7
$\begingroup$

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

5 Answers 5

11

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 $[-1,1]$.

  • 0
    @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.2012-10-24
11

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

3

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 $n \in \mathbb R [1, 10]$:Bessel function amplitudes

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.

2

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.

1

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 $\sin(t)$.

  • 0
    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 $\pi$.2012-10-24