Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
up vote 11 down vote accepted

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 $[-1,1]$.

share|cite|improve this answer
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 $\color{#00A000}{\sin}(t)$.

share|cite|improve this answer

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.

share|cite|improve this answer

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.

share|cite|improve this answer

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 $\sin(t)$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.