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 $\frac{30}{4}$
Best Answer
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.)