Question

方程在闭区间 $[0, \pi]$ 中有多少个解？ $\sin\left(\frac{\pi}{2}\cos x\right) = \cos\left(\frac{\pi}{2}\sin x\right)$

How many solutions does the equation $\sin\left(\frac{\pi}{2}\cos x\right) = \cos\left(\frac{\pi}{2}\sin x\right)$ have in the closed interval $[0, \pi]?$

$0$

$1$

$2$

$3$

$4$

答案：C

知识点：三角恒等式三角学

难度评级：2300

解答：

将右边写为 $\cos\left(\tfrac{\pi}{2}\sin x\right) = \sin\left(\tfrac{\pi}{2} - \tfrac{\pi}{2}\sin x\right)$ 。两个正弦相等要求或 $\begin{aligned} &\frac{\pi}{2}\cos x \\ &= \frac{\pi}{2}(1 - \sin x) + 2\pi k \end{aligned}$ $\begin{aligned} &\frac{\pi}{2}\cos x \\ &= \pi - \frac{\pi}{2}(1 - \sin x) + 2\pi k. \end{aligned}$

第一种化为 $\cos x + \sin x = 1 + 4k$ ；因为 $\cos x + \sin x \in [-\sqrt2, \sqrt2]$ ，只有 $k = 0$ 可行，得到 $\cos x + \sin x = 1$ ，在 $[0, \pi]$ 中的解为 $x = 0$ 和 $x = \tfrac{\pi}{2}$ 。第二种化为 $\cos x - \sin x = 1$ ，在 $[0, \pi]$ 中唯一解为 $x = 0$ 。

不同的解是 $x = 0$ 和 $x = \tfrac{\pi}{2}$ ，共 $2$ 个。

因此，正确答案是 C。

Write the right side as $\cos\left(\tfrac{\pi}{2}\sin x\right) = \sin\left(\tfrac{\pi}{2} - \tfrac{\pi}{2}\sin x\right).$ Equal sines require either $\begin{aligned} &\frac{\pi}{2}\cos x \\ &= \frac{\pi}{2}(1 - \sin x) + 2\pi k \end{aligned}$ or $\begin{aligned} &\frac{\pi}{2}\cos x \\ &= \pi - \frac{\pi}{2}(1 - \sin x) + 2\pi k. \end{aligned}$

The first reduces to $\cos x + \sin x = 1 + 4k;$ since $\cos x + \sin x \in [-\sqrt2, \sqrt2],$ only $k = 0$ works, giving $\cos x + \sin x = 1,$ with solutions $x = 0$ and $x = \tfrac{\pi}{2}$ in $[0, \pi].$ The second reduces to $\cos x - \sin x = 1,$ whose only solution in $[0, \pi]$ is $x = 0.$

The distinct solutions are $x = 0$ and $x = \tfrac{\pi}{2},$ for a total of $2.$

Thus, the correct answer is C.

2021 AMC 12A Spring 第 19 题

答案：C

难度评级：2300

解答：

其他年份的第 19 题