2023 AMC 12A Problem 25

Below is the professionally curated solution for Problem 25 of the 2023 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2023 AMC 12A solutions, or check the answer key.

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Concepts:De Moivre’s Theorem binomial theorem

Difficulty rating: 2650

25.

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

$-2023$

$-2022$

$-1$

$1$

$2023$

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: $a_{2023}=(-1)^{(2023-1)/2}\binom{2023}{2023}=(-1)^{1011}=-1.$

Thus, the correct answer is C.

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