How do I mathematically prove $ \sin^3(3t)$ is periodic?

Question

I am given $$x(t) = \sin(3t)$$ and asked to mathematically show that $x^3(t)$ is periodic, how exactly would I do this?

Answer 1

Using the fact that $\sin\left(3\left(x+\frac{2\pi}3\right)\right)=\sin(3x)$. It follow from this that$$\sin^3\left(3\left(x+\frac{2\pi}3\right)\right)=\sin^3(3x)$$and that therefore your function is periodic wih period $\frac{2\pi}3$.

Answer 2

Since $t\mapsto\sin(t)$ has period $2\pi$, the function $t\mapsto\sin(3t)$ has period $2\pi/3$. If you "speed up" by factor three, you expect the period to get shorter by the same factor. The third power has the same period as the original function.

You don't have to approve of the argument above as a proof. It does give you a reason to suspect that the period of $x^3$ might be $2\pi/3$. To check whether that is the case, start computing $x^3(t+2\pi/3)$ and see whether after simplification it is $x^3(t)$. Can you take it from here?