I was asked the following question two days ago, but I couldn't completely resolve it.

Here is the claim: $\mathcal R = (\mathbb R,+,\cdot)$ is the real field.

Let $I$ be an open interval (perhaps unbounded) in $\mathbb R$ and let $f: I \rightarrow \mathbb R$ be $C^1$ and such that $f'$ has no zeros. Then the structures:

$$(\mathcal R, sin (f)) = (\mathcal R, cos (f))$$ are interdefinable.

Notes: Start with $(\mathcal R, sin (f))$:

1) We know that the absolute value of $cos(f)$ is definable, simply by the Pythagorean theorem.

2) The fact that the derivative of $f$ has no zeros, along with the fact that the derivative of $sin(f)$ is $cos(f) \cdot f'$ and is definable means that we can identify all of the places at which $cos(f)$ switches signs. That is the zeros of $cos(f) \cdot f'$ are the same as $cos(f)$. The fact that $f'$ does not vanish assures that $cos(f)$ actually switches signs at these points.

So, here is one approach: identify the sign of $cos(f)$ on some interval near zero. Now we must simply identify, in a first order way, the number of zeros of the function $cos(f)$ between the interval on which you are defining the value and the first interval. I don't see exactly how to do this. Do you?

Of course, perhaps someone has a better way to do this.