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

2006 AMC 12A Problem 15 — Solution

2006 AMC 12A Problem 15

Below is the professionally curated solution for Problem 15 of the 2006 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 12A solutions, or check the answer key.

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Concepts:trigonometry

Difficulty rating: 1590

15.

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

$\dfrac{\pi}{6}$

$\dfrac{\pi}{3}$

$\dfrac{\pi}{2}$

$\dfrac{5\pi}{6}$

$\dfrac{7\pi}{6}$

Solution:

Since $\cos x = 0,$ we have $x = \tfrac{\pi}{2} + k\pi.$ Since $\cos(x + z) = \tfrac{1}{2},$ we have $x + z = 2n\pi \pm \tfrac{\pi}{3}.$

Taking $x = -\tfrac{\pi}{2}$ and $x + z = -\tfrac{\pi}{3}$ gives $z = -\tfrac{\pi}{3} + \tfrac{\pi}{2} = \tfrac{\pi}{6},$ the smallest positive value.

Thus, the correct answer is A.

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