# Does a bijective map from $(−π,π)→\mathbb R$ exist?

I'm having trouble proving that $\mathbb R$ is equinumerous to $(-π,π)$. I'm think about using a trigonometric function such as $\cos$ or $\sin$, but there are between the interval of $(0,1)$. Could someone help me define a bijective map from $(−π,π)→\mathbb R$?

You could use a trigonometric function such as $\tan$, although you must first divide the number by $2$ to instead get $\tan$ applied to a number in the interval $(-\pi/2,\pi/2)$.
For any open interval $(a,b)$, you can use:
$$f(x) = \frac{x -(a+b)/2}{(x-a)(x-b)}$$