Is the intersection of $\sin(\mathbb{N})$ and $\cos(\mathbb{N})$ empty?

Question

My guess is that the intersection is empty and this is as far as I got in an attempt to prove this by contradiction:

$\exists n,m \in \mathbb{N}, \cos(n)=\sin(m) \land n \neq m \quad (1)$

$\cos^2(n)=1-\cos^2(m) \iff \cos^2(n)+\cos^2(m)=1 \quad (2)$

I'm almost certain that the last equation can't be satisfied but I'm not sure how to proceed.

Answer 1

$$\cos x=\sin y\iff \cos x=\cos\left(\frac\pi2-y\right)\iff \frac\pi2-y=\pm x+2k\pi,\ k\in\Bbb Z$$

In particular, if $\cos x =\sin y$, then either $\dfrac{x+y}\pi$ or $\dfrac{y-x}\pi$ is in the form $\dfrac{1+4k}{2}$ for some $k\in\Bbb Z$.