Two equations involving $\sin$ and $\cos$.

Question

Let $w$ be a value which is known to be somewhere in the open interval $(a,b) \subset [0,2\pi]$.

If $x$ is a value in $[0,1)$, and we know that $$\cos{2\pi x} = \cos{w} \\ \sin 2\pi x = \sin w$$ can we prove that this happens if and only if $x \in (a/2\pi, b/2\pi)$?

This would be a convenient result in a larger proof I am writing, but I am unsure if this is true, and unsure how to prove if it is or isn't since my trig skills are very sloppy.

Answer 1

Yes, given that you are within one circle the sine and cosine values are sufficient to define an angle, so you have $2\pi x = w$. Given the restriction on $w$, the restriction on $x$ follows.

Answer 2

The system of trigonometric equations is satisfied if and only if $$2\pi x\equiv w\mod 2\pi\iff x\equiv \frac w{2\pi} \mod 1\enspace(\text{i.e} \bmod\mathbf Z).$$ So yes, $\;x\in\Bigl(\dfrac a{2\pi},\dfrac b{2\pi}\Bigr)$.