Question: I need a set-builder notation for $\{1 , 3 , 5 , -7 , -9 , -11 , 13 , 15 , 17 , \ldots\}$. It is an infinite set of odd positive integers, which are multiplied by $-1$ three by three. One of the terms of this question is not to use floor function.

Answer: 
If you are allowed to use square roots, then something like: $$\left\{(2n+1) \cdot(-1)^{\tfrac{1}{\sqrt{5}} \left(\left(\tfrac{1+\sqrt{5}}{2}\right)^{n-1} - \left(\tfrac{1-\sqrt{5}}{2}\right)^{n-1}\right) - (n-1)} \big| n \in \mathbb{N}\right\}$$
will work. (Here, we use the convention that $0 \in \mathbb{N}$.)

This answer has the advantage that it doesn't use the floor/ceiling functions, modular arithmetic, or trigonometric functions (as OP requested), but the downside is that it's complicated.
For $n\in\mathbb{N}$, $\cos(n\pi/3)$ is never $0$. It oscillates from being positive for three consecutive terms, then negative for three. Using this, one way to express your sequence is $$a\_n=\frac{\cos\left(\frac{(n-2)\pi}{3}\right)}{\left\lvert\cos\left(\frac{(n-2)\pi}{3}\right)\right\rvert}(2n-1)$$ As a set in set-builder notation, you have $$\left\{\frac{\cos\left(\frac{(n-2)\pi}{3}\right)}{\left\lvert\cos\left(\frac{(n-2)\pi}{3}\right)\right\rvert}(2n-1)\middle| n\in\mathbb{N}\right\}$$

Of course there is nothing more fundamental about the absolute value function than the signum function, so another way to express this is:$$\left\{\operatorname{sign}\left(\cos\left(\frac{(n-2)\pi}{3}\right)\right)(2n-1)\middle| n\in\mathbb{N}\right\}$$