How does one prove that if $\cos(t) = \cos(t')$ and $\sin(t) = \sin(t')$ then $t = t' + 2k\pi$? 
How does one prove that if $\cos(t) = \cos(t')$ and $\sin(t) = \sin(t')$ then $t = t' + 2k\pi$ ?

I've tried proving the above statement, which I think is valid.
I know $\sin(t)$ is injective on $[-\pi/2; \pi/2]$ and $\cos(t)$ is injective on $[0; \pi]$, but until now I've not been able to use this to prove the statement rigorously.
 A: also can 
$$\cos{x}-\cos{y}=0\Longrightarrow 2\sin{\dfrac{x-y}{2}}\sin{\dfrac{x+y}{2}}=0\tag{1}$$
$$\sin{x}-\sin{y}=0\Longrightarrow 2\sin{\dfrac{x-y}{2}}\cos{\dfrac{x+y}{2}}=0\tag{2}$$
with $(1),(2)$ must 
$$\sin{\dfrac{x-y}{2}}=0$$
since we kown 
$$\sin{(k\pi)}=0,k\in Z$$
so
$$\dfrac{x-y}{2}=k\pi,\Longrightarrow x=y+2k\pi,k\in Z$$
A: \begin{align*}
 \cos(t-t') &= \cos(t)\cos(t') + \sin(t)\sin(t') \\
            &= \cos^2(t) + \sin^2(t) = 1 \\
 \sin(t-t') &= \sin(t)\cos(t') - \cos(t)\sin(t') \\
            &= \sin(t)\cos(t) - \cos(t)\sin(t) = 0.
\end{align*}
As $\sin(t-t')=0$ we have $t-t'=k\pi$ for some $k\in\mathbb{Z}$, and so $\cos(t-t')=\cos(k\pi)=(-1)^k$.  However, we have also seen that $\cos(k\pi)=1$, so $k$ is even, so $t-t'=2m\pi$ for some $m\in\mathbb{Z}$.
A: If you want to do it rigorously, you could use the fact that $\sin$ and $\cos$ are orthonormal in an inner product space. See: http://www.jimworthey.com/orthoquestions.html
A: if $\cos(t) = \cos(t^\prime), $ then $t=\pm t^\prime + 2k\pi $ and if $\sin t = \sin t^\prime,$ then $t = t^\prime + 2k\pi, \pi - t^\prime + 2k\pi$ these follow form $\cos$ being an even function while $\sin$ is an odd function and both are $2\pi$-periodic. that is all you need. 
A: The point $P_t=(\cos t,\sin t)$ is the point where the ray at angle $t$ meets the unit  circle. Your statements are saying that $P_t=P_{t'}$. This means the two rays coincide, so their angles must differ by a multiple of a full rotation. That is, $t-t'=2k\pi$ for some integral $k$.
A: i dont know if this help but.
we have that for $k\in\mathbb{Z}$
$$\begin{align}
\sin t=\sin t'&\iff t'=2k\pi+\pi-t\vee t'=2k\pi+t\\
\cos t=\cos t'&\iff t'=2k\pi-t\vee t'=2k\pi+t
\end{align}$$
then to have $\sin t=\sin t'$ and $\cos t=\cos t'$ at same time you get $t'=2k\pi+t$ because
if $t'=2k\pi+\pi-t$ you have $\sin t'=\sin(2k\pi+\pi-t)=\sin(\pi-t)=\sin t$ but $\cos t'=\cos(2k\pi+\pi-t)=\cos(\pi-t)=-\cos t$
if $t'=2k\pi-t$ you have $\cos t'=\cos(2k\pi-t)=\cos(-t)=\cos t$ but $\sin t'=\sin(2k\pi-t)=\sin(-t)=-\sin t$
A: Cos(x) and sin(x) are periodic functions of period 2k(pi).hence sin(x)=sin(x') and cos(x)=cos(x').
