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2023 AMC 12A Problem 25

Problem 25 of 25HarderAlgebra

There is a unique sequence of integers a1,a_1, a2,a_2, a2023\cdots a_{2023} such that tan2023x=a1tanx+a3tan3x+a5tan5x++a2023tan2023x1+a2tan2x+a4tan4x+a2022tan2022x \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 tan2023x\tan 2023x is defined. What is a2023?a_{2023}?

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Solution

By De Moivre, (cosx+isinx)2023(\cos x+i\sin x)^{2023} =cos2023x+isin2023x.=\cos 2023x+i\sin 2023x. Expanding the left side and taking the ratio of imaginary to real parts gives tan2023x\tan 2023x as the stated rational function of tanx\tan x after dividing numerator and denominator by cos2023x.\cos^{2023}x. The coefficient a2023a_{2023} is the coefficient of tan2023x\tan^{2023}x in the numerator, which comes from the k=2023k=2023 term: a2023=(1)202312(20232023)=(1)1011=1. \begin{gathered} a_{2023}=(-1)^{\frac{2023-1}{2}}\binom{2023}{2023}\\ {}=(-1)^{1011}\\ {}=-1. \end{gathered} Thus, the correct answer is C.

More practice

Concepts: De Moivre’s Theorem · binomial theorem

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.