Does $\sin(t)$ have the same frequency as $\sin(\sin(t))$? 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
 A: 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.
A: In short, the answer is no if you look at instantaneous frequency.
The instantaneous frequency of a sinusoid is the derivative of the argument. That is, the frequency of $\sin(f(t))$ is $\frac{d}{dt}f(t)$. Thus the frequency of $\sin(\sin(t))$ is $\frac{d}{dt}\sin(t)=\cos(t)$. On the other hand, the frequency of $\sin(t)$ is $\frac{d}{dt}t = 1$.
So they do have the same period, but their spectral content is different.
A: For every function $f$, $f(\sin(t))$ is going to be $2\pi$-periodic, because $\sin(t)$ is $2\pi$-periodic. At every $2\pi$-interval, $\sin(t)$ is simply ranging over the values $[-1,1]$, so $f(\sin(t))$ is simply $f$ being evaluated over and over in the domain $[-1,1]$.
A: $\color{#C00000}{\sin}(x)$ is injective on $[-1,1]$ which is the range of $\color{#00A000}{\sin}(x)$. Thus,
$$
\color{#C00000}{\sin}(\color{#00A000}{\sin}(x))=\color{#C00000}{\sin}(\color{#00A000}{\sin}(y))\Leftrightarrow\color{#00A000}{\sin}(x)=\color{#00A000}{\sin}(y)
$$
Therefore, the period of $\color{#C00000}{\sin}(\color{#00A000}{\sin}(t))$ is the same as that of $\color{#00A000}{\sin}(t)$.
A: You also need to argue that the period is not shorter than $2\pi$ to be able to conclude that it is exactly two pi, even though it follows more or less directly from considering the graph of $\sin(t)$.
