<s>[INST] Verifying the period of $f(x)=\sin(x)+\cos(x/2)$ It seems clear from the graph of $f(x)=\sin(x)+\cos(x/2)$ that the period $p$ of the function is equal to $4\pi$.

To verify that $4\pi$ is a period of $f(x)$, note that
\begin{align}
\sin(x + 4\pi) + \cos\left(\frac{x + 4\pi}{2}\right) & =\sin(x)\cos(4\pi)+\cos(x)\sin(4\pi)+\cos(x/2)\cos(4\pi/2)-\sin(x/2)\sin(4\pi/2) \\
& =\sin(x)+\cos(x/2)
\end{align}
Thus $4\pi$ is indeed a period of $f$.  My question is, how would one go about trying to prove that $4\pi$ is the smallest $p>0$ such that $f(x+p)=f(x)$?
 [/INST] Let $$\sin(x + T) + \cos(\frac{x+T}{2}) = \sin(x) + \cos(\frac{x}{2})$$And $T\gt 0$. Then we have $$\sin(x+T) - \sin(x) = \cos(\frac{x}{2}) - \cos(\frac{x+T}{2}) \implies$$
$$2\sin(\frac{T}{2})\cos(\frac{2x+T}{2}) = -2\sin(\frac{2x + T}{4})\sin(\frac{-T}{4}) $$ So then $$\sin(\frac{T}{4}) = 0$$ Or $$2\cos(\frac{T}{4})\cos(\frac{2x+T}{2}) = \sin(\frac{2x + T}{4}) \tag{1}$$ For all $x\in \mathbb{R}$. It can be shown that it's not possible $(1)$ holds for all $x\in \mathbb{R}$. So we have $$T = 4k\pi$$ It implies that the fundamental period is $T = 4\pi$.
One way for proving the mentioned statement is using differentiation. For all $x\in \mathbb{R}$ $$2\cos(\frac{T}{4})\cos(\frac{2x+T}{2}) = \sin(\frac{2x + T}{4}) \implies$$ $$-2\cos(\frac{T}{4})\sin(\frac{2x+T}{2}) = \frac{1}{2}\cos(\frac{2x + T}{4}) \implies$$ $$-2\cos(\frac{T}{4})\cos(\frac{2x+T}{2}) = \frac{-1}{4}\sin(\frac{2x + T}{4}) \implies$$ $$\sin(\frac{2x + T}{4}) = \frac{1}{4}\sin(\frac{2x + T}{4}) \implies$$ $$\sin(\frac{2x + T}{4}) = 0 \tag{2}$$ No matter what's the value of $T$, it's not possible $(2)$ holds for all $x\in \mathbb{R}$.
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