How to prove periodicity of a trigonometric function 
$f(x)= \sin(2x)+3\cos(8x)$
Is the function periodic ?

What I did is equalize $f(x)=f(x+T)$ and after noting that $\sin(2x)=\sin(2T)=\sin(8x)=0$ and $\cos(2x)=\cos(2T)=\cos(8x)=1$ we get that both sides of the equation equal 3.
Is that enough to show that the function is periodic ?
Thanks.
 A: $$\begin{align}f(x+\pi)&=\sin(2(x+\pi))+3\cos(8(x+\pi))\\
&=\sin(2x+2\pi)+3\cos(8x+8\pi)\\
&=\sin(2x)\cos(2\pi)+\cos(2x)\sin(2\pi)+3\cos(8x)\cos(8\pi)-3\sin(8x)\sin(8\pi)\\
&=\sin(2x)+3\cos(8x)\\
&=f(x)\end{align}$$
A: no need to be too complicated here:
both $sin \;x$ and $cos \;x$ are perodic, each with period $2\pi$
so $sin \; 2x$ is periodic, with period $\frac{2\pi}{2} = \pi$
and $cos \; 8x$ is periodic, with period $\frac{2\pi}{8} = \frac{\pi}4$
can you find the least common multiple of $\pi$ and $\frac{\pi}4$?
do you see why that must be the period of $\sin(2x)+3\cos(8x)$?
A: periodic says that for some $T$  ,given function repeats its value after $T$,or  
$f(x)=f(x+T)$
now in your case  we know that $sin(x)=sin(x+2*\pi)$ and $cos(x)=cos(x+2*\pi)$
but for some  of periodic function,please read this
If the periods of two periodic functions do not have a common multiple, then their sum is not periodic. Perhaps the simplest example is $\sin(x) + \sin(\pi x),$ whose terms have least periods 2π and 2 respectively.
in your case  period of first function is   $2*\pi/2=\pi$ and  period of second function is $2*\pi/8=\pi/4$
can you continue?
