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.
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
$$\dfrac{x-y}{2}=k\pi,\Longrightarrow x=y+2k\pi,k\in Z$$
Cos(x) and sin(x) are periodic functions of period 2k(pi).hence sin(x)=sin(x') and cos(x)=cos(x').