Are $\sin$ and $\cos$ the only functions whose derivatives are equal to each other up to a sign? [closed]

The relationships $x'(t) = -y(t)$ and $y'(t) = x(t)$ imply $$x''(t) = -y'(t) = -x(t)$$ i.e. $$x''(t) = -x(t)$$ which only has solutions $x(t) = A \cos t + B \sin t$ for some constants $A$, $B$. For a given choice of the constants we then get $y(t) = -x'(t) = A \sin t - B \cos t$.

Basically, yes, they are. More precisely: if $x,y\colon\mathbb{R}\longrightarrow\mathbb{R}$ are differentiable functions such that $x'=-y$ and that $y'=x$, then there are numbers $k$ and $\omega$ such that$$(\forall t\in\mathbb{R}):x(t)=k\cos(t+\omega)\text{ and }y(t)=k\sin(t+\omega).$$

Your system is also satisfied by $$ x(t)=y(t)=0 $$

We also have that relationship for $2\sin$ and $2\cos$

I suppose you can set: $$ x\left(t\right) = \exp\left(-i \cdot t\right)\\ y\left(t\right) = i\cdot\exp\left(-i \cdot t\right)\\ i = \sqrt{-1} $$