Is it possible to fit cosine (or sine) graph into an algebraic curve?

Question

A set of the form $\{(x,y)\in \mathbb{R}^2\,:\,P(x,y)=0\}$ for a nonzero polynomial $P\in \mathbb{R}[x,y]$ is said to be an algebraic curve. This set can be denoted simply by "$P(x,y)=0$".

Question: Can we fit the graph of cosine, namely, $\{(t,\cos t)\in \mathbb{R}^2\,:\,t\in\mathbb{R}\}$, into an algebraic curve?

Intuition says "no" and that's correct.

Indeed, suppose there is some nonzero polynomial $P$ such that $P(t,\cos t)\equiv 0$. Consider the polynomial $Q_\lambda(x,y):=y-\lambda$, for $\lambda\in [-1,1]$. Due to cosine's periodicity, the algebraic curve $Q_\lambda(x,y)=0$ intersects the graph of cosine infinitely many times. This tells us that the algebraic curves $P(x,y)=0$ and $Q_\lambda(x,y)=0$ have an infinite intersection and, by an Algebraic Geometry theorem, $P$ and $Q_\lambda$ must have a nonconstant common factor. Since $Q_\lambda$ is irreducible, this factor must be $Q_\lambda$ itself. So we are saying that $y-\lambda$ is a factor of $P$ for every $\lambda \in [-1,1]$... Since $P$ is a polynomial, this is impossible, unless $P\equiv 0$, which we discarded by hypothesis.

Very good. But what if we wanted to answer the "local" version of the question?

Question (local): Can we fit a piece of the graph of cosine, say, $\{(t,\cos t)\in\mathbb{R}^2\,:\,t\in (a,b)\}$, into an algebraic curve?

Intuition still says that we can't, but I don't know how to prove this.

Since we're limited to a piece of the graph, we lose the argument of that infinite intersection...

Answer 1

General case (for $P\in\mathbb{R}[X]$):

This is a direct implication of the fact that any trigonometric function is transcendental in any open interval of $\mathbb{R}$. One such proof goes as follows. A function $f$ is real-algebraic iff the set $\{x^kf(x)^l\}$ is $\mathbb{R}$-linearly dependent. What we do then is find a ODE which annihilates this set (on said interval) and check if the wronskian is identically zero. This is a very powerful technique for these types of proofs. See this link for more details.

Special case (for $P\in\mathbb{Q}[X]$):

If $P[t,\cos(t)]=0$ on an open set, then there exists a non-zero rational s.t. $P[q,\cos(q)]=0$ meaning $\cos(q)$ is algebraic, but $\cos(x)$ is transcendental for $x\neq0$ algebraic.

Answer 2

If the level set $F(x,y)=0$ of a generic $C^1$ function $F:\mathbb{R}^2\to \mathbb{R}$ admits a local parameterization $(t,g(t))$ in some open subset of $\mathbb{R}^2$ containing $(a,b)$ where $F(a,b)=0$ and $\partial F/\partial y > 0$, then by the implicit function theorem (Theorem 1 here, with $n=1$) such a $g(t)$ is unique and must satisfy $$-g'(t)=\left.\frac{\partial F/\partial x}{\partial F/\partial y}\right|_{(t,g(t))}$$ in some interval $(t_1,t_2)$ containing $a$.

If $(t,\cos{t})$ were such an admissible parameterization on an interval where $\cos{t}$ is decreasing, then $F$ would have to satisfy $\sqrt{1-\cos^2{t}}=\left.\frac{\partial F/\partial x}{\partial F/\partial y}\right|_{(t,\cos{t})}$. Likewise on an interval where $\cos{t}$ is increasing, the same holds with a minus sign.

Assuming $F$ is a polynomial in $\mathbb{R}[x,y]$ derives a contradiction. The right-hand side is an element of the field of fractions $\mathbb{R}(t,\cos{t})$, that is, the field of rational functions $\left.\frac{P(x,y)}{Q(x,y)}\right|_{(t,\cos{t})}$ with $P,Q\in \mathbb{R}[x,y]$. The left-hand side cannot be equal to any such rational function evaluated at $(t,\cos{t})$ in any open interval $(t_1,t_2)$ because of the irreducible square root.

The condition $\partial F/\partial y > 0$ doesn't exclude the case $\partial F/\partial y < 0$ from the same result, because the level set $F(x,y)=0$ is the same set as the level set $-F(x,y)=0$. However, $\partial F/\partial y = 0$ is a separate case. If this is true everywhere in an open set containing $(a,b)$ with $F(a,b)=0$, then $F\in \mathbb{R}[x,y]\to F\in \mathbb{R}[x]$, and $F(x)=0$ in the whole open set means $F$ must be the zero polynomial.