Can $\sin(\frac{n\pi}{m})$ with $n,m \in \mathbb{Z}$ always be represented using only algebraic functions?
In other words can $sin$ of a rational multiple of $\pi$ always be represented using only only algebraic functions (e.g. addition, subtraction, division, multiplication, and roots)?
This question came after failing to find a calculator that could give me the exact value of inputs that I knew had exact values (things like $sin(\frac{\pi}{8})$ and $sin(\frac{\pi}{180})$). As a side question, is there a calculator that can do this? I tried WolframAlpha but it wouldn't always give me an exact value.
You can write $$ \sin\left(\frac{n\pi}{m}\right) = \frac {e^{n\pi i/m}-e^{-n\pi i/m}}{2i} = \frac {t^n-t^{-n}}{2i} $$ with $t = e^{\pi i/m}.$ $t$ is one of the $m$th roots of $-1.$
To be specific, it is the one with the smallest positive argument among the $m$ roots.
With a lot of hand-waving, you could write $t = \sqrt[m]{-1}$ to make it look more like an "algebraic function", as long as it is clear from the context which of the roots this is meant to be.
This would result in $$ \sin\left(\frac{n\pi}{m}\right) = \frac {\left(\sqrt[m]{-1}\right)^n-\left(\sqrt[m]{-1}\right)^{-n}}{2i} $$ However, as mentioned above, this requires additional explanations about the usage of the $\sqrt[m]{\;.\;}$ and I would not recommend to write it down in that way.
I do not think that there is a way of expressing $\sin\left(\frac{n\pi}{m}\right)$ in terms of functions that produce real results, only.