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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>
 +
\begin{align*}
 +
\cos 2023 x + i \sin 2023 x
 +
&= (\cos x + i \sin x)^{2023}\\
 +
&= \cos^{2023} x + \binom{2023}{1} \cos^{2022} x (i\sin x) + \binom{2023}{2} \cos^{2021} x (i \sin x)^{2} +\binom{2020}{3} \cos^{2020} x (i \sin x)^{3}\\
 +
&+ \dots + \binom{2023}{2022} \cos x (i \sin x)^{2022} + (i \sin x)^{2023}\\
 +
&= \cos^{2023} x + i \binom{2023}{1} \cos^{2022} x \sin x - \binom{2023}{2} \cos^{2021} x \sin^{2} x - i\binom{2023}{3} \cos^{2020} x \sin^{3} x + \dots\\
 +
&- \binom{2023}{2022} \cos x \sin^{2022} x - i \sin^{2023} x\\
 +
\end{align*}</cmath>
 +
 
 +
By equating real and imaginary parts:
 +
 
 +
<cmath>\cos 2023 x = \cos^{2023} x - \binom{2023}{2} \cos^{2021} x \sin^{2} x + \dots - \binom{2023}{2022} \cos x \sin^{2022} x</cmath>
 +
 
 +
<cmath>\sin 2023 x = \binom{2023}{1} \cos^{2022} x \sin x - \binom{2023}{3} \cos^{2020} x \sin^{3} x + \dots - \sin^{2023} x</cmath>
 +
 
 +
<cmath>
 +
\begin{align*}
 +
\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{ \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 }{ 1 - \binom{2023}{2} \tan^{2}x + \dots - \binom{2023}{2022} \tan^{2022} x }\\
 +
\end{align*}</cmath>
 +
 
 +
<cmath>a_{2023} = \boxed{\textbf{(C)}-1}</cmath>
 +
 
 +
This problem is the same as 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]
 +
 
 +
==Solution 2 (Formula of tanx)==
 
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>.
  
  
−
Notice: If you have time and don't know <math>tan{3x}</math> and <math>tan{4x}</math>, you'd have to keep deriving tan{kx} until you see the pattern.
+
Notice: If you have time and don't know <math>\tan{3x}</math> and <math>\tan{4x}</math>, you'd have to keep deriving <math>\tan{kx}</math> until you see the pattern.
  
 
~lprado
 
~lprado
 +
 +
 +
==Solution 3==
 +
 +
For odd <math>n</math>, we have
 +
<cmath>
 +
\begin{align*}
 +
\tan nx
 +
& = \frac{\sin nx}{\cos nx} \\
 +
& = \frac{\frac{1}{2i} \left( e^{i n x} - e^{-i n x} \right)}
 +
{\frac{1}{2} \left( e^{i n x} + e^{-i n x} \right)} \\
 +
& = - i \frac{e^{i n x} - e^{-i n x}}{e^{i n x} + e^{-i n x}} \\
 +
& = - i  \frac{\left( \cos x + i \sin x \right)^n - \left( \cos x - i \sin x \right)^n}
 +
{\left( \cos x + i \sin x \right)^n + \left( \cos x - i \sin x \right)^n} \\
 +
& = \frac{ - 2 i \sum_{m=0}^{(n-1)/2} \binom{n}{2m + 1} \left( \cos x \right)^{n - 2m - 1} \left( i \sin x \right)^{2m + 1}}
 +
{2 \sum_{m=0}^{(n-1)/2} \binom{n}{2m}
 +
\left( \cos x \right)^{n - 2m} \left( i \sin x \right)^{2m}} \\
 +
& = \frac{ \frac{1}{\left( \cos x \right)^n} \sum_{m=0}^{(n-1)/2} \binom{n}{2m + 1} \left( \cos x \right)^{n - 2m - 1} \left( i \sin x \right)^{2m + 1}}
 +
{i \frac{1}{\left( \cos x \right)^n} \sum_{m=0}^{(n-1)/2} \binom{n}{2m}
 +
\left( \cos x \right)^{n - 2m} \left( i \sin x \right)^{2m}} \\
 +
& = \frac{ \sum_{m=0}^{(n-1)/2} \binom{n}{2m + 1} \left( i \tan x \right)^{2m + 1}}
 +
{i \sum_{m=0}^{(n-1)/2} \binom{n}{2m}
 +
\left( i \tan x \right)^{2m}} \\
 +
& = \frac{ \sum_{m=0}^{(n-1)/2} \binom{n}{2m + 1} \left( \tan x \right)^{2m + 1}
 +
i^{2m + 1}}
 +
{\sum_{m=0}^{(n-1)/2} \binom{n}{2m}
 +
\left( \tan x \right)^{2m} i^{2m + 1}} \\
 +
& = \frac{ \sum_{m=0}^{(n-1)/2} \binom{n}{2m + 1} \left( \tan x \right)^{2m + 1}
 +
\left( -1 \right)^m}
 +
{\sum_{m=0}^{(n-1)/2} \binom{n}{2m}
 +
\left( \tan x \right)^{2m} \left( -1 \right)^m}
 +
.
 +
\end{align*}
 +
</cmath>
 +
 +
Thus, for <math>n = 2023</math>, we have
 +
<cmath>
 +
\begin{align*}
 +
a_{2023} & = \binom{2023}{2023} \left( -1 \right)^{(2023-1)/2} \\
 +
& = \left( -1 \right)^{1011} \\
 +
& = \boxed{\textbf{(C) -1}}.
 +
\end{align*}
 +
</cmath>
 +
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 +
==Solution 4==
 +
 +
 +
We can use recursion to find a pattern for this problem. Notice that,
 +
<cmath>
 +
\begin{align*}
 +
\tan x &= \tan x\\
 +
\tan 2x &= \dfrac{\tan x + \tan x}{1 - \tan^2 x} = \dfrac{2\tan x}{1 - \tan^2 x}\\
 +
\tan 3x &= \dfrac{\tan 2x + \tan x}{1 - \tan 2x \tan x} = \dfrac{3\tan x - \tan^3 x}{1 - 3\tan^2 x}
 +
\end{align*}
 +
</cmath>
 +
The coefficient of the highest degree term seems to be always <math>\pm 1</math>. Now, we prove this by an imcomplete  mathematical induction.
 +
 +
* Firstly, we suppose <math>n</math> is odd, and <math>\tan nx = \dfrac{a_1\tan x + a_3\tan^3 x + \cdots + a_n\tan^n x}{a_0 + a_2\tan^2 x + \cdots + a_{n - 1}\tan^{n - 1} x}</math>, then
 +
 +
<cmath>
 +
\begin{align*}
 +
\tan(n + 1)x &= \dfrac{\tan nx + \tan x}{1 - \tan nx \tan x}\\
 +
&= \dfrac{\dfrac{a_1\tan x + a_3\tan^3 x + \cdots + a_n\tan^n x}{a_0 + a_2\tan^2 x + \cdots + a_{n - 1}\tan^{n - 1} x} + \tan x}{1 - \dfrac{a_1\tan x + a_3\tan^3 x + \cdots + a_n\tan^n x}{a_0 + a_2\tan^2 x + \cdots + a_{n - 1}\tan^{n - 1} x}\tan x}\\
 +
&= \dfrac{a_1\tan x + a_3\tan^3 x + \cdots + a_n\tan^n x + a_0\tan x + a_2\tan^3 x + \cdots + a_{n - 1}\tan^n x}{a_0 + a_2\tan^2x + \cdots + a_{n - 1}\tan^{n - 1}x - a_1\tan^2 x - a_3\tan^4 x - \cdots \boxed{-a_n\tan^{n + 1} x}}
 +
\end{align*}
 +
</cmath>
 +
 +
The coefficient of the highest degree term becomes <math>-a_n</math>.
 +
 +
* Secondly, we suppose <math>n</math> is even, and <math>\tan nx = \dfrac{a_1\tan x + a_3\tan^3 x + \cdots + a_{n - 1}\tan^{n - 1} x}{a_0 + a_2\tan^2 x + \cdots + a_n\tan^n x}</math>, then
 +
<cmath>
 +
\begin{align*}
 +
\tan(n + 1)x &= \dfrac{\tan nx + \tan x}{1 - \tan nx \tan x}\\
 +
&= \dfrac{\dfrac{a_1\tan x + a_3\tan^3 x + \cdots + a_{n - 1}\tan^{n - 1} x}{a_0 + a_2\tan^2 x + \cdots + a_n\tan^n x} + \tan x}{1 - \dfrac{a_1\tan x + a_3\tan^3 x + \cdots + a_{n - 1}\tan^{n - 1} x}{a_0 + a_2\tan^2 x + \cdots + a_n\tan^n x}\tan x}\\
 +
&= \dfrac{a_1\tan x + a_3\tan^3 x + \cdots + a_{n - 1}\tan^{n - 1} x + a_0\tan x + a_2\tan^3 x + \cdots \boxed{+ a_n\tan^{n + 1} x}}{a_0 + a_2\tan^2 x + \cdots + a_n\tan^n x - a_1\tan^2 x - a_3\tan^4 x - \cdots - a_{n - 1}\tan^n x}
 +
\end{align*}
 +
</cmath>
 +
The coefficient of the highest degree term remains <math>a_n</math>.
 +
 +
When <math>n = 1</math>, <math>a_n = 1</math>. During the process of <math>n</math> increasing to 2023, <math>a_n</math> changed its sign a total of <math>2022 \div 2 = 1011</math> times.
 +
 +
Hence, <math>a_{2023} = 1 \times (-1)^{1011} = \boxed{\textbf{(C)}-1}</math>
 +
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Reda_mandymath reda_mandymath]
 +
 +
==Video Solution 1 by OmegaLearn==
 +
https://youtu.be/4KJR_1Kg4A4
 +
 +
==Video Solution==
 +
 +
https://youtu.be/0KH554CLayE
 +
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2023|ab=A|num-b=24|after=Last Problem}}
 
{{AMC12 box|year=2023|ab=A|num-b=24|after=Last Problem}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:38, 28 October 2024

Problem

There is a unique sequence of integers $a_1, a_2, \cdots a_{2023}$ such that \[\tan2023x = \frac{a_1 \tan x + a_3 \tan^3 x + a_5 \tan^5 x + \cdots + a_{2023} \tan^{2023} x}{1 + a_2 \tan^2 x + a_4 \tan^4 x \cdots + a_{2022} \tan^{2022} x}\]whenever $\tan 2023x$ is defined. What is $a_{2023}?$

$\textbf{(A) } -2023 \qquad\textbf{(B) } -2022 \qquad\textbf{(C) } -1 \qquad\textbf{(D) } 1 \qquad\textbf{(E) } 2023$

Solution 1

\begin{align*} \cos 2023 x + i \sin 2023 x &= (\cos x + i \sin x)^{2023}\\ &= \cos^{2023} x + \binom{2023}{1} \cos^{2022} x (i\sin x) + \binom{2023}{2} \cos^{2021} x (i \sin x)^{2} +\binom{2020}{3} \cos^{2020} x (i \sin x)^{3}\\ &+ \dots + \binom{2023}{2022} \cos x (i \sin x)^{2022} + (i \sin x)^{2023}\\ &= \cos^{2023} x + i \binom{2023}{1} \cos^{2022} x \sin x - \binom{2023}{2} \cos^{2021} x \sin^{2} x - i\binom{2023}{3} \cos^{2020} x \sin^{3} x + \dots\\ &- \binom{2023}{2022} \cos x \sin^{2022} x - i \sin^{2023} x\\ \end{align*}

By equating real and imaginary parts:

\[\cos 2023 x = \cos^{2023} x - \binom{2023}{2} \cos^{2021} x \sin^{2} x + \dots - \binom{2023}{2022} \cos x \sin^{2022} x\]

\[\sin 2023 x = \binom{2023}{1} \cos^{2022} x \sin x - \binom{2023}{3} \cos^{2020} x \sin^{3} x + \dots - \sin^{2023} x\]

\begin{align*} \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{ \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 }{ 1 - \binom{2023}{2} \tan^{2}x + \dots - \binom{2023}{2022} \tan^{2022} x }\\ \end{align*}

\[a_{2023} = \boxed{\textbf{(C)}-1}\]

This problem is the same as problem 7.64 in the Art of Problem Solving textbook Precalculus chapter 7 that asks to prove $\tan{nx} = \frac{\binom{n}{1}\tan{x} - \binom{n}{3}\tan^{3}{x} + \binom{n}{5}\tan^{5}{x} - \binom{n}{7}\tan^{7}{x} + \dots}{1 - \binom{n}{2}\tan^{2}{x} + \binom{n}{4}\tan^{4}{x} - \binom{n}{6}\tan^{6}{x} + \dots}$

~isabelchen

Solution 2 (Formula of tanx)

Note that $\tan{kx} = \frac{\binom{k}{1}\tan{x} - \binom{k}{3}\tan^{3}{x} + \cdots \pm \binom{k}{k}\tan^{k}{x}}{\binom{k}{0}\tan^{0}{x} - \binom{k}{2}\tan^{2}{x} + \cdots + \binom{k}{k-1}\tan^{k-1}{x}}$, 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 $\tan{2x}, \tan{3x},$ and $\tan{4x}$, and can notice the pattern from that. The expression given essentially matches the formula of $\tan{kx}$ exactly. $a_{2023}$ is evidently equivalent to $\pm\binom{2023}{2023}$, 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 $\binom{k}{k}\tan^{k}{x}$ is $\boxed{\textbf{(C) } -1}$.


Notice: If you have time and don't know $\tan{3x}$ and $\tan{4x}$, you'd have to keep deriving $\tan{kx}$ until you see the pattern.

~lprado


Solution 3

For odd $n$, we have \begin{align*} \tan nx & = \frac{\sin nx}{\cos nx} \\ & = \frac{\frac{1}{2i} \left( e^{i n x} - e^{-i n x} \right)} {\frac{1}{2} \left( e^{i n x} + e^{-i n x} \right)} \\ & = - i \frac{e^{i n x} - e^{-i n x}}{e^{i n x} + e^{-i n x}} \\ & = - i \frac{\left( \cos x + i \sin x \right)^n - \left( \cos x - i \sin x \right)^n} {\left( \cos x + i \sin x \right)^n + \left( \cos x - i \sin x \right)^n} \\ & = \frac{ - 2 i \sum_{m=0}^{(n-1)/2} \binom{n}{2m + 1} \left( \cos x \right)^{n - 2m - 1} \left( i \sin x \right)^{2m + 1}} {2 \sum_{m=0}^{(n-1)/2} \binom{n}{2m} \left( \cos x \right)^{n - 2m} \left( i \sin x \right)^{2m}} \\ & = \frac{ \frac{1}{\left( \cos x \right)^n} \sum_{m=0}^{(n-1)/2} \binom{n}{2m + 1} \left( \cos x \right)^{n - 2m - 1} \left( i \sin x \right)^{2m + 1}} {i \frac{1}{\left( \cos x \right)^n} \sum_{m=0}^{(n-1)/2} \binom{n}{2m} \left( \cos x \right)^{n - 2m} \left( i \sin x \right)^{2m}} \\ & = \frac{ \sum_{m=0}^{(n-1)/2} \binom{n}{2m + 1} \left( i \tan x \right)^{2m + 1}} {i \sum_{m=0}^{(n-1)/2} \binom{n}{2m} \left( i \tan x \right)^{2m}} \\ & = \frac{ \sum_{m=0}^{(n-1)/2} \binom{n}{2m + 1} \left( \tan x \right)^{2m + 1} i^{2m + 1}} {\sum_{m=0}^{(n-1)/2} \binom{n}{2m} \left( \tan x \right)^{2m} i^{2m + 1}} \\ & = \frac{ \sum_{m=0}^{(n-1)/2} \binom{n}{2m + 1} \left( \tan x \right)^{2m + 1} \left( -1 \right)^m} {\sum_{m=0}^{(n-1)/2} \binom{n}{2m} \left( \tan x \right)^{2m} \left( -1 \right)^m} . \end{align*}

Thus, for $n = 2023$, we have \begin{align*} a_{2023} & = \binom{2023}{2023} \left( -1 \right)^{(2023-1)/2} \\ & = \left( -1 \right)^{1011} \\ & = \boxed{\textbf{(C) -1}}. \end{align*}

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 4

We can use recursion to find a pattern for this problem. Notice that, \begin{align*} \tan x &= \tan x\\ \tan 2x &= \dfrac{\tan x + \tan x}{1 - \tan^2 x} = \dfrac{2\tan x}{1 - \tan^2 x}\\ \tan 3x &= \dfrac{\tan 2x + \tan x}{1 - \tan 2x \tan x} = \dfrac{3\tan x - \tan^3 x}{1 - 3\tan^2 x} \end{align*} The coefficient of the highest degree term seems to be always $\pm 1$. Now, we prove this by an imcomplete mathematical induction.

  • Firstly, we suppose $n$ is odd, and $\tan nx = \dfrac{a_1\tan x + a_3\tan^3 x + \cdots + a_n\tan^n x}{a_0 + a_2\tan^2 x + \cdots + a_{n - 1}\tan^{n - 1} x}$, then

\begin{align*} \tan(n + 1)x &= \dfrac{\tan nx + \tan x}{1 - \tan nx \tan x}\\ &= \dfrac{\dfrac{a_1\tan x + a_3\tan^3 x + \cdots + a_n\tan^n x}{a_0 + a_2\tan^2 x + \cdots + a_{n - 1}\tan^{n - 1} x} + \tan x}{1 - \dfrac{a_1\tan x + a_3\tan^3 x + \cdots + a_n\tan^n x}{a_0 + a_2\tan^2 x + \cdots + a_{n - 1}\tan^{n - 1} x}\tan x}\\ &= \dfrac{a_1\tan x + a_3\tan^3 x + \cdots + a_n\tan^n x + a_0\tan x + a_2\tan^3 x + \cdots + a_{n - 1}\tan^n x}{a_0 + a_2\tan^2x + \cdots + a_{n - 1}\tan^{n - 1}x - a_1\tan^2 x - a_3\tan^4 x - \cdots \boxed{-a_n\tan^{n + 1} x}} \end{align*}

The coefficient of the highest degree term becomes $-a_n$.

  • Secondly, we suppose $n$ is even, and $\tan nx = \dfrac{a_1\tan x + a_3\tan^3 x + \cdots + a_{n - 1}\tan^{n - 1} x}{a_0 + a_2\tan^2 x + \cdots + a_n\tan^n x}$, then

\begin{align*} \tan(n + 1)x &= \dfrac{\tan nx + \tan x}{1 - \tan nx \tan x}\\ &= \dfrac{\dfrac{a_1\tan x + a_3\tan^3 x + \cdots + a_{n - 1}\tan^{n - 1} x}{a_0 + a_2\tan^2 x + \cdots + a_n\tan^n x} + \tan x}{1 - \dfrac{a_1\tan x + a_3\tan^3 x + \cdots + a_{n - 1}\tan^{n - 1} x}{a_0 + a_2\tan^2 x + \cdots + a_n\tan^n x}\tan x}\\ &= \dfrac{a_1\tan x + a_3\tan^3 x + \cdots + a_{n - 1}\tan^{n - 1} x + a_0\tan x + a_2\tan^3 x + \cdots \boxed{+ a_n\tan^{n + 1} x}}{a_0 + a_2\tan^2 x + \cdots + a_n\tan^n x - a_1\tan^2 x - a_3\tan^4 x - \cdots - a_{n - 1}\tan^n x} \end{align*} The coefficient of the highest degree term remains $a_n$.

When $n = 1$, $a_n = 1$. During the process of $n$ increasing to 2023, $a_n$ changed its sign a total of $2022 \div 2 = 1011$ times.

Hence, $a_{2023} = 1 \times (-1)^{1011} = \boxed{\textbf{(C)}-1}$

~reda_mandymath

Video Solution 1 by OmegaLearn

https://youtu.be/4KJR_1Kg4A4

Video Solution

https://youtu.be/0KH554CLayE

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

See Also

2023 AMC 12A (Problems • Answer Key • Resources)
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