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

It's common to know the triple angle formula, tan⁡(3x)=tan⁡(x)⋅tan⁡(x−π3)⋅tan⁡(x+π3). \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⁡(2π5−x)tan⁡(π5−x)tan⁡(x)tan⁡(π5+x)tan⁡(2π5+x). \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.n. Can anyone prove it? tan⁡[(2n+1)x]=(−1)n∏k=12n+1tan⁡(x+k−n−12n+1π)\tan [(2n + 1)x] = (-1)^n \prod_{k=1}^{2n + 1} \tan\left(x + \frac{k - n - 1}{2n + 1}\pi \right)

#Geometry

Note by Pi Han Goh
7 months, 1 week ago

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Comments

We have the identities sin⁡nx=2n−1∏k=0n−1sin⁡(x+kπn)\sin nx = 2^{n - 1} \prod_{k = 0}^{n - 1} \sin \left( x + \frac{k \pi}{n} \right) and cos⁡nx=2n−1∏k=1nsin⁡(x+(2k−1)π2n).\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

https://math.stackexchange.com/questions/2095330/product-identity-multiple-angle-or-sinnx-2n-1-prod-k-0n-1-sin-left

Jon Haussmann - 7 months, 1 week ago

Summoning @Mark Hennings

Pi Han Goh - 7 months, 1 week ago

I found the tan⁡5x\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.

David Vreken - 7 months, 1 week ago

Here's a start to a proof:

Since tan⁡x=eix−e−ixi(eix+e−ix)=e2ix−1i(e2ix+1)\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)=e2ixe2ia−1i(e2ixe2ia+1)\tan(x + a) = \cfrac{e^{2ix}e^{2ia} - 1}{i(e^{2ix}e^{2ia} + 1)}. Therefore,

tan⁡(π3−x)⋅tan⁡x⋅tan⁡(π3+x)\tan(\cfrac{\pi}{3} - x) \cdot \tan x \cdot \tan(\cfrac{\pi}{3} + x)

=−tan⁡(x−π3)⋅tan⁡x⋅tan⁡(x+π3)= -\tan(x - \cfrac{\pi}{3}) \cdot \tan x \cdot \tan(x + \cfrac{\pi}{3})

=−e2ixe−2π3i−1i(e2ixe−2π3i+1)⋅e2ix−1i(e2ix+1)⋅e2ixe2π3i−1i(e2ixe2π3i+1)= -\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)}

=−e6ix−e4ix(e2π3i+e−2π3i+1)+e2ix(e2π3i+e−2π3i+1)−1i3(e6ix+e4ix(e2π3i+e−2π3i+1)+e2ix(e2π3i+e−2π3i+1)+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)}

=e6ix−e4ix(0)+e2ix(0)−1i(e6ix+e4ix(0)+e2ix(0)+1)= \cfrac{e^{6ix} - e^{4ix}(0) + e^{2ix}(0) - 1}{i(e^{6ix} + e^{4ix}(0) + e^{2ix}(0) + 1)}

=e6ix−1i(e6ix+1)= \cfrac{e^{6ix} - 1}{i(e^{6ix} + 1)}

=tan⁡3x= \tan 3x

Likewise,

tan⁡(2π5−x)⋅tan⁡(π5−x)⋅tan⁡x⋅tan⁡(π5+x)⋅tan⁡(2π5+x)\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−2π5)⋅tan⁡(x−π5)⋅tan⁡x⋅tan⁡(x+π5)⋅tan⁡(x+2π5)= \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})

=e2ixe−4π5i−1i(e2ixe−4π5i+1)⋅e2ixe−2π5i−1i(e2ixe−2π5i+1)⋅e2ix−1i(e2ix+1)⋅e2ixe2π5i−1i(e2ixe2π5i+1)⋅e2ixe4π5i−1i(e2ixe4π5i+1)= \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)}

=e10ix−e8ix(e4π5i+e−4π5i+e2π5i+e−2π5i+1)+e6ix(e6π5i+e−6π5i+e4π5i+e−4π5i+2e2π5i+2e−2π5i+2)−e4ix(e6π5i+e−6π5i+e4π5i+e−4π5i+2e2π5i+2e−2π5i+2)+e2ix(e4π5i+e−4π5i+e2π5i+e−2π5i+1)−1i5(e10ix+e8ix(e4π5i+e−4π5i+e2π5i+e−2π5i+1)+e6ix(e6π5i+e−6π5i+e4π5i+e−4π5i+2e2π5i+2e−2π5i+2)+e4ix(e6π5i+e−6π5i+e4π5i+e−4π5i+2e2π5i+2e−2π5i+2)+e2ix(e4π5i+e−4π5i+e2π5i+e−2π5i+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) + 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)}

=e10ix−e8ix(e4π5i+e−4π5i+e2π5i+e−2π5i+1)+2e6ix(e4π5i+e−4π5i+e2π5i+e−2π5i+1)−2e4ix(e4π5i+e−4π5i+e2π5i+e−2π5i+1)+e2ix(e4π5i+e−4π5i+e2π5i+e−2π5i+1)−1i(e10ix+e8ix(e4π5i+e−4π5i+e2π5i+e−2π5i+1)+2e6ix(e4π5i+e−4π5i+e2π5i+e−2π5i+1)+2e4ix(e4π5i+e−4π5i+e2π5i+e−2π5i+1)+e2ix(e4π5i+e−4π5i+e2π5i+e−2π5i+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)}

=e10ix−e8ix(0)+2e6ix(0)−2e4ix(0)+e2ix(0)−1i(e10ix+e8ix(0)+2e6ix(0)+2e4ix(0)+e2ix(0)+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)}

=e10ix−1i(e10ix+1)= \cfrac{e^{10ix} - 1}{i(e^{10ix} + 1)}

=tan⁡5x= \tan 5x

Since e−2nπ2n+1i+e−2(n−1)π2n+1i+...+e−2π2n+1i+e0π2n+1i+e2π2n+1i+...+e2(n−1)πn+1i+e2nπ2n+1i=0e^{\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 nn, the above middle terms will always cancel out for the general case (more work needed here) and will prove the given generalization.

David Vreken - 7 months, 1 week ago
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