# 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

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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 $[-1,1]$.

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The frequency requires the prime period, though. Just because a function has period x doesn't mean it doesn't also have period x/2 or x/3. In an extreme example, what if f is arcsin? – ex0du5 Oct 24 '12 at 19:28
@ex0du5: With $f=\arcsin$, you get a triangular sawtooth function - of period $2\pi$. But with $f(x)=x^2$, you are right: This would produce period $\pi$ and not just $2\pi$. Not to mention $f(x)=$const. – Hagen von Eitzen Oct 24 '12 at 19:50
Sure, the function could also be $T$ periodic for some other $T$, depending on the $f$. But it should be easy to check for nice $f$ whether there would be smaller periods. – Christopher A. Wong Oct 24 '12 at 19:51
@ChristopherA.Wong: Right. In this special case, $x\in\[-1,1]$ and $\sin x=0$ implies $x=0$, therefore $\sin(\sin(x))=0$ iff $x=k\pi$. So apart from $2\pi$, only $\pi$ would be a possible period. But $\sin(\sin(\pi/2))>0$, $\sin(\sin(-\pi/2))<0$ rules this out. – Hagen von Eitzen Oct 24 '12 at 19:55

$\color{#C00000}{\sin}(x)$ 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 $\color{#00A000}{\sin}(t)$.

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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.

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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.

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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 $\sin(t)$.

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I do not understand "the zeros of $\sin(\sin(t))$ is not evenly spaced in the interval $[−1,1]$". The zeros of $\sin(\sin(t))$ are the integer multiples of $\pi$. – robjohn Oct 24 '12 at 21:09