How to describe the ring $\mathbb{C}[\cos{x},\sin{x}]$ as a quotient ring of $\mathbb{C}[X,Y]$? Here, $\sin{x}$ and $\cos{x}$ are real functions $\mathbb{R} \to \mathbb{R}$.
First, let me explain what I mean by "as a quotient ring". It can be easily shown that
$$
\mathbb{R}[\cos{x},\sin{x}] \cong \mathbb{R}[X,Y]/(X^2+Y^2-1).
$$
Here $\mathbb{R}[\cos{x},\sin{x}]$ is a quotient ring of $\mathbb{R}[X,Y]$. So when the field is changed to $\mathbb{C}$, I immediately think about the ring
$$
\mathbb{C}[X,Y]/(X^2+Y^2-1).
$$
As people have pointed out, my argument quoted below is questionable, so I redo it in update 2. This paragraph is left here only for the record. I elaborated it in update 3.

However, $\mathbb{C}[\cos{x},\sin{x}]$ is not a UFD ($\sin^2{x}=(1+\cos{x})(1-\cos{x})$), but $\mathbb{C}[X,Y]/(X^2+Y^2-1) \cong \mathbb{C}[T,T^{-1}]$ is a UFD. Therefore $\mathbb{C}[\cos{x},\sin{x}]$ can only be a proper subring quotient ring of $\mathbb{C}[X,Y]/(X^2+Y^2-1)$. An extra restriction is missing here. Is there anyway to find it? I think there is something derived from the fact that $\mathbb{C}$ is algebraically closed.

Update 1: It is not very obvious that $\mathbb{C}[X,Y]/(X^2+Y^2-1)$ is a PID and therefore UFD, but one can find proofs in this question post: Ring of trigonometric functions with real coefficients
Update 2: It seems my argument on being UFD is a little messed up, so I will redo it. First of all, $R=\mathbb{R}[X,Y]/(X^2+Y^2-1)$ is not a UFD (one can show that its ideal class group is $\mathbb{Z}/2\mathbb{Z}$) and $S=\mathbb{C}[X,Y]/(X^2+Y^2-1) \cong \mathbb{C}[T,T^{-1}]$ is a UFD (see the link in update 1). Extending the scalar here is not a trivial matter, so the relation between $\mathbb{C}[\cos{x},\sin{x}]$ and $\mathbb{C}[X,Y]/(X^2+Y^2-1)$ is not likely to be as immediate as the case of real scalar.
Update 3: In the ring $\mathbb{R}[\cos{x},\sin{x}]$, irreducible elements are of the form $a\sin{x}+b\cos{x}+c$ where $a^2+b^2 \ne 0$; meanwhile, in the ring $\mathbb{C}[\cos{x},\sin{x}]$, irreducible elements are of the form $\cos{x}+i\sin{x}+a$ where $a \in \mathbb{C}^\ast$. I found them on this book (section: The Trigonometric Polynomial Rings). Hence $\sin{x}$, $1-\cos{x}$ and $1+\cos{x}$ are irreducible in $\mathbb{R}[\cos{x},\sin{x}]$ but not irreducible in $\mathbb{C}[\cos{x},\sin{x}]$. This is the source of my mistake. According to the comments this question have received, this change is not easily spotted. So I think I shall leave it here because it may serve as a counter-example on irreducibility.
 A: First of all, $\mathbb{C}[\sin,\cos]$ certainly cannot be a proper subring of $\mathbb{C}[X,Y]/(X^2+Y^2-1)$ (with $X$ corresponding to $\cos$ and $Y$ corresponding to $\sin$).  As a $\mathbb{C}$-algebra, $\mathbb{C}[X,Y]/(X^2+Y^2-1)$ is generated by $X$ and $Y$, so there is not any proper $\mathbb{C}$-subalgebra that contains both $X$ and $Y$.  A priori, $\mathbb{C}[\sin,\cos]$ could be a proper quotient of $\mathbb{C}[X,Y]/(X^2+Y^2-1)$, since there could be more relations between $\sin$ and $\cos$ that are not generated by the single relation $X^2+Y^2-1$.  However, this is not the case (and the proof is basically the same as the proof over $\mathbb{R}$ that you say you know), so $\mathbb{C}[\sin,\cos]$ really is isomorphic to $\mathbb{C}[X,Y]/(X^2+Y^2-1)$.
So what's going on?  Well, the equation $\sin^2=(1+\cos)(1-\cos)$ simply does not imply that the ring is not a UFD.  After all, $6^2=4\cdot 9$ but $\mathbb{Z}$ is still a UFD.  To conclude that you have a failure of unique factorization, you would need to know something more about the factors $\sin,1-\cos,$ and $1+\cos$, such as that they are irreducible and not associate to each other.
In fact, none of these factors are irreducible.  To see this, let us use the isomorphism $\mathbb{C}[X,Y]/(X^2+Y^2-1)\cong\mathbb{C}[T,T^{-1}]$ which is given by mapping $T$ to $X+iY$ (and $T^{-1}$ to $X-iY$).  So in terms of $T$, $\sin$ (or $Y$) would be $\frac{T-T^{-1}}{2i}=\frac{T^2-1}{2iT}$.  This is not irreducible, because $\frac{1}{2iT}$ is a unit (so can be ignored) and $T^2-1$ factors as $(T+1)(T-1)$ (and neither factor is a unit).  Similarly, $1\pm\cos$ becomes $1\pm\frac{T+T^{-1}}{2}=\pm\frac{T^2\pm 2T+1}{2T}$ and the numerator factors as $(T\pm 1)^2$.  So up to units, $\sin$ is $(T+1)(T-1)$ and $1+\cos$ and $1-\cos$ are $(T+1)^2$ and $(T-1)^2$, so the factorization $\sin^2=(1+\cos)(1-\cos)$ is just like the factorization $6^2=4\cdot 9$ of integers (with $T+1$ corresponding to $2$ and $T-1$ corresponding to $3$).
