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

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==Solution==
 
==Solution==
<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>
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<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,</math> and <math>\cos(-\pi/2)=0</math>. So, <math>\cos(sin(x))</math> is <math>\frac{2\pi}{2} = \pi \boxed{\textbf{(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 21:41, 16 February 2017

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

See Also

2017 AMC 12B (ProblemsAnswer KeyResources)
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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