The function $\mathbb{R}\to\mathbb{R}$ defined by $x\mapsto sin(x)$ is the unique function $f:\mathbb{R}\to\mathbb{R}$ such that
- $f(0)=0$.
- $f'(0)=1$.
- $f''+f=0$.
How do we express these properties characterizing $\sin(x)$ as sentences using first-order predicates in an appropriate first-order language $\mathcal{L}$? Do we need to express additional properties, such as that $\sin(x)$ is everywhere differentiable?
This what I have so far: Consider $\mathbb{R}$ as the $\mathcal{L}$-structure $M:=(\mathbb{R},+,-,\times,0,1,<,=)$, i.e. the real ordered field. Then, the language $\mathcal{L}$ has ternary predicates $\dot{+}, \dot{-}, \dot{\times}$, a binary predicate $\dot{<}$, constants $\dot{0},\dot{1}$, and equality $\dot{=}$.
Viewing the function $\sin(x)$ as binary relation $r$ on the domain of $M$- i.e., $(a,b)\in r$ simply means $b=\sin(a)$; we introduce a new binary predicate $R$ in $\mathcal{L}$. I gather that we must now write that no matter how $R$ is interpreted as a relation $R^{M}$ in $M$, if 1-3 appropriately expressed are satisfied, then $r=R^{M}$.
Indeed, 1 should be the $\mathcal{L}$-sentence $R\dot{0}\dot{0}$ (or $\dot{0}R\dot{0}$ if you prefer to write it like that) since, when intepreted, it will mean $(0,0)\in R^{M}$.
However, I have a problem expressing the first and second derivatives. I guess we do need to use the $\varepsilon-\delta$ definition and perhaps add new symbols to the language?
First of all, if the graph of the function $f$ is defined by $\varphi_f(x,y)$, let's write down the definition of $\lim_{x\to a} f(x) = L$. Note that this is a formula $\psi_f(a,L)$ with two free variables $a$ and $L$.
\begin{align*} \forall \varepsilon\, (\varepsilon > 0\to \exists \delta\, (\delta > 0\land \forall x\,(((a-\delta < x) \land (x \neq a)\land (x < a+\delta)) \to\\ \exists y\, (\varphi_f(x,y)\land (L-\varepsilon < y) \land (y < L+\varepsilon)))) \end{align*}
Now to define the derivative, note that $f'(a) = L$ exactly when $\lim_{x\to a} \frac{f(x)-f(a)}{x-a} = L$. So we can define $f'$ by the formula $\psi_g(a,L)$ above, where $g$ is the function $g(x,a) = \frac{f(x)-f(a)}{x-a}$ (which depends on the parameter $a$). If the graph of $f$ is defined by $\varphi_f(x,y)$, then the graph of $g$ is defined by $\varphi_g(x,y,a)$: $$x\neq a\land \exists z\, \exists w\,(\varphi_f(x,z)\land \varphi_f(a,w)\land (y\times (x-a) = z - w)).$$ We can drop the $x\neq a$ clause at the beginning if we only use the value $y$ in situations where we know $x \neq a$.
Putting these observations together, we can express your second condition $f'(0) = 1$ (writing $R(x,y)$ for the graph of $f$, as in your question) by substituting $\varphi_g$ for $\varphi_f$ in the definition of $\psi_f$ and substituting $0$ and $1$ for $a$ and $L$.
\begin{align*} \forall \varepsilon\, (\varepsilon > 0\to \exists \delta\, (\delta > 0\land \forall x\,(((-\delta < x) \land (x \neq 0)\land (x < \delta))\\ \to \exists y\, ((\exists z\, \exists w\,(R(x,z)\land R(a,w))\land (y\times x = z - w))\\\land (1-\varepsilon < y) \land (y < 1+\varepsilon)))) \end{align*}
I'll leave it to you to write down a formula $\varphi_{f''}(x,y)$ for the graph of $f''$ (which involves nesting the above construction). But once you do, you can express $f''+f = 0$ by: $$\forall x\, \exists y\, \exists z\, (\varphi_{f''}(x,y)\land R(x,z)\land y+z = 0)$$