For given $t$ and $x$ and $y$, is there at least one $f$ such that $\cos ft = x, \sin ft =y$?

Question

Suppose that $t$, $x$ and $y$ are given and are all in $\mathbb{R}$. Is there always at least one $f$ such that $\cos ft = x, \sin ft =y$?

Edit: OK I forgot to add that given $x$ and $y$ are such that $x^2 + y^2 = 1$ and $t \neq 0$ and $t>0$.

Answer 1

The measure of the angle in the centre (in radians) is $cos^{-1}(x)$. So $f=cos^{-1}(x)/t$ will work.

Answer 2

As$$\sin^2(ft)+\cos^2(ft)=1$$ is true whatever the argument, any $f$ is a solution.