Let $R$ be a commutative ring. Write $A[\imath]=\tfrac{A[x]}{\langle x^2+1\rangle }$. Write also $R[\cos\theta,\sin\theta]=\tfrac{R[x,y]}{\langle x^2+y^2-1\rangle}$. Consider the $R[\imath]$-subalgebra $R[\cos\theta+\imath \sin\theta]\leq R[\imath,\cos\theta,\sin\theta]$.
Question. Are $R[\imath]$ algebra morphisms $R[\cos\theta+\imath\sin\theta]\to A$ in canonical bijection with elements $a+\imath b\in A$ where $a,b\in A$ satisfy $a^2+b^2=1$?
Something is strange because I think there are reciprocal $A[\imath]$-algebra isos$$A[\imath, \cos\theta,\sin\theta]\cong \tfrac{A[\tfrac12,\imath , z,\bar z]}{\langle z\bar z-1\rangle }\cong A[\tfrac 12,\imath][z,\tfrac1z]$$ to the punctured complex plane where the $\rightarrow$ direction acts by $\cos\theta\mapsto \tfrac{z+\bar z}{2}$ and $\sin\theta\mapsto \tfrac{z-\bar z}{2\imath}$.
That is strange because if this is indeed an $A[\imath]$-algebra iso then $e^{\imath\theta}:=\cos\theta+\imath \sin \theta$ just acts like a generic unit, so $A[\imath]$-algebra morphisms out of $A[e^{\imath \theta}]$ just pick out units not necessarily of the desired form. For instance if $A=\mathbb R$ is a field then $10\in \mathbb R[\imath]$ is a unit not on the circle.
In the latter case, is the functor on $A[\imath]$-algebras that returns the "circle" representable in some other way?
Edit. I think it's hopeless since the "circle" is not functorial: $A[\imath]$-algebra morphisms need not take circle elements to circle elements.
This line is ambiguous; $R[\cos \theta + i \sin \theta]$ is not an $R[i]$-subalgebra. If you're talking about $R[i, \cos \theta + i \sin \theta]$ then as an $R[i]$-algebra this is just $R[i][z, z^{-1}]$ (assuming for simplicity that $2$ is invertible in $R$) so $z$ can be sent to any unit, as you've observed.
You are also correct that elements $a + ib$ in $R[i]$-algebras such that $a^2 + b^2 = 1$ aren't functorial (and this is an example).
There is a "circle functor" but it takes an arbitrary commutative ring $A$ to the set of pairs $(a, b) \in A^2$ such that $a^2 + b^2 = 1$, and it is represented by $\mathbb{Z}[a, b]/(a^2 + b^2 - 1)$. This is even an affine group scheme, namely $SO_2$. It can also be thought of as consisting of elements $a + bi \in A[i]$, so it is the group of norm $1$ elements in $A[i]$ with respect to the norm $N(a + bi) = a^2 + b^2$. So we do not assume that $A$ contains $i$ but instead we freely adjoin it.