Difference between revisions of "2023 AMC 12A Problems/Problem 25"
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| + | ==Problem== | ||
| + | There is a unique sequence of integers <math>a_1, a_2, \cdots a_{2023}</math> such that | ||
| + | <cmath> | ||
| + | \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} | ||
| + | </cmath>whenever <math>\tan 2023x</math> is defined. What is <math>a_{2023}?</math> | ||
| + | <math>\textbf{(A) } -2023 \qquad\textbf{(B) } -2022 \qquad\textbf{(C) } -1 \qquad\textbf{(D) } 1 \qquad\textbf{(E) } 2023</math> | ||
| + | |||
| + | ==Solution== | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC12 box|year=2023|ab=A|num-b=24|after=Last Problem}} | ||
| + | {{MAA Notice}} | ||
Revision as of 23:01, 9 November 2023
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
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
