Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2017 AMC 12B Problems/Problem 7"
(Add solution 2)
 
(12 intermediate revisions by 9 users not shown)
Line 1: Line 1:
−
==Problem 7==
+
==Problem==
 
The functions <math>\sin(x)</math> and <math>\cos(x)</math> are periodic with least period <math>2\pi</math>. What is the least period of the function <math>\cos(\sin(x))</math>?
 
The functions <math>\sin(x)</math> and <math>\cos(x)</math> are periodic with least period <math>2\pi</math>. What is the least period of the function <math>\cos(\sin(x))</math>?
  
 
<math>\textbf{(A)}\ \frac{\pi}{2}\qquad\textbf{(B)}\ \pi\qquad\textbf{(C)}\ 2\pi \qquad\textbf{(D)}\ 4\pi \qquad\textbf{(E)} </math> The function is not periodic.
 
<math>\textbf{(A)}\ \frac{\pi}{2}\qquad\textbf{(B)}\ \pi\qquad\textbf{(C)}\ 2\pi \qquad\textbf{(D)}\ 4\pi \qquad\textbf{(E)} </math> The function is not periodic.
  
−
==Solution==
+
==Solution 1==
−
<math>\sin(x)</math> has values <math>0, 1, 0, -1</math> at its peaks and x-intercepts. Increase them to <math>0, \pi/2, 0, -\pi/2</math>. Then we plug them into <math>\cos(x)</math>. <math>\cos(0)=1, \cos(\pi/2)=0, \cos(0)=1, and \cos(-\pi/2)=0</math>. So, <math>\cos(sin(x))</math> is <math>\frac{2\pi}{2} = \pi \boxed{\textbf{(B)}}</math>
+
Start by noting that <math>\cos(-x)=\cos(x)</math>. Then realize that under this function, the negative sine values yield the same as their positive value, so take the absolute value of the sine function to get the new period. This has period <math>\pi</math>, so the answer is <math>\boxed{(B)}</math>.
 +
 
 +
==Solution 2 (Brute Force)==
 +
First, we can eliminate D and E, as both <math>\sin</math> and <math>\cos</math> are periodic with period <math>2\pi</math>, as stated. Therefore the nested function must be periodic with period <math>p \le 2\pi</math>. Next, we know <math>\sin</math> and <math>\cos</math> have ranges of <math>[-1, 1]</math> with turning points/zeroes at values of <math>x</math> in the set <math>S = \{0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}\}</math>. These 4 values are easy to compute and should convey the relevant information about our function, so let's consider these.
 +
 
 +
First compute the values of <math>\sin(x)</math> to be <math>(0, 1, 0, -1)</math> over the set <math>S</math>. Therefore, the values of <math>\cos(\sin(x))</math> are <math>(1, \cos(1), 1, \cos(-1))</math>. You may notice that one appears twice: at the values <math>x=0</math> and <math>x=\pi</math>, and guess that the period is <math>\pi</math>. Alternatively, you can use the fact that <math>\cos(-x)=\cos(x)</math> as in solution 1, which can be shown by considering the unit circle definition of cosine. You then get that the values of <math>\cos(\sin(x))</math> are <math>(1, \cos(1), 1, \cos(1))</math>, which is more obviously periodic with period <math>\pi</math>, therefore the answer is <math>\boxed{(B)}</math>.
 +
 
 +
NOTE: This method is not strict proof that the function has period <math>\pi</math> as we only consider 4 values in the interval.
 +
 
 +
~rawr3507
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2017|ab=B|num-b=6|num-a=8}}
 
{{AMC12 box|year=2017|ab=B|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
[[Category:Introductory Trigonometry Problems]]

Latest revision as of 05:42, 25 October 2024

Problem

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

$\textbf{(A)}\ \frac{\pi}{2}\qquad\textbf{(B)}\ \pi\qquad\textbf{(C)}\ 2\pi \qquad\textbf{(D)}\ 4\pi \qquad\textbf{(E)}$ The function is not periodic.

Solution 1

Start by noting that $\cos(-x)=\cos(x)$. Then realize that under this function, the negative sine values yield the same as their positive value, so take the absolute value of the sine function to get the new period. This has period $\pi$, so the answer is $\boxed{(B)}$.

Solution 2 (Brute Force)

First, we can eliminate D and E, as both $\sin$ and $\cos$ are periodic with period $2\pi$, as stated. Therefore the nested function must be periodic with period $p \le 2\pi$. Next, we know $\sin$ and $\cos$ have ranges of $[-1, 1]$ with turning points/zeroes at values of $x$ in the set $S = \{0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}\}$. These 4 values are easy to compute and should convey the relevant information about our function, so let's consider these.

First compute the values of $\sin(x)$ to be $(0, 1, 0, -1)$ over the set $S$. Therefore, the values of $\cos(\sin(x))$ are $(1, \cos(1), 1, \cos(-1))$. You may notice that one appears twice: at the values $x=0$ and $x=\pi$, and guess that the period is $\pi$. Alternatively, you can use the fact that $\cos(-x)=\cos(x)$ as in solution 1, which can be shown by considering the unit circle definition of cosine. You then get that the values of $\cos(\sin(x))$ are $(1, \cos(1), 1, \cos(1))$, which is more obviously periodic with period $\pi$, therefore the answer is $\boxed{(B)}$.

NOTE: This method is not strict proof that the function has period $\pi$ as we only consider 4 values in the interval.

~rawr3507

See Also

2017 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_7&oldid=230616"