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> | <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> | ||
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| + | Solution by TheUltimate123 (Eric Shen) | ||
==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 21:42, 16 February 2017
Problem 7
The functions 
and 
are periodic with least period 
. What is the least period of the function 
?
The function is not periodic.
Solution
has values 
at its peaks and x-intercepts. Increase them to 
. Then we plug them into . 
and 
. So, 
is 
Solution by TheUltimate123 (Eric Shen)
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. 
