Question

求下式的值： $\begin{aligned} &\tan^2\frac{\pi}{16}\cdot\tan^2\frac{3\pi}{16} \\ &\quad {}+\tan^2\frac{\pi}{16}\cdot\tan^2\frac{5\pi}{16} \\ &\quad {}+\tan^2\frac{3\pi}{16}\cdot\tan^2\frac{7\pi}{16} \\ &\quad {}+\tan^2\frac{5\pi}{16}\cdot\tan^2\frac{7\pi}{16}? \end{aligned}$

What is the value of $\begin{aligned} &\tan^2\frac{\pi}{16}\cdot\tan^2\frac{3\pi}{16} \\ &\quad {}+\tan^2\frac{\pi}{16}\cdot\tan^2\frac{5\pi}{16} \\ &\quad {}+\tan^2\frac{3\pi}{16}\cdot\tan^2\frac{7\pi}{16} \\ &\quad {}+\tan^2\frac{5\pi}{16}\cdot\tan^2\frac{7\pi}{16}? \end{aligned}$

$28$

$68$

$70$

$72$

$84$

答案：B

知识点：三角恒等式因式分解

难度评级：2370

解答：

令 $a=\tan^2\tfrac{\pi}{16},\$ $b=\tan^2\tfrac{3\pi}{16},\$ $c=\tan^2\tfrac{5\pi}{16},\$ $d=\tan^2\tfrac{7\pi}{16},$ 原式为 $ab+ac+bd+cd$ $=(a+d)(b+c)$ 。

因为 $\tfrac{7\pi}{16}=\tfrac{\pi}{2}-\tfrac{\pi}{16}$ ，所以 $d=\cot^2\tfrac{\pi}{16}$ ，从而 $a+d=\tan^2\tfrac{\pi}{16}+\cot^2\tfrac{\pi}{16}$ $=\tfrac{4}{\sin^2(\pi/8)}-2$ $=14+8\sqrt2$ 。同理 $b+c=\tfrac{4}{\sin^2(3\pi/8)}-2$ $=14-8\sqrt2$ 。两者乘积为 $14^2-(8\sqrt2)^2=196-128=68$ 。

因此正确答案是 B。

With $a=\tan^2\tfrac{\pi}{16},\$ $b=\tan^2\tfrac{3\pi}{16},\$ $c=\tan^2\tfrac{5\pi}{16},\$ $d=\tan^2\tfrac{7\pi}{16},$ the expression is $ab+ac+bd+cd$ $=(a+d)(b+c).$

Since $\tfrac{7\pi}{16}=\tfrac{\pi}{2}-\tfrac{\pi}{16},$ we have $d=\cot^2\tfrac{\pi}{16},$ so $a+d=\tan^2\tfrac{\pi}{16}+\cot^2\tfrac{\pi}{16}$ $=\tfrac{4}{\sin^2(\pi/8)}-2$ $=14+8\sqrt2.$ Likewise $b+c=\tfrac{4}{\sin^2(3\pi/8)}-2$ $=14-8\sqrt2.$ Their product is $14^2-(8\sqrt2)^2=196-128=68.$

Thus, the correct answer is B.

2024 AMC 12A 第 23 题

答案：B

难度评级：2370

解答：

其他年份的第 23 题