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Difference between revisions of "2006 AMC 12A Problems/Problem 15"

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== Solution ==
 
== Solution ==
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*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]].
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*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>.
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<!-- 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>.
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== Solution 2 ==
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Notice that <math>\cos (x+z) = \cos x \cos z - \sin x \sin z</math>, we will obtain <math>\cos (x+z) = - \sin x \sin z</math> after plugging <math>\cos x = 0</math>. Knowing that <math>\cos (x+z) = \frac{1}{2}</math>, we have <math>\sin x \sin z = - \frac{1}{2}</math>.
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Now we can try plugging the answer choices back in to see if each works:
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When <math>z = \frac{\pi}{6}</math>, <math>\sin z = \frac{1}{2}</math> and <math>\sin x = - 1</math>. Since <math>\sin x</math> is obtainable when <math>x = \frac{3\pi}{2}</math>, <math>z = \frac{\pi}{6}</math> works.
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The question asks for the smallest positive value of <math>z</math>, and <math>\frac{\pi}{6}</math> is the smallest among the choices. Thus, the answer is <math>\boxed{\textbf{(A) } \frac{\pi}{6}}</math>
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~[https://artofproblemsolving.com/wiki/index.php/User:Andy_li0805 Andy_li0805]
  
 
== See also ==
 
== See also ==
 
* [[2006 AMC 12A Problems]]
 
* [[2006 AMC 12A Problems]]
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*[[2006 AMC 12A Problems/Problem 14|Previous Problem]]
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*[[2006 AMC 12A Problems/Problem 16|Next Problem]]
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{{AMC12 box|year=2006|ab=A|num-b=14|num-a=16}}
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{{MAA Notice}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Latest revision as of 21:36, 19 August 2025

Problem

Suppose $\cos x=0$ and $\cos (x+z)=1/2$. What is the smallest possible positive value of $z$?

$\mathrm{(A) \ } \frac{\pi}{6}\qquad \mathrm{(B) \ } \frac{\pi}{3}\qquad \mathrm{(C) \ } \frac{\pi}{2}\qquad \mathrm{(D) \ } \frac{5\pi}{6} \quad \mathrm{(E) \ } \frac{7\pi}{6}$

Solution

  • For $\cos x = 0$, x must be in the form of $\frac{\pi}{2} + \pi n$, where $n$ denotes any integer.
  • For $\cos (x+z) = 1 / 2$, $x + z = \frac{\pi}{3} +2\pi n, \frac{5\pi}{3} + 2\pi n$.

The smallest possible value of $z$ will be that of $\frac{5\pi}{3} - \frac{3\pi}{2} = \frac{\pi}{6} \Rightarrow \mathrm{(A)}$.

Solution 2

Notice that $\cos (x+z) = \cos x \cos z - \sin x \sin z$, we will obtain $\cos (x+z) = - \sin x \sin z$ after plugging $\cos x = 0$. Knowing that $\cos (x+z) = \frac{1}{2}$, we have $\sin x \sin z = - \frac{1}{2}$.

Now we can try plugging the answer choices back in to see if each works:

When $z = \frac{\pi}{6}$, $\sin z = \frac{1}{2}$ and $\sin x = - 1$. Since $\sin x$ is obtainable when $x = \frac{3\pi}{2}$, $z = \frac{\pi}{6}$ works.

The question asks for the smallest positive value of $z$, and $\frac{\pi}{6}$ is the smallest among the choices. Thus, the answer is $\boxed{\textbf{(A) } \frac{\pi}{6}}$

~Andy_li0805

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

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