# How does one calculate the product of $\tan 1^{\circ} … \tan 45^{\circ}?$

I have seen a question asked on yahoo asking to find the value of

$\tan 1^{\circ} \cdot \tan 2^{\circ} \cdot \dots \cdot \tan 45^{\circ}$ (in degrees)

I have seen various results concerning products of trigonometric values such as

$(1 + \tan1^{\circ}) (1 + \tan2^{\circ}) ... (1 + \tan45^{\circ}) = 2^{23} \\$ $\prod_{k=1}^{89} \sin(k \pi/180) = -\prod_{k=-89}^{89} (\sin(k \pi/180)-1) = 360/2^{180}$

I have tried doing various things with roots of unity but nothing much that hasn't been done before. Does anyone have any ideas?

Thanks

Edit: Including some other statements which I consider equivalent to finding this answer

$\cos 1^{\circ} ... \cos 45^{\circ}\\ \sin 1^{\circ} ... \sin45^{\circ} \\ (1 + \tan 46^{\circ})...(1+\tan 89^{\circ}) \\ (\sin 1^{\circ} + \cos 1^{\circ})...(\sin 45^{\circ} + \cos45^{\circ}) \\ (\sin 2^{\circ} + 1)(\sin 4^{\circ} + 1)....(\sin 88^{\circ} + 1) \\ (1 + \tan^2 1^{\circ})(1 + \tan^2 2^{\circ}) .. (1 + \tan^2 45^{\circ})$

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Whatever the exact answer is, it appears to be within an order of magnitude of $\pi^{-45}$, which seems like an interesting fact in its own right... –  Micah Feb 4 '13 at 1:35
(Though, I suppose, not an encouraging one, if it's actually meaningful; the 45th term of a sequence of algebraic numbers which is $O(\pi^{-n})$ probably doesn't have a nice closed form.) –  Micah Feb 4 '13 at 2:12
to correct myself one of the links was already there –  Maesumi Feb 4 '13 at 15:03
$(1 + \tan 46^{\circ})(1 + \tan 47^{\circ})...(1 + \tan 89^{\circ}) = 2^{22} (\tan 1^{\circ} \tan2^{\circ} ... \tan 45^{\circ})^{-1}$ –  muzzlator Feb 4 '13 at 15:17
@Micah: if $\Pi_n=\prod_{j=1}^n \tan(\frac{45j}n^\circ)=\prod_{j=1}^n \tan(\frac{\pi j}{4n})$, then $\frac1n \log\Pi_n$ is a Riemann sum for the integral $\frac4\pi \int_0^{\pi/4} \log\tan x\,dx$, which equals $−4G/\pi$, where $G$ is Catalan's constant. Therefore $\Pi_n$ is approximately $e^{−4Gn/\pi} \approx 0.311535n$, rather than $\pi^{−n} \approx 0.31831^n$. –  Greg Martin Feb 15 '13 at 21:44

I considered the more general formula : $$T(m):=\prod_{k=1}^m \tan\left(\frac{k\pi}{4m}\right)$$ and noticed that the result for small values of $m$ was solution of a polynomial of degree $\le m$. For $m=45$ I found that the answer was solution of this irreducible polynomial of degree $24$ : $$1\\- 3256701697315828896312\,x^1 - 325994294876282580655116\,x^2 + 7220097128841103979624568\,x^3 + 112578453555034444841119842\,x^4 + 493898299320136273975435032\,x^5 + 649061666980531722406164708\,x^6 - 840700351973464244018822232\,x^7 -2457988129238279755530778353\,x^8 - 138286882106888055215208624\,x^9 +2474525072938192662606171624\,x^{10}+326024084648343835216068912\,x^{11} - 1088043811994145989051965476\,x^{12} + 5147738954805237173669808\,x^{13} +182273284200850360076819304\,x^{14} - 33045263177263307887100976\,x^{15} + 677463542076505961377071\,x^{16} +170537100491574073221480\,x^{17} - 6714674580553776884700\,x^{18} - 128584156182235814952\,x^{19} - 339010000890501150\,x^{20}\\ +776030507612856\,x^{21} - 397610115660\,x^{22}\\ + 37004040\,x^{23} + 6561\,x^{24}$$
The product $x$ is the smallest positive root of this polynomial $\,\displaystyle P(x):=\sum_{i=0}^{24} a_i\,x^i\$ and $\ x\approx \dfrac {a_1}{a_2-(a_1)^2}\,$ (to $37$ digits). –  Raymond Manzoni Feb 18 '13 at 0:34
Your “experimental” result can be shown completely rigorously using Galois theory in ${\mathbb Q}(\zeta_{360})$. See this related question –  Ewan Delanoy Feb 18 '13 at 4:57