Why is the fundamental period of $\sin^3(2t)$ given by $1/\pi$ rather than $\pi$?

Question

Take a look at the following function: $$ x(t) = \sin^3(2t) $$ In order to show the periodicity of the signal, we need to prove the following equality $$ x(t) = x(t+T) $$ We first use some trigonometric identities: $$ x(t) = \sin^3(2t) = \frac{3}{4} \sin(2t)-\frac{1}{4}\sin(6t) \tag{1} $$ What I've done is as follows:

$$ \begin{align} x(t+T) &= \sin^3(2(t+T)) \\ &=\frac{3}{4}\sin(2(t+T)) - \frac{1}{4}\sin(6(t+T)) \\ &=\frac{3}{4}\sin(2(t+T)) - \frac{1}{4}\sin(6(t+T)) \\ &=\frac{3}{4}\sin(2t+2T) - \frac{1}{4}\sin(6t+6T) \\ &= \frac{3}{4}\Big[(\sin(2t)\cos(2T))+(\cos(2t)\sin(2T))\Big] - \\ &\frac{1}{4}\Big[(\sin(6t)\cos(6T))+(\cos(6t)\sin(6T))\Big] \tag{2} \end{align} $$ In order for the condition to hold, we choose $2T=2\pi \implies T=\pi$. Substituting $T$ in (2), we get

$$ \begin{align} x(t+T) &= \frac{3}{4}\Big[\sin(2t)\Big] - \frac{1}{4}\Big[\sin(6t)\Big] \\ &= x(t) \end{align} $$

So, the fundamental period is $T=\pi$ sec. The author states that the fundamental period is $T=1/\pi$ sec. Is it correct?

Answer 1

$T=1/\pi$ is definitely not the period of $s(t)$ as, for example,

$$x(2\cdot 0) = \sin^3(0) = 0 \neq x(2(0 + 1/\pi)) = \sin^3(2/\pi) \approx 0.2101$$

As noted in the comments, it's possible it was a typo or misprint or something since the period of $x(t)$ is indeed $\pi$, but we have the relation that frequency $f = 1/T$ so that might have been what is being referred to (as far as intention goes).

Whether that's the intention, I wouldn't know without having the book in front of me. You might have to just figure that bit out for yourself.

Answer 2

The fundamental period is clearly $\pi$. See the figure below. According to the formula

$ \sin^3(2t) = \frac{1}{4}[3\sin(2t)-\sin(6t)]$.

The final signal, i.e., $\sin^3(2t)$ completes one cycle in t=3.14 or $\pi$ sec.