Trig functions of rational multiples of π are algebraic numbers. Are all trig functions of irrational multiples of π transcendental?
Yes. If the number itself extends forever a multiple of it is also going to trend the same fashion. Like infinity + 1 for example. The thing with a sin, cos type function is the Pi is understood in the definition of the function so you can have a whole number.
Hi IncogNegro, and welcome to Physics and Maths. Your post above is the opposite of maths. Please don't do that again. If you're going to attempt an answer to a question in a field that you don't have any education in, then please add a suitable disclaimer to your posts, something like this: Hi mathman, This is a harder question than I can handle rigorously, but if I had to guess, I'd guess no, because I think we should be able to find a counterexample by working backward. All we need is a non-transcendental number x such that: \(\frac{\arcsin(x)}{\pi}\) is irrational. Now, consider all possible values for \(0 < x < 1\) There are uncountably infinite possibilities, right? This means there are uncountably infinite possible values for \(\frac{\arcsin(x)}{\pi}\), which means that they can't all be rational, and some (most) must be irrational.
It appears that there is some confusion in what I was trying to ask. Fact: If angle is rational multiple of π, trig function is algebraic number. Fact: Almost all other trig functions are transcendental. Question: Are ALL other trig functions transcendental?
\(\sin \tan^{\tiny -1} \frac{1}{2} = \frac{1}{\sqrt{5}}\) which is not transcendental but \(\frac{\tan^{\tiny -1} \frac{1}{2}}{\pi}\) is probably not rational. So while one can construct a triangle with sides \(1 : 2 : \sqrt{5}\) -- the related angle does rationally relate to pi. The sequence \(a_n = \tan \left( n \tan^{\tiny -1} \frac{1}{2} \right)\) is given by the recurrence relation: \(a_0 = 0 \\ a_{n+1} = \frac{1 + 2 a_n }{2 - a_n} = -2 + { 5 }{2 - a_n } \) which is happily invertible and gives \(a_{n-1} = 2 - { 5 }{2 + a_n } \) which correctly gives \(a_{-n} = -a_{n}\). We also have \(a_{n \pm m} = \frac{a_n \pm a_m}{1 \mp a_n a_m }\). And solving the recurrence we also have: \(a_n = -i \frac{(2 + i)^n - (2 - i)^n}{(2 + i)^n + (2 - i)^n}\) But I don't yet see a way to prove \(a_n\) never repeats so I'm not sure how you prove \(\frac{\tan^{\tiny -1} \frac{1}{2}}{\pi}\) is irrational.
Because there are only countably infinite rationals. But, my argument is wrong anyway, because there are also only countably infinite algebraics.
Expressing a trig function in terms of a multiple of pi, then proving it is transcendental when the multiple is irrational..
I understand where you're coming from. While I may not know the math, there are still areas which I can address. I explained it and wanted to make it clear, sometimes it may come across as being overly assertive. Anyway mathman will suss it out.
Hmm, this statement seems incorrect, counterexample : \(cotan(k \pi)=\pm \infty\) for \(k=\) integer. Can you put the above in a mathematical formalism? Are you saying that: \(f(k\pi)=\) transcendental if \(k=\) irrational ? where \(f={sin,cos,tan,cotan,arcsin,.....\)
We were told by our number theory lecturer that there are more transcendental numbers than irrational or rational ones. I guess that makes sense because only countably infinite roots of countably infinite polynomials over the rationals exist. But I'm wingin' it here.
Sorry, I was a little confused. Anyway, i'm guessing that the second fact probably won't contribute much to answering the open question. Hmm..
The same for tan 90. @mathman Maybe it would be easier to solve the problem if you narrowed the definition of "trig function".
There exists \(0<x<\pi\) such that \( sin (x)=\frac{2}{\pi}\). If the first part of your conjecture is true, \(x\) cannot be rational. In other words, there are irrational values of \(0<x<\pi\) for which \(sin(x)\) is transcendental. Likewise, There exists \(-\pi/2<x'<\pi/2\) such that \( cos (x')=\frac{2}{\pi}\). If the first part of your conjecture is true, \(x'\) cannot be rational. In other words, there are irrational values of \(-\pi/2<x'<\pi/2\) for which \(cos(x')\) is transcendental.
Pi is greater than one and so there is NO real argument that makes either sine or cosine equal to the square root of pi since both are bounded above by 1.
Thank you for altering your post to change \(\sqrt{\pi}\) to \(\frac{2}{\pi}\). Thank you also for removing the conjecture that this was weaker. But you don't offer a proof that either \(\frac{\sin^{\tiny-1}\,\frac{2}{\pi}}{\pi}\) or \(\sin^{\tiny-1}\,\frac{2}{\pi}\) is irrational.