Period of $f(x)=\sin{2x}-\sin{\frac{x}{2}}$ I know the periods of the two functions $f(x)=\sin{2x}$ and $g(x)=\sin{\frac{x}{2}}$, which are $\pi$ and $4\pi$ respectively. But what is their period when you subtract them? 
More generally, if $f(x)=\sin{(ax+c)}$ and $g(x)=\sin{(bx+d)},$ what is the period of $f(x)-g(x)?$ 
 A: The period is the smallest value $>0$ that you can add to the function parameter so that the function value will be unchanged.
Combining two periodic functions means they will combine to the same value when you can add a value to the parameter that is a multiple of both periods. In this case the smallest such value is $4\pi$, since this is the least common multiple of $\pi$ and $4\pi$. 
If the periods of contributing functions were, say, $6$ and $10$, the combined function would be periodic on an interval of $\text{lcm}(6,10) = 30$. If the periods had no common integer multiple - say one period was $2$ and the other was $\pi$ - the combined function would not be periodic.
There may be shorter periods in some cases (which would have to divide this value) but I'm not sure if that is possible with simple addition of functions.
A: In short, period of $\sin 2x$ is $\pi$ and period of $\sin \dfrac{x}{2}$ is $4\pi$. Then answer will be $\text{lcm} (1,4)\cdot\pi=4\pi$

Now, elaborately:
$\sin 2x-\sin \dfrac{x}{2}=2\sin (x-\dfrac{x}{4})\cos(x+\dfrac{x}{4})=2\sin \dfrac{3x}{4}\cos\dfrac{5x}{4}$
If $x=4k\pi$ and $k\in \mathbb{Z}$, either $\cos\dfrac{5x}{4}$ or $\sin \dfrac{3x}{4}$ is $0$. Hence period is $4\pi$.
