TITLE: Does $\sin(t)$ have the same frequency as $\sin(\sin(t))$?
QUESTION [7 upvotes]: I plotted $\sin(t)$ and below it $\sin(\sin(t))$ on my computer and it looks as if they have the same frequency.
That led me to wonder about the following statement:

$\sin(t)$ has the same frequency as $\sin(\sin(t))$

Is this statement true or false, and how to prove it? Many thanks

REPLY [3 votes]: In general, it is possible to express trigonometric functions of trigonometric functions via the Jacobi-Anger expansion.  In the case of $\sin(\sin(t))$, we have:
$\sin(\sin(t)) = 2 \sum_{n=1}^{\infty} J_{2n-1}(1) \sin\left[\left(2n-1\right) t\right]$,
where $J_{2n-1}$ is the Bessel function of the first kind of order $2n-1$.  It is clear from this expansion that the zeroes of $\sin(\sin(t))$ are the same as that of $\sin(t)$, since any even multiple of $\pi$ for the argument $t$ will also lead to an even multiple of $\pi$ for the $\sin[(2n-1)t]$ term in the expansion.  
As MrMas mentioned, though the functions have the same period, their spectral content is different.  The expansion can viewed as a Fourier series for the spectral components of $\sin(\sin(t))$, the amplitudes of which are governed by the amplitude of the $2n-1$-th Bessel function.  Here is a plot of $|J_n(1)|$ for $n \in \mathbb R [1, 10]$:
For z = 1 in $\sin(z\sin(t))$, there is very little harmonic content, and in the time domain $\sin(\sin(t))$ doesn't look terribly different from an ordinary sine wave.