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 $f(x+p)=f(x)$?
 A: 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 $x\in \mathbb{R}$.
A: If $f$ is periodic with period $T$ then so is $f'$. This means that
$$ f'(x) = c \implies f'(x+T) = c.$$
When looking at the graph of $f$ it looks like the solutions of the equation
$$ f'(x) = f'(\pi)$$
are exactly the points $S = \{  \pi + 4k \pi : k \in \mathbb Z\}$ . If we manage to prove this then we are finished since for a smaller period $\tilde T,$ $\pi + \tilde T$ would not be a solution of the equation which contradicts the periodicity of $f'$.
We now solve $f'(x) = f'(\pi).$ By definition of $f$ we have $f'(x) = \cos(x) - \frac 12 \sin \frac x2$ and $f'(\pi) = -1.5$. Let $x = 2u$ and write $-1.5$ as $ -1.5 = -1 - 1/2$ and we have
$$ f'(x) = -1.5 \iff \cos(2u)- \frac 12\sin(u) = -1 - 1/2 \iff \cos(2u)+1 -\frac 1 2 \sin(u)  + 1/2 = 0.$$
Using the identity
$$ \cos(2u) + 1 = 2 -2 \sin^2(u)$$
(which can be deduce from the double angle formula and the $\cos^2 u + \sin^2 u =1)$ we have
$$ -2 \sin^2 u - \frac 1 2 \sin u + 2.5 = 0 \iff -4 \sin^2(u)- \sin(u) + 5 = 0$$
This is a quadratic equation in $\sin(u)$ whose solutions are
$$ \sin(u) = 1, \sin(u) = - 5/4.$$
Since $-5/4 < -1$ we have
$$ f'(x) = -1.5 \iff \sin(x/2) = 1 \iff x = \pi + 4 k \pi, k \in \mathbb Z$$
Therefore $4\pi$ is the smallest possible period of $f$.
A: If you already have a $p$ such that $f(x+p)=f(x)$ for all $x$, then you can look at the function on the segment $[0,p)$ and check if it can be written as several copies (after proving that any other period is $p/n$ for integer $n$). You can look at the intersections with the origin, for instance. If there is more than one, you can check if $f'(x_1)=f'(x_2)$. if not, then there is no smaller period. Otherwise, you have to keep checking.
