Difference between revisions of "2023 AMC 12A Problems/Problem 25"
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\tan2023x | \tan2023x | ||
&= \frac{ \sin2023x }{ \cos2023x } = \frac{ \binom{2023}{1} \cos^{2022} x \sin x - \binom{2023}{3} \cos^{2020} x \sin^{3} x + \dots - \sin^{2023} x }{ \cos^{2023} x - \binom{2023}{2} \cos^{2021} x \sin^{2} x + \dots - \binom{2023}{2022} \cos x \sin^{2022} x }\\ | &= \frac{ \sin2023x }{ \cos2023x } = \frac{ \binom{2023}{1} \cos^{2022} x \sin x - \binom{2023}{3} \cos^{2020} x \sin^{3} x + \dots - \sin^{2023} x }{ \cos^{2023} x - \binom{2023}{2} \cos^{2021} x \sin^{2} x + \dots - \binom{2023}{2022} \cos x \sin^{2022} x }\\ | ||
− | &= \frac{ \binom{2023}{1} \frac{\cos^{2022} x \sin x}{\ | + | &= \frac{ \binom{2023}{1} \frac{\cos^{2022} x \sin x}{\cos^{2023} x} - \binom{2023}{3} \frac{\cos^{2020} x \sin^{3} x}{\cos^{2023} x} + \dots - \frac{\sin^{2023} x}{\cos^{2023} x} }{ \frac{\cos^{2023} x}{\cos^{2023} x} - \binom{2023}{2} \frac{\cos^{2021} x \sin^{2} x}{\cos^{2023} x} + \dots - \binom{2023}{2022} \frac{\cos x \sin^{2022} x}{\cos^{2023} x} |
}\\ | }\\ | ||
&= \frac{ \binom{2023}{1} \tan x - \binom{2023}{3} \tan^{3}x + \dots - \tan^{2023}x }{ \tan^{2023}x - \binom{2023}{2} \tan^{2021}x + \dots - \binom{2023}{2022} \tan x }\\ | &= \frac{ \binom{2023}{1} \tan x - \binom{2023}{3} \tan^{3}x + \dots - \tan^{2023}x }{ \tan^{2023}x - \binom{2023}{2} \tan^{2021}x + \dots - \binom{2023}{2022} \tan x }\\ | ||
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<cmath>a_{2023} = \boxed{\textbf{(C)}-1}</cmath> | <cmath>a_{2023} = \boxed{\textbf{(C)}-1}</cmath> | ||
+ | |||
+ | This problem is highly similar to problem 7.64 in the Art of Problem Solving textbook Precalculus chapter 7 that asks to prove <math>\tan{nx} = \frac{\binom{n}{1}\tan{x} - \binom{n}{3}\tan^{3}{x} + \binom{n}{5}\tan^{5}{x} - \binom{n}{7}\tan^{7}{x} + \dots}{1 - \binom{n}{2}\tan^{2}{x} + \binom{n}{4}\tan^{4}{x} - \binom{n}{6}\tan^{6}{x} + \dots}</math> | ||
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] |
Revision as of 11:17, 10 November 2023
Contents
Problem
There is a unique sequence of integers such that
whenever
is defined. What is
Solution 1
By equating real and imaginary parts:
This problem is highly similar to problem 7.64 in the Art of Problem Solving textbook Precalculus chapter 7 that asks to prove
Solution 2 (Formula of tanx)
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.