Proving $f(x) = \sin(x) + \sin(\sqrt{x})$ is not periodic without using derivation

Question

Hello I come across a dilemma in my school, I need to prove that $f(x) = \sin(x) + \sin(\sqrt{x})$ is not periodic without using derivation

Answer 1

Suppose there is $T>0$ such that, for all $\alpha\ge 0$,

\begin{align}0=&\sin(\alpha+T)+\sin\sqrt{\alpha+T}-\sin\alpha-\sin\sqrt\alpha=\\=&2\cos\frac{\alpha+T}{2}\sin\frac T2+2\sin\frac{T}{2\sqrt{\alpha+T}+2\sqrt{\alpha}}\cos\frac{\sqrt{\alpha+T}+\sqrt\alpha}{2}\end{align}

Notice that $$\lim_{\alpha\to\infty}2\sin\frac{T}{2\sqrt{\alpha+T}+2\sqrt{\alpha}}\cos\frac{\sqrt{\alpha+T}+\sqrt\alpha}{2}=0$$

Thus, by hypothesis, it must hold $$\lim_{\alpha\to\infty}2\cos\frac{\alpha+T}{2}\sin\frac T2=0$$

However, $2\cos\frac{\alpha_k+T}{2}\sin\frac T2=2(-1)^k\sin\frac T2$ for $\alpha_k=2k\pi-T$. Thus, it must hold $$\sin\frac T2=0$$ Id est, $$T=2h\pi$$ for some integer $h>0$. Thus, it must hold $$2\sin\frac{2h\pi}{2\sqrt{\alpha+2h\pi}+2\sqrt{\alpha}}\cos\frac{\sqrt{\alpha+2h\pi}+\sqrt\alpha}{2}=0\qquad\forall\alpha\ge0$$ Evaulate in $\alpha=0$ $$\sin\frac{h\pi}{\sqrt{2h\pi}}\cos\frac{\sqrt{2h\pi}}{2}=\sin\sqrt{\frac{h\pi}2}\cos\sqrt\frac{h\pi}{2}=0$$

But this would mean that $\sqrt\pi$ is the root of some rational number: famously false.