Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2021 AMC 12A Problems/Problem 19"

(Condensed Sol 2 a bit.)
(Condensed Sol 3 considerably.)
Line 42: Line 42:
  
 
==Solution 3 (Graphs and Analyses)==
 
==Solution 3 (Graphs and Analyses)==
−
Let <math>f(x)=\sin\left(\frac{\pi}{2}\cos x\right)</math> and <math>g(x)=\cos \left( \frac{\pi}2 \sin x\right).</math> This problem is equivalent to counting the intersections of the graphs of <math>y=f(x)</math> and <math>y=g(x)</math> in the closed interval <math>[0,\pi].</math> We make a table of values, as shown below:
+
This problem is equivalent to counting the intersections of the graphs of <math>y=\sin\left(\frac{\pi}{2}\cos x\right)</math> and <math>y=\cos\left(\frac{\pi}{2}\sin x\right)</math> in the closed interval <math>[0,\pi].</math> We construct a table of values, as shown below:
 
<cmath>\begin{array}{c|ccc}
 
<cmath>\begin{array}{c|ccc}
 
& & & \\ [-2ex]
 
& & & \\ [-2ex]
Line 50: Line 50:
 
\boldsymbol{\cos x} & 1 & 0 & -1 \\ [1.5ex]
 
\boldsymbol{\cos x} & 1 & 0 & -1 \\ [1.5ex]
 
\boldsymbol{\frac{\pi}{2}\cos x} & \frac{\pi}{2} & 0 & -\frac{\pi}{2} \\ [1.5ex]
 
\boldsymbol{\frac{\pi}{2}\cos x} & \frac{\pi}{2} & 0 & -\frac{\pi}{2} \\ [1.5ex]
−
\boldsymbol{f(x)} & 1 & 0 & -1 \\ [1.5ex]
+
\boldsymbol{\sin\left(\frac{\pi}{2}\cos x\right)} & 1 & 0 & -1 \\ [1.5ex]
 
\hline  
 
\hline  
 
& & & \\ [-1ex]
 
& & & \\ [-1ex]
 
\boldsymbol{\sin x} & 0 & 1 & 0 \\ [1.5ex]
 
\boldsymbol{\sin x} & 0 & 1 & 0 \\ [1.5ex]
 
\boldsymbol{\frac{\pi}{2}\sin x} & 0 & \frac{\pi}{2} & 0 \\ [1.5ex]
 
\boldsymbol{\frac{\pi}{2}\sin x} & 0 & \frac{\pi}{2} & 0 \\ [1.5ex]
−
\boldsymbol{g(x)} & 1 & 0 & 1 \\ [1ex]
+
\boldsymbol{\cos\left(\frac{\pi}{2}\sin x\right)} & 1 & 0 & 1 \\ [1ex]
 
\end{array}</cmath>
 
\end{array}</cmath>
−
For the graphs of <math>y=f(x)</math> and <math>y=g(x),</math> we will analyze their increasing/decreasing behaviors in <math>[0,\pi]:</math>  
+
For <math>x\in[0,\pi],</math> note that:
  
−
*The graph of <math>y=f(x)</math> in <math>[0,\pi]</math> (from left to right) has the same behavior as the graph of <math>y=\sin x</math> in <math>\left[-\frac{\pi}{2},\frac{\pi}{2}\right]</math> (from right to left): The output is from <math>1</math> to <math>-1</math> (from left to right), inclusive, and strictly decreasing.
+
* <math>\frac{\pi}{2}\cos x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right],</math> so <math>\sin\left(\frac{\pi}{2}\cos x\right)\in[-1,1].</math>
  
−
*The graph of <math>y=g(x)</math> in <math>[0,\pi]</math> (from left to right) has two parts:
+
* <math>\frac{\pi}{2}\sin x\in\left[0,\frac{\pi}{2}\right],</math> so <math>\cos\left(\frac{\pi}{2}\sin x\right)\in[0,1].</math>
−
<ol style="margin-left: 3em;">
 
−
  <li>The graph of <math>y=g(x)</math> in <math>\left[0,\frac{\pi}{2}\right]</math> has the same behavior as the graph of <math>y=\cos x</math> in <math>\left[0,\frac{\pi}{2}\right]</math> (from left to right): The output is from <math>1</math> to <math>0</math> (from left to right), inclusive, and strictly decreasing.</li><p>
 
−
  <li>The graph of <math>y=g(x)</math> in  <math>\left[\frac{\pi}{2},\pi\right]</math> has the same behavior as the graph of <math>y=\cos x</math> in <math>\left[0,\frac{\pi}{2}\right]</math> (from right to left): The output is from <math>0</math> to <math>1</math> (from left to right), inclusive, and strictly increasing.</li><p>
 
−
</ol>
 
−
If <math>x\in\left(\frac{\pi}{2},\pi\right],</math> then <math>f(x)<0</math> and <math>g(x)>0.</math> So, their graphs do not intersect.
 
  
−
If <math>x\in\left[0,\frac{\pi}{2}\right],</math> then <math>0\leq f(x),g(x)\leq1.</math> Clearly, the graphs intersect at <math>x=0</math> and <math>x=\frac{\pi}{2}</math> (at points <math>(0,1)</math> and <math>\left(\frac{\pi}{2},0\right),</math> respectively), but we will determine whether they are the <i><b>only</b></i> points of intersection:
+
For the graphs to intersect, we need <math>\sin\left(\frac{\pi}{2}\cos x\right)\in[0,1].</math> This occurs when <math>\frac{\pi}{2}\cos x\in\left[0,\frac{\pi}{2}\right].</math>
  
−
Let <math>A=\frac{\pi}{2}\cos x</math> and <math>B=\frac{\pi}{2}\sin x.</math> It follows that <math>A,B\in\left[0,\frac{\pi}{2}\right].</math> Since <math>\sin A = \cos B,</math> we know that <math>A+B=\frac{\pi}{2}</math> by the cofunction identity:
+
By the Cofunction Identity <math>\cos\theta=\sin\left(\frac{\pi}{2}-\theta\right),</math> we have
 +
<cmath>\sin\left(\frac{\pi}{2}\cos x\right) = \sin\left(\frac{\pi}{2}-\frac{\pi}{2}\sin x\right).</cmath>
 +
Since <math>\frac{\pi}{2}\cos x\in\left[0,\frac{\pi}{2}\right],</math> we can apply the arcsine function to both sides and then simplify:
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
\frac{\pi}{2}\cos x + \frac{\pi}{2}\sin x &= \frac{\pi}{2} \\
+
\frac{\pi}{2}\cos x &= \frac{\pi}{2}-\frac{\pi}{2}\sin x \\
−
\cos x + \sin x &=1.
+
\cos x &= 1 - \sin x \\
 +
\sin x + \cos x &= 1.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 
From the last block of equations in Solution 2, we conclude that <math>(0,1)</math> and <math>\left(\frac{\pi}{2},0\right)</math> are the only points of intersection. So, the answer is <math>\boxed{\textbf{(C) }2}.</math>
 
From the last block of equations in Solution 2, we conclude that <math>(0,1)</math> and <math>\left(\frac{\pi}{2},0\right)</math> are the only points of intersection. So, the answer is <math>\boxed{\textbf{(C) }2}.</math>

Revision as of 00:32, 17 September 2021

Problem

How many solutions does the equation $\sin \left( \frac{\pi}2 \cos x\right)=\cos \left( \frac{\pi}2 \sin x\right)$ have in the closed interval $[0,\pi]$?

$\textbf{(A) }0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3\qquad \textbf{(E) }4$

Solution 1 (Inverse Trigonometric Functions)

The ranges of $\frac{\pi}2 \sin x$ and $\frac{\pi}2 \cos x$ are both $\left[-\frac{\pi}2, \frac{\pi}2 \right],$ which is included in the range of $\arcsin,$ so we can use it with no issues. \begin{align*} \frac{\pi}2 \cos x &= \arcsin \left( \cos \left( \frac{\pi}2 \sin x\right)\right) \\ \frac{\pi}2 \cos x &= \arcsin \left( \sin \left( \frac{\pi}2 - \frac{\pi}2 \sin x\right)\right) \\ \frac{\pi}2 \cos x &= \frac{\pi}2 - \frac{\pi}2 \sin x \\ \cos x &= 1 - \sin x \\ \cos x + \sin x &= 1. \end{align*} This only happens at $x = 0, \frac{\pi}2$ on the interval $[0,\pi],$ because one of $\sin$ and $\cos$ must be $1$ and the other $0.$ Therefore, the answer is $\boxed{\textbf{(C) }2}.$

~Tucker

Solution 2 (Cofunction Identity)

By the Cofunction Identity $\cos\theta=\sin\left(\frac{\pi}{2}-\theta\right),$ we simplify the given equation: \begin{align*} \sin \left( \frac{\pi}2 \cos x\right) &= \sin \left(\frac{\pi}2 - \frac{\pi}2 \sin x\right) \\ \frac{\pi}2 \cos x &= \frac{\pi}2 - \frac{\pi}2 \sin x + 2n\pi \end{align*} for some integer $n.$

We rearrange and simplify: \[\sin x + \cos x = 1 + 4n.\] By rough constraints, we know that $-2 < \sin x + \cos x < 2,$ from which $-2 < 1 - 4n < 2.$ The only possibility is $n=0,$ so we get \begin{align*} \sin x + \cos x &= 1 && (*) \\ \sin^2 x + \cos^2 x + 2\sin x \cos x &= 1 \\ 2\sin x \cos x &= 0 \\ \sin(2x) &= 0 \\ 2x &= k\pi \\ x &= \frac{k\pi}{2} \end{align*} for some integer $k.$

The possible solutions in $[0,\pi]$ are $x=0,\frac{\pi}{2},\pi.$ However, $x=\pi$ is an extraneous solution by squaring $(*).$ Therefore, the answer is $\boxed{\textbf{(C) }2}.$

~MRENTHUSIASM

Solution 3 (Graphs and Analyses)

This problem is equivalent to counting the intersections of the graphs of $y=\sin\left(\frac{\pi}{2}\cos x\right)$ and $y=\cos\left(\frac{\pi}{2}\sin x\right)$ in the closed interval $[0,\pi].$ We construct a table of values, as shown below: \[\begin{array}{c|ccc} & & & \\ [-2ex] & \boldsymbol{x=0} & \boldsymbol{x=\frac{\pi}{2}} & \boldsymbol{x=\pi} \\ [1.5ex] \hline & & & \\ [-1ex] \boldsymbol{\cos x} & 1 & 0 & -1 \\ [1.5ex] \boldsymbol{\frac{\pi}{2}\cos x} & \frac{\pi}{2} & 0 & -\frac{\pi}{2} \\ [1.5ex] \boldsymbol{\sin\left(\frac{\pi}{2}\cos x\right)} & 1 & 0 & -1 \\ [1.5ex] \hline & & & \\ [-1ex] \boldsymbol{\sin x} & 0 & 1 & 0 \\ [1.5ex] \boldsymbol{\frac{\pi}{2}\sin x} & 0 & \frac{\pi}{2} & 0 \\ [1.5ex] \boldsymbol{\cos\left(\frac{\pi}{2}\sin x\right)} & 1 & 0 & 1 \\ [1ex] \end{array}\] For $x\in[0,\pi],$ note that:

  • $\frac{\pi}{2}\cos x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right],$ so $\sin\left(\frac{\pi}{2}\cos x\right)\in[-1,1].$
  • $\frac{\pi}{2}\sin x\in\left[0,\frac{\pi}{2}\right],$ so $\cos\left(\frac{\pi}{2}\sin x\right)\in[0,1].$

For the graphs to intersect, we need $\sin\left(\frac{\pi}{2}\cos x\right)\in[0,1].$ This occurs when $\frac{\pi}{2}\cos x\in\left[0,\frac{\pi}{2}\right].$

By the Cofunction Identity $\cos\theta=\sin\left(\frac{\pi}{2}-\theta\right),$ we have \[\sin\left(\frac{\pi}{2}\cos x\right) = \sin\left(\frac{\pi}{2}-\frac{\pi}{2}\sin x\right).\] Since $\frac{\pi}{2}\cos x\in\left[0,\frac{\pi}{2}\right],$ we can apply the arcsine function to both sides and then simplify: \begin{align*} \frac{\pi}{2}\cos x &= \frac{\pi}{2}-\frac{\pi}{2}\sin x \\ \cos x &= 1 - \sin x \\ \sin x + \cos x &= 1. \end{align*} From the last block of equations in Solution 2, we conclude that $(0,1)$ and $\left(\frac{\pi}{2},0\right)$ are the only points of intersection. So, the answer is $\boxed{\textbf{(C) }2}.$

~MRENTHUSIASM (credit given to TheAMCHub)

Remark

The graphs of $f(x)=\sin \left( \frac{\pi}2 \cos x\right)$ (in red) and $g(x)=\cos \left( \frac{\pi}2 \sin x\right)$ (in blue) are shown below.

File:Problem 19 Diagram.png

Graph in Desmos: https://www.desmos.com/calculator/brjh3gybcc

~MRENTHUSIASM

Video Solution by OmegaLearn (Using Triangle Inequality & Trigonometry)

https://youtu.be/wJxN1YPuyCg

~ pi_is_3.14

See also

2021 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_19&oldid=162339"