Difference between revisions of "2023 AMC 12A Problems/Problem 25"
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Revision as of 10:02, 10 November 2023
Contents
Problem
There is a unique sequence of integers 
such that
![\[\tan2023x = \frac{a_1 \tan x + a_3 \tan^3 x + a_5 \tan^5 x + \cdots + a_{2023} \tan^{2023} x}{1 + a_2 \tan^2 x + a_4 \tan^4 x \cdots + a_{2022} \tan^{2022} x}\]](http://latex.artofproblemsolving.com/0/c/4/0c431ac36edd847d292077bee7205bbf3f2c3649.png)
whenever 
is defined. What is 
Solution 1
By equating real and imaginary parts:
![\[\cos 2023 x = \cos^{2023} x - \binom{2023}{2} \cos^{2021} x \sin^{2} x + \dots - \binom{2023}{2022} \cos x \sin^{2022} x\]](http://latex.artofproblemsolving.com/5/6/0/560853fd66769364759b09232bb4b348e9f4f2dc.png)
![\[\sin 2023 x = \binom{2023}{1} \cos^{2022} x \sin x - \binom{2023}{3} \cos^{2020} x \sin^{3} x + \dots - \sin^{2023} x\]](http://latex.artofproblemsolving.com/f/7/a/f7a6d4d85f661be5a354a8b14acf1e8055a72c84.png)
![\[a_{2023} = \boxed{\textbf{(C)}-1}\]](http://latex.artofproblemsolving.com/3/5/3/3532c241bd86352a92f8563fe965756dcb792f86.png)
Solution 2 (Formula of tan(x))
Note that 
, where k is odd and the sign of each term alternates between positive and negative. To realize this during the test, you should know the formulas of 
and 
, and can notice the pattern from that. The expression given essentially matches the formula of 
exactly. 
is evidently equivalent to 
, or 1. However, it could be positive or negative. Notice that in the numerator, whenever the exponent of the tangent term is congruent to 1 mod 4, the term is positive. Whenever the exponent of the tangent term is 3 mod 4, the term is negative. 2023, which is assigned to k, is congruent to 3 mod 4. This means that the term of 
is 
.
Notice: If you have time and don't know 
and , you'd have to keep deriving until you see the pattern.
~lprado
Video Solution 1 by OmegaLearn
See Also
| 2023 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 24 |
Followed by Last Problem |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
