# Interdefinability of two expansions of the Real Field

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.

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Let's assume $f'>0$ on $I$. The function $s$ that sends $x$ to $\frac{f'(x)\cos(f(x))}{|f'(x)\cos(f(x))|}$ if $f'(x)\cos(f(x))\neq 0$ and to $0$ else is definable in the first structure and, as you noted, $\cos(f(x))=s(x)|cos(f(x))|$ is therefore also definable.
You didn't miss anything assuming that $f'>0$ on $I$, I agree. Now I can't remember if the person said that the function was actually $C^1,$ as I wrote or if it was supposed to just differentiable. Anyway, since you answered the question I actually wrote down, I accept your answer. –  James Freitag Nov 19 '11 at 18:13