) + e^{2ix}(e^{\frac{2\pi}{3}i} + e^{-\frac{2\pi}{3}i} + 1) - 1}{i^3(e^{6ix} + e^{4ix}(e^{\frac{2\pi}{3}i} + e^{-\frac{2\pi}{3}i} + 1) + e^{2ix}(e^{\frac{2\pi}{3}i} + e^{-\frac{2\pi}{3}i} + 1) + 1)}$

$= \cfrac{e^{6ix} - e^{4ix}(0) + e^{2ix}(0) - 1}{i(e^{6ix} + e^{4ix}(0) + e^{2ix}(0) + 1)}$

$= \cfrac{e^{6ix} - 1}{i(e^{6ix} + 1)}$

$= \tan 3x$

Likewise,

$\tan(\cfrac{2\pi}{5} - x) \cdot \tan(\cfrac{\pi}{5} - x) \cdot \tan x \cdot \tan(\cfrac{\pi}{5} + x) \cdot \tan(\cfrac{2\pi}{5} + x)$

$= \tan(x - \cfrac{2\pi}{5}) \cdot \tan(x - \cfrac{\pi}{5}) \cdot \tan x \cdot \tan(x + \cfrac{\pi}{5}) \cdot \tan(x + \cfrac{2\pi}{5})$

$= \cfrac{e^{2ix}e^{-\frac{4\pi}{5}i} - 1}{i(e^{2ix}e^{-\frac{4\pi}{5}i} + 1)} \cdot \cfrac{e^{2ix}e^{-\frac{2\pi}{5}i} - 1}{i(e^{2ix}e^{-\frac{2\pi}{5}i} + 1)} \cdot \cfrac{e^{2ix} - 1}{i(e^{2ix} + 1)} \cdot \cfrac{e^{2ix}e^{\frac{2\pi}{5}i} - 1}{i(e^{2ix}e^{\frac{2\pi}{5}i} + 1)} \cdot \cfrac{e^{2ix}e^{\frac{4\pi}{5}i} - 1}{i(e^{2ix}e^{\frac{4\pi}{5}i} + 1)}$

$= \cfrac{e^{10ix} - e^{8ix}(e^{\frac{4\pi}{5}i} + e^{-\frac{4\pi}{5}i} + e^{\frac{2\pi}{5}i} + e^{-\frac{2\pi}{5}i} + 1) + e^{6ix}(e^{\frac{6\pi}{5}i} + e^{-\frac{6\pi}{5}i} + e^{\frac{4\pi}{5}i} + e^{-\frac{4\pi}{5}i} + 2e^{\frac{2\pi}{5}i} + 2e^{-\frac{2\pi}{5}i} + 2) - e^{4ix}(e^{\frac{6\pi}{5}i} + e^{-\frac{6\pi}{5}i} + e^{\frac{4\pi}{5}i} + e^{-\frac{4\pi}{5}i} + 2e^{\frac{2\pi}{5}i} + 2e^{-\frac{2\pi}{5}i} + 2) + e^{2ix}(e^{\frac{4\pi}{5}i} + e^{-\frac{4\pi}{5}i} + e^{\frac{2\pi}{5}i} + e^{-\frac{2\pi}{5}i} + 1) - 1}{i^5(e^{10ix} + e^{8ix}(e^{\frac{4\pi}{5}i} + e^{-\frac{4\pi}{5}i} + e^{\frac{2\pi}{5}i} + e^{-\frac{2\pi}{5}i} + 1) + e^{6ix}(e^{\frac{6\pi}{5}i} + e^{-\frac{6\pi}{5}i} + e^{\frac{4\pi}{5}i} + e^{-\frac{4\pi}{5}i} + 2e^{\frac{2\pi}{5}i} + 2e^{-\frac{2\pi}{5}i} + 2) + e^{4ix}(e^{\frac{6\pi}{5}i} + e^{-\frac{6\pi}{5}i} + e^{\frac{4\pi}{5}i} + e^{-\frac{4\pi}{5}i} + 2e^{\frac{2\pi}{5}i} + 2e^{-\frac{2\pi}{5}i} + 2) + e^{2ix}(e^{\frac{4\pi}{5}i} + e^{-\frac{4\pi}{5}i} + e^{\frac{2\pi}{5}i} + e^{-\frac{2\pi}{5}i} + 1) + 1)}$

$= \cfrac{e^{10ix} - e^{8ix}(e^{\frac{4\pi}{5}i} + e^{-\frac{4\pi}{5}i} + e^{\frac{2\pi}{5}i} + e^{-\frac{2\pi}{5}i} + 1) + 2e^{6ix}(e^{\frac{4\pi}{5}i} + e^{-\frac{4\pi}{5}i} + e^{\frac{2\pi}{5}i} + e^{-\frac{2\pi}{5}i} + 1) - 2e^{4ix}(e^{\frac{4\pi}{5}i} + e^{-\frac{4\pi}{5}i} + e^{\frac{2\pi}{5}i} + e^{-\frac{2\pi}{5}i} + 1) + e^{2ix}(e^{\frac{4\pi}{5}i} + e^{-\frac{4\pi}{5}i} + e^{\frac{2\pi}{5}i} + e^{-\frac{2\pi}{5}i} + 1) - 1}{i(e^{10ix} + e^{8ix}(e^{\frac{4\pi}{5}i} + e^{-\frac{4\pi}{5}i} + e^{\frac{2\pi}{5}i} + e^{-\frac{2\pi}{5}i} + 1) + 2e^{6ix}(e^{\frac{4\pi}{5}i} + e^{-\frac{4\pi}{5}i} + e^{\frac{2\pi}{5}i} + e^{-\frac{2\pi}{5}i} + 1) + 2e^{4ix}(e^{\frac{4\pi}{5}i} + e^{-\frac{4\pi}{5}i} + e^{\frac{2\pi}{5}i} + e^{-\frac{2\pi}{5}i} + 1) + e^{2ix}(e^{\frac{4\pi}{5}i} + e^{-\frac{4\pi}{5}i} + e^{\frac{2\pi}{5}i} + e^{-\frac{2\pi}{5}i} + 1) + 1)}$

$= \cfrac{e^{10ix} - e^{8ix}(0) + 2e^{6ix}(0) - 2e^{4ix}(0) + e^{2ix}(0) - 1}{i(e^{10ix} + e^{8ix}(0) + 2e^{6ix}(0) + 2e^{4ix}(0) + e^{2ix}(0) + 1)}$

$= \cfrac{e^{10ix} - 1}{i(e^{10ix} + 1)}$

$= \tan 5x$

Since $e^{\frac{-2n\pi}{2n + 1}i} + e^{\frac{-2(n - 1) \pi}{2n + 1}i} + ... + e^{\frac{-2 \pi}{2n + 1}i} + e^{\frac{0 \pi}{2n + 1}i} + e^{\frac{2\pi}{2n + 1}i} + ... + e^{\frac{2(n - 1)\pi}{n + 1}i} + e^{\frac{2n\pi}{2n + 1}i} = 0$ for any integer $n$, the above middle terms will always cancel out for the general case (more work needed here) and will prove the given generalization.

- 4 weeks ago