0
$\begingroup$

Question 7 of the AMC 12B goes as follows:

The functions $\sin(x)$ and $\cos(x)$ are periodic with least period $2\pi$. What is the least period of the function $\cos(\sin(x))$?

The solution provided on AoPS is:

$\sin(x)$ has values $0, 1, 0, -1$ at its peaks and x-intercepts. Increase them to $0, \pi/2, 0, -\pi/2$. Then we plug them into $\cos(x)$. $\cos(0)=1, \cos(\pi/2)=0, \cos(0)=1,$ and $\cos(-\pi/2)=0$. So, $\cos(sin(x))$ is $\frac{2\pi}{2} = \pi \boxed{\textbf{(B)}}$

I'm not familiar with finding periods of more complex functions (i.e., outside the basic $y=A(sin|cos|tan)(Bx+C)+d$ sinusoidal function form), and I don't understand:

  • What it means to "increase them" by multiplying them by $\frac{\pi}{2}$ and why these are plugged into the cosine function.
  • Why and how the calculation $\frac{2\pi}{2}$ gives the periodicity of the entire function.

Can someone explain this to me? If necessary/relevant, can you explain a general method for finding periodicity of composite functions?

$\endgroup$

1 Answer 1

1
$\begingroup$

That "solution" is nonsense and you should not try to make sense of it. Here's a different argument. Since $\sin(x+\pi)=-\sin x$ and $\cos$ is even, $\cos(\sin(x+\pi))=\cos(-\sin x)=\cos(\sin x)$ so $\pi$ is a period. On the other hand, $\cos(\sin x)=1$ implies $\sin x=0$ since the only $y\in[-1,1]$ such that $\cos y=1$ is $y=0$. If $p$ is a period, then $\cos(\sin p)=\cos(\sin 0)=1$, and thus $\sin p=0$ so $p$ is a multiple of $\pi$. This proves $\pi$ is the minimal period.

In general, there's no simple way to find the minimal period of a composition. If $p$ is a period of $f$ then $p$ is also a period of $g\circ f$ for any $g$, since $g(f(x+p))=g(f(x))$. But there may be coincidences which cause $g\circ f$ to have a smaller minimal period than $f$ (or no minimal period at all), as in this case.

$\endgroup$
2
  • $\begingroup$ That took me a moment to digest but it makes sense. I'm just a little unsure how you "proved" that pi is the minimal period. $\endgroup$ Nov 6, 2017 at 0:33
  • $\begingroup$ Well, $\pi$ is a period, and every other period is a multiple of $\pi$. That means $\pi$ is the minimal period. $\endgroup$ Nov 6, 2017 at 0:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .