First time posting here so please forgive any lack of adherence to best practices.
sin(x) is a transcendental function. However most common values for the angle x will yield an algebraic number result. In particular, sin(q) (in degrees) or sin(qπ) (in radians) will always be algebraic.
I can reverse-engineer a transcendental result by making y=sin(x), assigning a transcendental number to y in [-1, 1], and then solving for x. For example, sin(arcsin(1/e)) must be transcendental. But x=(arcsin(1/e)) is not what I would call a "niece value of x".
After a couple of hours researching the internet I didn't find much.
I have the intuition that sin(q) (q rational, not zero and in radians), as well as sin(k) (k irrational and in degrees, or in radians but not of the form qπ) would most of the times (or always?) be transcendental. And these would be "nice". But I know it can be extremely difficult to prove.
So the question is, are there "nice" values of x such as sin(x) is confirmed to be transcendental? I tried sin(1 radian) in Wolfram Alpha and it said it is transcendental. Is that true? How does it even know? What about other integers or π (in degrees)? Or 1/π, e, Φ, ln(2) or √2 (either in radians or degrees)? Would the sine of these numbers be transcendental?
As a consequence of the Lindemann–Weierstrass theorem, $\sin(x)$ is transcendental for any nonzero algebraic number $x$. The same happen with all the other trigonometric functions.
Here's the proof: Let $x$ a nonzero algebraic complex number. Assume $a = \sin(x)$ is algebraic.
Then $$a = \frac{e^{ix}-e^{-ix}}{2i}$$ so
$$(e^{ix})^2-2ai e^{ix}-1 = 0$$
Since the polynomial $z^2-2aiz-1$ has algebraic coefficients, its roots are algebraic, that is $e^{ix}$ is algebraic, but this contradicts this particular case of the Lindemann–Weierstrass theorem: If $z$ is a nonzero algebraic number, then $e^z$ is transcendental.