Here is my signal
Cos(n/2)*cos(pi*n/4)
cos(n/2) has period 4pi and cos(pi*n/4) has period 8
Now, the question is will the signal be periodic for fundamental period 32pi ?
Here is my signal
Cos(n/2)*cos(pi*n/4)
cos(n/2) has period 4pi and cos(pi*n/4) has period 8
Now, the question is will the signal be periodic for fundamental period 32pi ?
$\cos\left(\dfrac{n}{2}\right)*\cos\left(\pi*\dfrac{n}{4}\right)$
Now finding out period of $\cos\left(\dfrac{n}{2}\right)$
$\cos\left(\dfrac{n}{2}\right)=\cos\left(2*\pi*\dfrac{n}{N1}\right) \implies \dfrac{1}{2}=2*\dfrac{\pi}{N1} \implies N1=4*\pi$
Now finding out period of second function $\cos\left(\pi*\dfrac{n}{4}\right)$
$\cos\left(\pi*\dfrac{n}{4}\right) = \cos\left(2*\pi*\dfrac{n}{N2}\right)\implies \dfrac{\pi}{4}=2*\dfrac{\pi}{N2} \implies N2=8$.
Now for total function period is LCM of $(N1,N2)$ i.e, $\mathrm{LCM}(4*\pi,8) \implies32*\pi$.
Therefore, the period of the function is $\boxed{32*\pi}$.
You can try using this:
(1) cos(a+b) = cos(a)*cos(b)+sin(a)*sin(b)
(2) cos(a-b) = cos(a)*cos(b)-sin(a)*sin(b)
Then you can do (1)+(2) and get that
cos(a)*cos(b)=1/2(cos(a+b)+cos(a-b))
Then you can identify the period of cos(a+b)-cos(a-b) by using this for example http://fourier.eng.hmc.edu/e101/lectures/Fundamental_Frequency.pdf
Hope this helps