Question

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

$-2023$

$-2022$

$-1$

$1$

$2023$

Answer: C

Concepts:De Moivre’s Theorem binomial theorem

Difficulty rating: 2650

Solution:

By De Moivre, $(\cos x+i\sin x)^{2023}$ $=\cos 2023x+i\sin 2023x.$ Expanding the left side and taking the ratio of imaginary to real parts gives $\tan 2023x$ as the stated rational function of $\tan x$ after dividing numerator and denominator by $\cos^{2023}x.$

The coefficient $a_{2023}$ is the coefficient of $\tan^{2023}x$ in the numerator, which comes from the $k=2023$ term: $\begin{gathered} a_{2023}=(-1)^{(2023-1)/2}\binom{2023}{2023}\\ {}=(-1)^{1011}\\ {}=-1. \end{gathered}$

Thus, the correct answer is C.

2023 AMC 12A Problem 25

Answer: C

Difficulty rating: 2650

Solution:

Problem 25 in Other Years