Let $f: \mathbb{R}\rightarrow \mathbb{R}$
$$f(x)=\frac{\sin(\sin{(x)}+x)}{2+\cos{(\lvert x\rvert}+\cos{(x)})}$$
I am having trouble showing whether or not this function is monotonic and/or periodic.
Monotonic: The function is only monotonic if it is entirely non-increasing or entirely non-decreasing.
Monotonically increasing: $\forall x,y$ such that $x\le y$ it follows that $f(x) \le f(y)$. Choose $x=\pi,y=\frac{3\pi}{2}$: $$\implies f(\pi)=\frac{\sin{(\sin{\pi}+\pi)}}{2+\cos{(\pi+\cos{\pi}})}=\frac{\cos{1}}{2} \\f(\frac{3\pi}{2})=\frac{\sin{(\sin{\frac{3 \pi}{2}}+\frac{3\pi}{2})}}{2+\cos(\frac{3\pi}{2}+\cos{\frac{3\pi}{2}})}=\frac{-\cos{1}}{2} \\ \frac{\cos{1}}{2} > -\frac{\cos{1}}{2} \implies \text{not monotonically increasing}$$
Monotonically decreasing $\forall x,y$ such that $x \le y$ it follows that $f(x) \ge f(y)$. Choose $x=0, y=\frac{\pi}{2}$:
$$\implies f(0)=0 \\ f(\frac{\pi}{2})=\frac{\sin{(\sin{\frac{\pi}{2}}+\frac{\pi}{2})}}{2+\cos{(\frac{\pi}{2}+\cos{\frac{\pi}{2}}})}=\frac{\cos{1}}{2} \\ 0 < \frac{\cos{1}}{2} \implies \text{not monotonically decreasing}$$
Is it possible to argue this way? Am I getting the definitions for monotonicity right?
Periodic: This is where I am less sure and quite confused. I know that a function $f$ is periodic if $f(x+T)=f(x)$. I know that $\sin{(x)}$ and $\cos{(x)}$ are $2 \pi$ periodic functions so I my attempt was to just plug in $f(x+2 \pi)$ and see where it leads me.
$$\implies f(x+2 \pi)=\frac{\sin(\sin{(x+2 \pi)}+2 \pi)}{2+\cos{(\lvert x+2 \pi\rvert}+\cos{(x+2 \pi)})}=\frac{\sin{x}}{2+\cos{(\lvert x+2\pi\rvert+\cos{x})}}$$
If $x \ge 0$ I get the desired solution and $f(x+2 \pi)=f(x)$. However, for negative $x$, $\lvert x+2 \pi \rvert \not=x+2 \pi$. So I cannot show periodicity for all negative $x$. How can I resolve this? Can a function be periodic for positive $x$ and not for negative?