2006 AMC 12A Problems/Problem 15

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

2006 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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All AMC 12 Problems and Solutions

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2006 AMC 12A Problems/Problem 15

Contents

Problem

Solution

Solution 2

See also