Context :
i'm unable to understand the underlined portion
and when i was solving this question for finding it's period i'm not getting final answer
question:
the fundamental period of first product term is $\dfrac{16\times 15}{gcd(16,15)}=240$ and fundamental period of second product term is $\dfrac{8\times 6}{gcd(8,6)}=24$.so,fundamental period of sum of these product terms i.e, $x[n]$ will be $\dfrac{240\times 24}{gcd(240,24)}=240$ but answer given is 120 and it is however smaller as written in the textbook . but how do i know when it will be smaller
note:
all functions are discrete sequences means $n$ can take only integer values
The fundamental period is the smallest $k>0$ such that $u(n+k)=u(n)$.
Actually the first result stated is wrong: even for a sum you have no such guarantee on the fundamental period.
Take for instance:
$$u(n)=0,1,0,1,\dots$$ $$v(n)=1,0,1,0,\dots$$
They have period $N_1=N_2=2$. The fundamental period of the sum should be
$$N=\frac{N_1N_2}{gcd(N_1,N_2)}=2$$
But $u(n)+v(n)=1,1,1,1,\dots$
Here the fundamental period of the sum is $1$.
Note that the fundamental period of $u(n)v(n)$ is also $1$, since $u(n)v(n)=0,0,0,0,\dots$.
Regarding the preent specific case,
$$u(n)=\cos(\frac{n\pi}{8})\cos(\frac{2n\pi}{15})+\sin(\frac{n\pi}{3})\sin(\frac{n\pi}{4})$$
The trig functions have period resp. 16,15,6,8, hence the lcm of these numbers is certainly a period of $u$, that is $240$.
Is it the smallest?
The prime divisors of $240$ are $2$, $3$ and $5$. The simplest may be to check $u(240/2)$, $u(240/3)$ and $u(240/5)$ and compare to $u(0)=1$.
We have
$$u(120)=\cos(15\pi)\cos(16\pi)+\sin(40\pi)\sin(3\pi)=-1$$ $$u(80)=\cos(10\pi)\cos(10\pi+2\pi/3)+\sin(26\pi+2\pi/3)\sin(20\pi)=-1/2$$ $$u(48)=\cos(6\pi)\cos(6\pi+2\pi/5)+\sin(16\pi)\sin(12\pi)=\cos(2\pi/5)\ne1$$
Conclusion: no divisor of $240$ may be a smaller period. Hence $240$ is the fundamental period.