Difference between revisions of "2023 AMC 12A Problems/Problem 25"
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<math>\textbf{(A) } -2023 \qquad\textbf{(B) } -2022 \qquad\textbf{(C) } -1 \qquad\textbf{(D) } 1 \qquad\textbf{(E) } 2023</math> | <math>\textbf{(A) } -2023 \qquad\textbf{(B) } -2022 \qquad\textbf{(C) } -1 \qquad\textbf{(D) } 1 \qquad\textbf{(E) } 2023</math> | ||
| − | ==Solution== | + | ==Solution 1== |
| + | |||
| + | <cmath>\tan2023x = \frac{ \sin2023x }{ \cos2023x }</cmath> | ||
| + | |||
| + | <cmath>(\cos x + i \sin x)^{2023}</cmath> | ||
| + | |||
| + | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
| + | |||
| + | ==Solution 2 (Formula of tan(x))== | ||
Note that <math>\tan{kx} = \frac{\binom{k}{1}\tan{x} - \binom{k}{3}\tan^{3}{x} + \cdots \pm \binom{k}{k}\tan^{k}{x}}{\binom{k}{0}\tan^{0}{x} - \binom{k}{2}\tan^{2}{x} + \cdots + \binom{k}{k-1}\tan^{k-1}{x}}</math>, 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 <math>\tan{2x}, \tan{3x},</math> and <math>\tan{4x}</math>, and can notice the pattern from that. The expression given essentially matches the formula of <math>\tan{kx}</math> exactly. <math>a_{2023}</math> is evidently equivalent to <math>\pm\binom{2023}{2023}</math>, 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 <math>\binom{k}{k}\tan^{k}{x}</math> is <math>\boxed{\textbf{(C) } -1}</math>. | Note that <math>\tan{kx} = \frac{\binom{k}{1}\tan{x} - \binom{k}{3}\tan^{3}{x} + \cdots \pm \binom{k}{k}\tan^{k}{x}}{\binom{k}{0}\tan^{0}{x} - \binom{k}{2}\tan^{2}{x} + \cdots + \binom{k}{k-1}\tan^{k-1}{x}}</math>, 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 <math>\tan{2x}, \tan{3x},</math> and <math>\tan{4x}</math>, and can notice the pattern from that. The expression given essentially matches the formula of <math>\tan{kx}</math> exactly. <math>a_{2023}</math> is evidently equivalent to <math>\pm\binom{2023}{2023}</math>, 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 <math>\binom{k}{k}\tan^{k}{x}</math> is <math>\boxed{\textbf{(C) } -1}</math>. | ||
Revision as of 09:18, 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
![\[\tan2023x = \frac{ \sin2023x }{ \cos2023x }\]](http://latex.artofproblemsolving.com/3/5/e/35e3f0557376c59650a24979dafe3e72295425f8.png)
![\[(\cos x + i \sin x)^{2023}\]](http://latex.artofproblemsolving.com/6/8/6/68692646634cbd9510fbf10f7c8bbb7940a5ca2b.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. 
