2023 AMC 12A Problem 25

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

Expand

(\cos x+i\sin x)^{2023}

and use

\tan 2023x=\dfrac{\operatorname{Im}}{\operatorname{Re}}

Dividing by

\cos^{2023}x,

the top coefficient of

\tan^{2023}x

is

(-1)^{\frac{2023-1}{2}}\binom{2023}{2023}

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)^{\frac{2023-1}{2}}\binom{2023}{2023}\\ {}=(-1)^{1011}\\ {}=-1. \end{gathered}

Thus, the correct answer is C.

Problem 25 in other years