Does there exist a $z\in \Bbb R$ such that $\sin z=t \in \Bbb T$?

Question

Does there exist a $z\in \Bbb R$ such that $\sin z=t \in \Bbb T$: the set of transcendental numbers?

I've had this doubt and I didn't know how to tackle it...

Edit: Changed my domain to reals only, as complex argument was trivial.

I see, that was fairly easy to see... What about real arguments? — YoTengoUnLCD, Jul 30 '15 at 05:11
There are no real numbers that satisfy this, simply because $\sin$ is a continuous function for all real numbers. — SalmonKiller, Jul 30 '15 at 05:11
Why does the continuity on $\Bbb R$ implies that it won't take a transcendental value? $\sin z =1/e$ should have a real z, for example, I think. — YoTengoUnLCD, Jul 30 '15 at 05:13
I know it is cosine specific, but what of the value $\theta$ such that $\cos\theta=\theta$? I am certain there is a corresponding angle whose sine works. specifically, $(\pi/2)-\theta$. I think that number is transcendental. It at least can't be written nicely. — Terra Hyde, Jul 30 '15 at 05:15
http://math.stackexchange.com/questions/958239/
That question is pretty relevant. — Terra Hyde, Jul 30 '15 at 05:18
How about $\sin 1$, or in fact $\sin x$ for any $x\in\mathbb Q\setminus{0}$? — Hagen von Eitzen, Jul 30 '15 at 05:53

score 2 · Answer 1 · answered Jul 30 '15 at 05:23

Nonconstructively, it's not too hard to see that $\sin z$ is transcendental for almost all $z \in \mathbb{C}$. Indeed, even restricting to the real numbers, almost all real numbers in $[-1,1]$ are transcendental. And thus almost every $x \in \mathbb{R}$ is such that $\sin x$ is transcendental.

To see that almost all real numbers are transcendental, note that the algebraic numbers are countable (since their polynomials are countable). But the number of points in $[-1, 1]$ (or, indeed, all of $\mathbb{R}$) is uncountable.

Does there exist a $z\in \Bbb R$ such that $\sin z=t \in \Bbb T$?

1 Answers1