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2024 AMC 12B Problems/Problem 11: Difference between revisions

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Let \( f(n) = \sin^2(n^\circ) \)
Let \( f(n) = \sin^2(n^\circ) \)


The function \( f(n) = \sin^2(n) \) is continuous for all real values of \( n \). We want to find the average value of \( f(n) \) over the interval: [<math>\frac{\pi}{180}</math>, <math>\frac{\pi}{2}</math>]
The function \( f(n) = \sin^2(n) \) is continuous for all real values of \( n \). We want to find the average value of \( f(n) \) over the interval: <math>\left[\frac{\pi}{180}, \frac{\pi}{2}\right]</math>
 
 
The average value of \( f(n) \) is <math>\frac{1}{\frac{\pi}{2} - \frac{\pi}{180}} \int_{1\cdot\frac{\pi}{180}}^{90\cdot\frac{\pi}{180}} \sin^{2}(x) \, dx = \frac{180}{89\pi} \int_{\pi/180}^{\pi/2} \sin^2(x) \, dx</math>
 
 
We can solve this integral using the power-reduction identity: <math>\sin^2(x) = \frac{1 - \cos(2x)}{2}</math>
 
<math>\frac{180}{89\pi} \int_{\pi/180}^{\pi/2} \frac{1 - \cos(2x)}{2} \, dx</math>  
<math>= \frac{180}{89\pi} \left.\left(\frac{x}{2} - \frac{\sin(2x)}{4}\right)\right|_{\pi/180}^{\pi/2}</math>  
<math>= \frac{180}{89\pi} \left[ \left(\frac{\pi}{4} - 0\right) - \left(\frac{\pi}{360} - \frac{\sin\left(\frac{\pi}{90}\right)}{4}\right) \right]</math>
<math>= \frac{180}{89\pi} \left( \frac{89\pi}{360} + \frac{\sin\left(\frac{\pi}{90}\right)}{4} \right)</math>
 
 
The taylor series of <math>\sin(\theta) = \theta - \frac{\theta^3}{6} + \frac{\theta^5}{120} - \cdots</math>
 
Because <math>\sin\left(\frac{\pi}{90}\right)</math> is a small angle, we can use a first-order taylor approximation and approximate <math>\sin\left(\frac{\pi}{90}\right) \approx \frac{\pi}{90}</math>
 
Hence, our result is <math>\frac{180}{89\pi}\left(\frac{89\pi}{360} + \frac{\pi}{360}\right) = \frac{180}{89\pi}\cdot\frac{90\pi}{360} = \frac{180}{89\pi}\cdot\frac{\pi}{4} = \frac{45}{89}</math>
 
 
While <math>\frac{45}{89}</math> isn't an option, it is greater than <math>\frac{1}{2}</math>, and out of the given options, only <math>\frac{91}{180}</math> is greater than <math>\frac{1}{2}</math> (we don't get <math>\frac{91}{180}</math> exactly because we approximated <math>\sin\left(\frac{\pi}{90}\right)</math> via first-order Taylor approximation, but it's close enough to <math>\frac{91}{180}</math>).
 
So the answer is


==Video Solution 1 by SpreadTheMathLove==
==Video Solution 1 by SpreadTheMathLove==

Revision as of 18:52, 2 August 2025

Problem

Let $x_n = \sin^2(n^{\circ})$. What is the mean of $x_1,x_2,x_3,\dots,x_{90}$?

$\textbf{(A) } \frac{11}{45} \qquad\textbf{(B) } \frac{22}{45} \qquad\textbf{(C) } \frac{89}{180} \qquad\textbf{(D) } \frac{1}{2} \qquad\textbf{(E) } \frac{91}{180}$

Solution 1

Add up $x_1$ with $x_{89}$, $x_2$ with $x_{88}$, and $x_i$ with $x_{90-i}$. Notice \[x_i+x_{90-i}=\sin^2(i^{\circ})+\sin^2((90-i)^{\circ})=\sin^2(i^{\circ})+\cos^2(i^{\circ})=1\] by the Pythagorean identity. Since we can pair up $1$ with $89$ and keep going until $44$ with $46$, we get \[x_1+x_2+\dots+x_{90}=44+x_{45}+x_{90}=44+\left(\frac{\sqrt{2}}{2}\right)^2+1^2=\frac{91}{2}\] Hence the mean is $\boxed{\textbf{(E) }\frac{91}{180}}$

~kafuu_chino

Solution 2

We can add a term $x_0$ into the list, and the total sum of the terms won't be affected since $x_0=0$. Once $x_0$ is added into the list, the average of the $91$ terms is clearly $\frac{1}{2}$. Hence the total sum of the terms is $\frac{91}{2}$. To get the average of the original $90$, we merely divide by $90$ to get $\frac{91}{180}$. Hence the mean is $\boxed{\textbf{(E) }\frac{91}{180}}$

This method is called constructing a variable (although most of you already know).

~tsun26, ShortPeopleFartalot

Solution 3 (Inductive Reasoning)

If we use radians to rewrite the question, we have: $x_n=\sin^2\left(\frac{n\pi}{2\times90}\right)$. Notice that $90$ have no specialty beyond any other integers, so we can use some inductive processes.

If we change $90$ to $2$: \[\frac{\sin^2\left(\frac{\pi}{4}\right)+\sin^2\left(\frac{2\pi}{4}\right)}{2}=\frac{\left(\frac{1}{\sqrt{2}}\right)^2+\left(1\right)^2}{2}=\frac{\frac{1}{2}+1}{2}=\frac{3}{4}\,.\]

If we change $90$ to $3$: \[\frac{\sin^2\left(\frac{\pi}{6}\right)+\sin^2\left(\frac{2\pi}{6}\right)+\sin^2\left(\frac{3\pi}{6}\right)}{3}=\frac{\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2+\left(1\right)^2}{3}=\frac{\frac{1}{4}+\frac{3}{4}+1}{3}=\frac{2}{3}=\frac{4}{6}\,.\]

By intuition, although not rigorous at all, we can guess out the solution if we change $90$ into $k$, we get $\frac{k+1}{2k}$. Thus, if we plug in $k=90$, we get $\frac{90+1}{2\times90}=\boxed{\mathbf{(E)}\,\frac{91}{180}}$

~Prof. Joker

Solution 4

~Kathan

Solution 4

Note that $\sin^2(x) = \frac{1 - \cos(2x)}{2}$. We want to determine $\frac{1}{90}\sum_{n = 1}^{90} \sin^2(n^{\circ})$.

\begin{align*} &= \frac{1}{90} \sum_{n = 1}^{90} \frac{1 - \cos(2n)}{2} \\ &= \frac{1}{2} -\frac{1}{180}\sum_{n = 1}^{90} \cos(2n) \\ \end{align*}

Graphing $\cos(x)$, we can pair $\cos(2^{\circ}) + \cos(178^{\circ}) = 0$ and so on. We are left with $\cos(90^{\circ}) + \cos(180^{\circ}) = -1$.

Our answer is $\frac{1}{2} + \frac{1}{180} = \boxed{\textbf{(E) }\frac{91}{180}}$

~vinyx


Solution 5 (Calculus)

Let \( f(n) = \sin^2(n^\circ) \)

The function \( f(n) = \sin^2(n) \) is continuous for all real values of \( n \). We want to find the average value of \( f(n) \) over the interval: $\left[\frac{\pi}{180}, \frac{\pi}{2}\right]$


The average value of \( f(n) \) is $\frac{1}{\frac{\pi}{2} - \frac{\pi}{180}} \int_{1\cdot\frac{\pi}{180}}^{90\cdot\frac{\pi}{180}} \sin^{2}(x) \, dx = \frac{180}{89\pi} \int_{\pi/180}^{\pi/2} \sin^2(x) \, dx$


We can solve this integral using the power-reduction identity: $\sin^2(x) = \frac{1 - \cos(2x)}{2}$

$\frac{180}{89\pi} \int_{\pi/180}^{\pi/2} \frac{1 - \cos(2x)}{2} \, dx$ $= \frac{180}{89\pi} \left.\left(\frac{x}{2} - \frac{\sin(2x)}{4}\right)\right|_{\pi/180}^{\pi/2}$ $= \frac{180}{89\pi} \left[ \left(\frac{\pi}{4} - 0\right) - \left(\frac{\pi}{360} - \frac{\sin\left(\frac{\pi}{90}\right)}{4}\right) \right]$ $= \frac{180}{89\pi} \left( \frac{89\pi}{360} + \frac{\sin\left(\frac{\pi}{90}\right)}{4} \right)$


The taylor series of $\sin(\theta) = \theta - \frac{\theta^3}{6} + \frac{\theta^5}{120} - \cdots$

Because $\sin\left(\frac{\pi}{90}\right)$ is a small angle, we can use a first-order taylor approximation and approximate $\sin\left(\frac{\pi}{90}\right) \approx \frac{\pi}{90}$

Hence, our result is $\frac{180}{89\pi}\left(\frac{89\pi}{360} + \frac{\pi}{360}\right) = \frac{180}{89\pi}\cdot\frac{90\pi}{360} = \frac{180}{89\pi}\cdot\frac{\pi}{4} = \frac{45}{89}$


While $\frac{45}{89}$ isn't an option, it is greater than $\frac{1}{2}$, and out of the given options, only $\frac{91}{180}$ is greater than $\frac{1}{2}$ (we don't get $\frac{91}{180}$ exactly because we approximated $\sin\left(\frac{\pi}{90}\right)$ via first-order Taylor approximation, but it's close enough to $\frac{91}{180}$).

So the answer is

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=gJq7DhLNnZ4&t=0s

See also

2024 AMC 12B (Problems • Answer Key • Resources)
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Problem 10 		Followed by
Problem 12
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