Question

已知 $\cos x = 0$ 且 $\cos(x + z) = \tfrac{1}{2}$ 。 $z$ 的最小正值是多少？

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

$\dfrac{\pi}{6}$

$\dfrac{\pi}{3}$

$\dfrac{\pi}{2}$

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

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

答案：A

知识点：三角学

难度评级：1590

解答：

因为 $\cos x = 0,$ 所以 $x = \tfrac{\pi}{2} + k\pi.$ 又因为 $\cos(x + z) = \tfrac{1}{2},$ 所以 $x + z = 2n\pi \pm \tfrac{\pi}{3}.$

在单位圆上， $\tfrac{\pi}{2}$ 的奇数倍与同余于 $\pm\tfrac{\pi}{3}$ 的角之间，最小正角距离为 $\tfrac{\pi}{6}.$ 取 $x = -\tfrac{\pi}{2}$ 和 $x + z = -\tfrac{\pi}{3},$ 可达到此值，此时 $z = -\tfrac{\pi}{3} + \tfrac{\pi}{2} = \tfrac{\pi}{6}.$

所以正确答案是 A。

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

On the unit circle, the smallest positive angular separation between an odd multiple of $\tfrac{\pi}{2}$ and an angle congruent to $\pm\tfrac{\pi}{3}$ is $\tfrac{\pi}{6}.$ It is attained by taking $x = -\tfrac{\pi}{2}$ and $x + z = -\tfrac{\pi}{3},$ which gives $z = -\tfrac{\pi}{3} + \tfrac{\pi}{2} = \tfrac{\pi}{6}.$

Thus, the correct answer is A.

2006 AMC 12A 第 15 题

答案：A

难度评级：1590

解答：

其他年份的第 15 题