$t\mapsto (\cos t,\sin t)$ is a bijection from $[0,2\pi )$ onto the real unit circle

The Pythagorean identity puts every image point on $S^1$.

Given $(x,y)\in S^1$, cosine is continuous on $[0,\pi ]$ with endpoint values $1,-1$; the intermediate value theorem and strict decrease give a unique $u\in [0,\pi ]$ with $\cos u=x$. Since $\sin u\ge 0$ and ${\sin }^{2}u=1-x^2=y^2$, one has $\sin u=\mathrm{\mid }y\mathrm{\mid }$. If $y\ge 0$ take $t=u$; if $y<0$, take $t=2\pi -u\in (\pi ,2\pi )$, for which periodicity and parity give $(\cos t,\sin t)=(x,y)$.

If $(\cos s,\sin s)=(\cos t,\sin t)$, the subtraction formulas and the Pythagorean identity give $\sin (s-t)=0$ and $\cos (s-t)=1$. The zero-set theorem makes $s-t=m\pi$ for some integer $m$; iterating the $\pi$-shift gives $\cos (m\pi )=(-1{)}^m$, so $m$ is even. Thus $s-t$ is a multiple of $2\pi$. Since $s,t\in [0,2\pi )$, $\mathrm{\mid }s-t\mathrm{\mid }<2\pi$, hence $s=t$.