<s>[INST] How to find the period of this sinusoid? I'm stuck trying to find the period of this sinusoid and would really like some pointers to different ways to approach this problem.
$$x(t) = \cos(\frac{4\pi t}{5})\sin^2(\frac{8\pi t}{3})$$
I would like to turn the product into a sum and do something about that $2$ in the exponent of $\sin$
$$x(t) = \cos(\frac{4\pi t}{5})\sin^2(\frac{8\pi t}{3}) = \cos(\frac{4\pi t}{5})\left[ 1 - \cos(\frac{16 \pi t}{3}) \right]\frac{1}{2} $$
$$ = \frac{1}{2}\left[ \cos(\frac{4\pi t}{5}) + \cos(\frac{4\pi t}{5}) \cos(\frac{16 \pi t}{3})\right]$$
$$ = \frac{1}{2}\left[ \cos(\frac{4\pi t}{5}) + \frac{1}{2}\cos(\frac{4\pi t}{5} + \frac{16 \pi t}{3}) \cos(\frac{16 \pi t}{3} - \frac{4\pi t}{5})\right]$$
So the sum of cosines will have a period that is the LCM of of the periods of the individuals terms we added. The angular frequencies are $\frac{4\pi}{5}, \frac{92\pi}{15}, \frac{68\pi}{15}$.
Converting this to periods -> we get $\frac{5}{2}, \frac{15}{46}, \frac{15}{34}$, but the answer is $\frac{30}{4}$
 [/INST] To get the LCM, we want the smallest $T$ which is an integer multiple of the three periods. That is, we want integers $n_1,n_2,n_3$ such that $$T=\frac{5}{2}n_1=\frac{15}{46}n_2=\frac{15}{34}n_3$$ with $T$ as small as possible.
To this end we first focus on the middle equality, which rearranges to $23n_1=3n_2$. Since $23$ and $3$ are prime, this requires that $23$ divides $n_2$ and $3$ divides $n_1$. Hence there must be an integer $k$ such that $n_1=3k$ and $n_2=23k$.
With this, the third quantity of the period equality becomes redundant and we have $$T=\frac{15}{2}k=\frac{15}{34}n_3.$$
From this we deduce the value of $n_3$ and therefore know $T$ in terms of $k$. From this we deduce the smallest such value and conclude. (I leave the reader to fill in the details.)
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