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,</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>
 
 
Solution by TheUltimate123 (Eric Shen)
 
 
==Solution II==
 
 
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>.
 
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>.
  

Revision as of 21:38, 18 September 2019

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

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 (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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