AMC12 2023 A

AMC12 2023 A · Q25

AMC12 2023 A · Q25. It mainly tests Logarithms (rare), Algebra misc.

There is a unique sequence of integers $a_1, a_2, \cdots a_{2023}$ such that \[\tan2023x = \frac{a_1 \tan x + a_3 \tan^3 x + a_5 \tan^5 x + \cdots + a_{2023} \tan^{2023} x}{1 + a_2 \tan^2 x + a_4 \tan^4 x \cdots + a_{2022} \tan^{2022} x}\]whenever $\tan 2023x$ is defined. What is $a_{2023}?$

存在唯一的整数序列 $a_1, a_2, \cdots a_{2023}$，使得只要 $\tan 2023x$ 定义，则 \[\tan2023x = \frac{a_1 \tan x + a_3 \tan^3 x + a_5 \tan^5 x + \cdots + a_{2023} \tan^{2023} x}{1 + a_2 \tan^2 x + a_4 \tan^4 x \cdots + a_{2022} \tan^{2022} x}\]。求 $a_{2023}$？

(A) -2023 -2023

(B) -2022 -2022

(D) 1 1

(E) 2023 2023

Answer

Correct choice: (C)

正确答案：（C）

Solution

\begin{align*} \cos 2023 x + i \sin 2023 x &= (\cos x + i \sin x)^{2023}\\ &= \cos^{2023} x + \binom{2023}{1} \cos^{2022} x (i\sin x) + \binom{2023}{2} \cos^{2021} x (i \sin x)^{2} +\binom{2023}{3} \cos^{2020} x (i \sin x)^{3}\\ &+ \dots + \binom{2023}{2022} \cos x (i \sin x)^{2022} + (i \sin x)^{2023}\\ &= \cos^{2023} x + i \binom{2023}{1} \cos^{2022} x \sin x - \binom{2023}{2} \cos^{2021} x \sin^{2} x - i\binom{2023}{3} \cos^{2020} x \sin^{3} x + \dots\\ &- \binom{2023}{2022} \cos x \sin^{2022} x - i \sin^{2023} x\\ \end{align*} By equating real and imaginary parts: \[\cos 2023 x = \cos^{2023} x - \binom{2023}{2} \cos^{2021} x \sin^{2} x + \dots - \binom{2023}{2022} \cos x \sin^{2022} x\] \[\sin 2023 x = \binom{2023}{1} \cos^{2022} x \sin x - \binom{2023}{3} \cos^{2020} x \sin^{3} x + \dots - \sin^{2023} x\] \begin{align*} \tan2023x &= \frac{ \sin2023x }{ \cos2023x } = \frac{ \binom{2023}{1} \cos^{2022} x \sin x - \binom{2023}{3} \cos^{2020} x \sin^{3} x + \dots - \sin^{2023} x }{ \cos^{2023} x - \binom{2023}{2} \cos^{2021} x \sin^{2} x + \dots - \binom{2023}{2022} \cos x \sin^{2022} x }\\ &= \frac{ \binom{2023}{1} \frac{\cos^{2022} x \sin x}{\cos^{2023} x} - \binom{2023}{3} \frac{\cos^{2020} x \sin^{3} x}{\cos^{2023} x} + \dots - \frac{\sin^{2023} x}{\cos^{2023} x} }{ \frac{\cos^{2023} x}{\cos^{2023} x} - \binom{2023}{2} \frac{\cos^{2021} x \sin^{2} x}{\cos^{2023} x} + \dots - \binom{2023}{2022} \frac{\cos x \sin^{2022} x}{\cos^{2023} x} }\\ &= \frac{ \binom{2023}{1} \tan x - \binom{2023}{3} \tan^{3}x + \dots - \tan^{2023}x }{ 1 - \binom{2023}{2} \tan^{2}x + \dots - \binom{2023}{2022} \tan^{2022} x }\\ \end{align*} \[a_{2023} = \boxed{\textbf{(C)}-1}\] This problem is the same as problem 7.64 in the Art of Problem Solving textbook Precalculus chapter 7 that asks to prove $\tan{nx} = \frac{\binom{n}{1}\tan{x} - \binom{n}{3}\tan^{3}{x} + \binom{n}{5}\tan^{5}{x} - \binom{n}{7}\tan^{7}{x} + \dots}{1 - \binom{n}{2}\tan^{2}{x} + \binom{n}{4}\tan^{4}{x} - \binom{n}{6}\tan^{6}{x} + \dots}$

使用复指数形式： \[\cos 2023 x + i \sin 2023 x = (\cos x + i \sin x)^{2023}\] 展开后，分离实部和虚部，得到 $\sin 2023x$ 和 $\cos 2023x$ 的表达式。然后， \[\tan2023x = \frac{\sin2023x}{\cos2023x}\] 化简后，分母和分子均为 $\cos^{2023} x$ 的倍数，消去后得到 \[\tan2023x = \frac{ \binom{2023}{1} \tan x - \binom{2023}{3} \tan^{3}x + \dots - \tan^{2023}x }{ 1 - \binom{2023}{2} \tan^{2}x + \dots - \binom{2023}{2022} \tan^{2022} x }\] 因此，$a_{2023} = \boxed{\textbf{(C)}-1}$。

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