If $\cos x=\cos y$ and $\sin x=\sin y$ is $x=y$ or $x=2\pi,y=0$ or $x=0,y=2\pi$? $x,y\in[0,2\pi]$

Question

I know that if $\sin x=\sin y$ then $x=n\pi+(-1)^ny$

and $\cos x=\cos y$ implies that $x=y+2k\pi$ or $x=-y+2l\pi$. Since the latter does not fit the form above $x=y+2k\pi$ which implies that $k=0,1,-1$. Hence, we have x=y or $x=2\pi,y=0$ or $x=0,y=2\pi$.

Answer 1

Since $e^{ix}=e^{iy}$ by Euler's identity, it follows $x-y=2n\pi$ for some integer $n$. If both $x,y$ belong to $[0,2\pi]$, only possible values are $x=y\in [0,2\pi]$, $(x,y)= (0,2\pi)$ or $(2\pi,0)$.

Answer 2

Use http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html,

$$0=\cos x-\cos y=?$$

$$0=\sin x-\sin y=?$$

Now $\sin\dfrac{x+y}2=\cos\dfrac{x+y}2=0$ is untenable, right?

So, $\sin\dfrac{x-y}2=0\implies\dfrac{x-y}2=m\pi$ for some integer $m$