Problem:
There is a unique sequence of integers a1,a2,a3,…,a2023 such that
tan2023x=1+a2tan2x+a4tan4x+⋯+a2022tan2022xa1tanx+a3tan3x+a5tan5x+⋯+a2023tan2023x
whenever tan2023x is defined. What is a2023?
Answer Choices:
A. −2023
B. −2022
C. −1
D. 1
E. 2023
Solution:
By de Moivre's Formula and the Binomial Theorem,
cosnx+isinnx==(cosx+isinx)ncosnx+i(1n)cosn−1x⋅sinx+i2(2n)cosn−2x⋅sin2x.
By equating the real and imaginary parts of the respective sides above, it follows that for odd n,
cosnx=cosnx−(2n)cosn−2x⋅sin2x+(4n)cosn−4x⋅sin4x−⋯±(n−1n)cosx⋅sinn−1x
=Cn(cosx,sinx)
and
sinnx=(1n)cosn−1x⋅sinx−(3n)cosn−3x⋅sin3x+⋯±sinnx=Sn(cosx,sinx),
where Cn(u,v) and Sn(u,v) are homogeneous polynomials of degree n. (A homogeneous polynomial is a polynomial for which all terms have the same total degree.) Hence
tannx=Cn(cosx,sinx)Sn(cosx,sinx).
By dividing the numerator and denominator above by cosnx, it follows that
tannx=Cn(1,tanx)Sn(1,tanx).
The general term in Sn(u,v) has the form
(−1)k(2k+1n)un−2k−1v2k+1,
so with n=2023, the last term occurs when k=1011. Hence the last term of Sn(1,tanx) is −1(20232023)v2023, and thus a2023=(C)−1.
The problems on this page are the property of the MAA's American Mathematics Competitions