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Difference between revisions of "2017 AMC 12B Problems/Problem 7"

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==Solution II==
 
==Solution II==
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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 surprisingly <math>\boxed{(B)}</math>!!!!
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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, surprisingly, <math>\boxed{(B)}</math>.
  
 
==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}}

Revision as of 14:38, 7 February 2018

Problem 7

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

$\sin(x)$ 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 $\frac{2\pi}{2} = \pi \boxed{\textbf{(B)}}$

Solution by TheUltimate123 (Eric Shen)

Solution II

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, surprisingly, $\boxed{(B)}$.

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

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