Difference between revisions of "2006 AMC 12A Problems/Problem 15"
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Suppose <math>\cos x=0</math> and <math>\cos (x+z)=1/2</math>. What is the smallest possible positive value of <math>z</math>? | Suppose <math>\cos x=0</math> and <math>\cos (x+z)=1/2</math>. What is the smallest possible positive value of <math>z</math>? | ||
| − | <math> \mathrm{(A) \ } \frac{\pi}{6}\qquad \mathrm{(B) \ } \frac{\pi}{3}\qquad \mathrm{(C) \ } \frac{\pi}{2}\qquad \mathrm{(D) \ } \frac{5\pi}{6}</math> | + | <math> \mathrm{(A) \ } \frac{\pi}{6}\qquad \mathrm{(B) \ } \frac{\pi}{3}\qquad \mathrm{(C) \ } \frac{\pi}{2}\qquad \mathrm{(D) \ } \frac{5\pi}{6} \quad \mathrm{(E) \ } \frac{7\pi}{6}</math> |
| − | <math>\ | + | == Solution == |
| + | *For <math>\cos x = 0</math>, x must be in the form of <math>\frac{\pi}{2} + \pi n</math>, where <math>n</math> denotes any [[integer]]. | ||
| + | *For <math>\cos (x+z) = 1 / 2</math>, <math>x + z = \frac{\pi}{3} +2\pi n, \frac{5\pi}{3} + 2\pi n</math>. | ||
| − | = | + | <!-- explanation needed-->The smallest possible value of <math>z</math> will be that of <math>\frac{5\pi}{3} - \frac{3\pi}{2} = \frac{\pi}{6} \Rightarrow \mathrm{(A)}</math>. |
== See also == | == See also == | ||
* [[2006 AMC 12A Problems]] | * [[2006 AMC 12A Problems]] | ||
| − | + | ||
| − | + | {{AMC12 box|year=2006|ab=A|num-b=14|num-a=16}} | |
| + | {{MAA Notice}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
Latest revision as of 11:45, 4 August 2017
Problem
Suppose 
and 
. What is the smallest possible positive value of 
?
Solution
- For
, x must be in the form of 
, where 
denotes any integer. - For
, 
.
, x must be in the form of 
, where 
denotes any 
, 
.The smallest possible value of will be that of 
.
See also
| 2006 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 14 |
Followed by Problem 16 |
| 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. 
