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

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<math>\textbf{(A) }0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3\qquad \textbf{(E) }4</math>
 
<math>\textbf{(A) }0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3\qquad \textbf{(E) }4</math>
  
−
==Solution==
+
==Solution 1 (Inverse Trigonometric Functions)==
−
<math>\sin \left( \frac{\pi}2 \cos x\right)=\cos \left( \frac{\pi}2 \sin x\right)</math>
+
The ranges of <math>\frac{\pi}2 \sin x</math> and <math>\frac{\pi}2 \cos x</math> are both <math>\left[-\frac{\pi}2, \frac{\pi}2 \right],</math> which is included in the range of <math>\arcsin,</math> so we can use it with no issues.
 +
<cmath>\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*}</cmath>
 +
This only happens at <math>x = 0, \frac{\pi}2</math> on the interval <math>[0,\pi],</math> because one of <math>\sin</math> and <math>\cos</math> must be <math>1</math> and the other <math>0.</math> Therefore, the answer is <math>\boxed{\textbf{(C) }2}.</math>
  
−
The interval is <math>[0,\pi]</math>, which is included in the range of both <math>\arccos</math> and <math>\arcsin</math>, so we can use them with no issues.
+
~Tucker
 +
 
 +
==Solution 2 (Cofunction Identity)==
 +
By the Cofunction Identity <math>\cos\theta=\sin\left(\frac{\pi}{2}-\theta\right),</math> we rewrite the given equation: <cmath>\sin \left(\frac{\pi}2 \cos x\right) = \sin \left(\frac{\pi}2 - \frac{\pi}2 \sin x\right).</cmath>
 +
Recall that if <math>\sin\theta=\sin\phi,</math> then <math>\theta=\phi+2n\pi</math> or <math>\theta=\pi-\phi+2n\pi</math> for some integer <math>n.</math> Therefore, we have two cases:
 +
<ol style="margin-left: 1.5em;">
 +
  <li><b><math>\boldsymbol{\frac{\pi}2 \cos x = \left(\frac{\pi}2 - \frac{\pi}2 \sin x\right) + 2n\pi}</math> for some integer <math>\boldsymbol{n}</math></b> <p>
 +
We rearrange and simplify: <cmath>\sin x + \cos x = 1 + 4n.</cmath>
 +
By rough constraints, we know that <math>-2 < \sin x + \cos x < 2,</math> from which <math>-2 < 1 - 4n < 2.</math> The only possibility is <math>n=0,</math> so
 +
<cmath>\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*}</cmath>
 +
for some integer <math>k.</math> <p>
 +
We get <math>x=0,\frac{\pi}{2}</math> for this case. Note that <math>x=\pi</math> is an extraneous solution by squaring <math>(*).</math></li><p>
 +
  <li><b><math>\boldsymbol{\frac{\pi}2 \cos x = \pi - \left(\frac{\pi}2 - \frac{\pi}2 \sin x\right) + 2n\pi}</math> for some integer <math>\boldsymbol{n}</math></b> <p>
 +
Similar to Case 1, we conclude that <math>n=0,</math> so <cmath>\cos x - \sin x = 1.</cmath>
 +
We get <math>x=0</math> for this case.</li><p>
 +
</ol>
 +
Together, we obtain <math>\boxed{\textbf{(C) }2}</math> solutions: <math>x=0,\frac{\pi}{2}.</math>
 +
 
 +
~MRENTHUSIASM
 +
 
 +
==Solution 3 (Graphs and Analyses)==
 +
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}
 +
& & & \\ [-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}</cmath>
 +
For <math>x\in[0,\pi],</math> note that:
  
−
<math>\frac{\pi}2 \cos x=\arcsin \left( \cos \left( \frac{\pi}2 \sin x\right)\right)</math>
+
* <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>
  
−
<math>\frac{\pi}2 \cos x=\frac{\pi}2 - \frac{\pi}2 \sin x</math>
+
* <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>
  
−
<math>\cos x = 1 - \sin x</math>
+
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>
  
−
<math>\cos x + \sin x = 1</math>
+
By the Cofunction Identity <math>\cos\theta=\sin\left(\frac{\pi}{2}-\theta\right),</math> we rewrite the given equation:
 +
<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> and <math>\frac{\pi}{2}\sin x\in\left[0,\frac{\pi}{2}\right],</math> it follows that <math>x\in\left[0,\frac{\pi}{2}\right]</math> and <math>\frac{\pi}{2}-\frac{\pi}{2}\sin x\in\left[0,\frac{\pi}{2}\right].</math>
  
−
This only happens at <math>x = 0, \frac{\pi}2</math> on the interval <math>[0,\pi]</math>, because one of <math>\sin</math> and <math>\cos</math> must be <math>1</math> and the other <math>0</math>. Therefore, the answer is <math>\boxed{C: 2}</math>
+
We can apply the arcsine function to both sides, then rearrange and simplify:
 +
<cmath>\begin{align*}
 +
\frac{\pi}{2}\cos x &= \frac{\pi}{2}-\frac{\pi}{2}\sin x \\
 +
\sin x + \cos x &= 1.
 +
\end{align*}</cmath>
 +
From Case 1 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, as shown below:
 +
<asy>
 +
/* Made by MRENTHUSIASM */
 +
size(600,200);
  
−
~Tucker
+
real f(real x) { return sin(pi/2*cos(x)); }
 +
real g(real x) { return cos(pi/2*sin(x)); }
 +
 
 +
draw(graph(f,0,pi),red,"$y=\sin\left(\frac{\pi}{2}\cos x\right)$");
 +
draw(graph(g,0,pi),blue,"$y=\cos\left(\frac{\pi}{2}\sin x\right)$");
 +
 
 +
real xMin = 0;
 +
real xMax = 5/4*pi;
 +
real yMin = -2;
 +
real yMax = 2;
 +
 
 +
//Draws the horizontal gridlines
 +
void horizontalLines()
 +
{
 +
  for (real i = yMin+1; i < yMax; ++i)
 +
  {
 +
    draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4));
 +
  }
 +
}
 +
 
 +
//Draws the vertical gridlines
 +
void verticalLines()
 +
{
 +
  for (real i = xMin+pi/2; i < xMax; i+=pi/2)
 +
  {
 +
    draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4));
 +
  }
 +
}
 +
 
 +
//Draws the horizontal ticks
 +
void horizontalTicks()
 +
{
 +
  for (real i = yMin+1; i < yMax; ++i)
 +
  {
 +
    draw((-1/8,i)--(1/8,i), black+linewidth(1));
 +
  }
 +
}
 +
 
 +
//Draws the vertical ticks
 +
void verticalTicks()
 +
{
 +
  for (real i = xMin+pi/2; i < xMax; i+=pi/2)
 +
  {
 +
    draw((i,-1/8)--(i,1/8), black+linewidth(1));
 +
  }
 +
}
 +
 
 +
horizontalLines();
 +
verticalLines();
 +
horizontalTicks();
 +
verticalTicks();
 +
draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
 +
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
 +
label("$x$",(xMax,0),(2,0));
 +
label("$y$",(0,yMax),(0,2));
 +
 
 +
pair A[];
 +
A[0] = (pi/2,0);
 +
A[1] = (pi,0);
 +
A[2] = (0,1);
 +
A[3] = (0,0);
 +
A[4] = (0,-1);
 +
 
 +
label("$\frac{\pi}{2}$",A[0],(0,-2.5));
 +
label("$\pi$",A[1],(0,-2.5));
 +
label("$1$",A[2],(-2.5,0));
 +
label("$0$",A[3],(-2.5,0));
 +
label("$-1$",A[4],(-2.5,0));
 +
 
 +
dot((0,1),linewidth(5));
 +
dot((pi/2,0),linewidth(5));
 +
 
 +
add(legend(),point(E),40E,UnFill);
 +
</asy>
 +
Therefore, the answer is <math>\boxed{\textbf{(C) }2}.</math>
 +
 
 +
~MRENTHUSIASM (credit given to TheAMCHub)
 +
 
 +
== Video Solution by OmegaLearn (Using Triangle Inequality & Trigonometry) ==
 +
https://youtu.be/wJxN1YPuyCg
 +
 
 +
~ pi_is_3.14
 +
 
 +
==Video Solution (Quick and Easy)==
 +
https://youtu.be/6AWb9cqFblU
 +
 
 +
~Education, the Study of Everything
  
 
==See also==
 
==See also==
 
{{AMC12 box|year=2021|ab=A|num-b=18|num-a=20}}
 
{{AMC12 box|year=2021|ab=A|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:10, 5 November 2022

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 rewrite the given equation: \[\sin \left(\frac{\pi}2 \cos x\right) = \sin \left(\frac{\pi}2 - \frac{\pi}2 \sin x\right).\] Recall that if $\sin\theta=\sin\phi,$ then $\theta=\phi+2n\pi$ or $\theta=\pi-\phi+2n\pi$ for some integer $n.$ Therefore, we have two cases:

  1. $\boldsymbol{\frac{\pi}2 \cos x = \left(\frac{\pi}2 - \frac{\pi}2 \sin x\right) + 2n\pi}$ for some integer $\boldsymbol{n}$

    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 \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 $k.$

    We get $x=0,\frac{\pi}{2}$ for this case. Note that $x=\pi$ is an extraneous solution by squaring $(*).$

  2. $\boldsymbol{\frac{\pi}2 \cos x = \pi - \left(\frac{\pi}2 - \frac{\pi}2 \sin x\right) + 2n\pi}$ for some integer $\boldsymbol{n}$

    Similar to Case 1, we conclude that $n=0,$ so \[\cos x - \sin x = 1.\] We get $x=0$ for this case.

Together, we obtain $\boxed{\textbf{(C) }2}$ solutions: $x=0,\frac{\pi}{2}.$

~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 rewrite the given equation: \[\sin\left(\frac{\pi}{2}\cos x\right) = \sin\left(\frac{\pi}{2}-\frac{\pi}{2}\sin x\right).\] Since $\frac{\pi}{2}\cos x\in\left[0,\frac{\pi}{2}\right]$ and $\frac{\pi}{2}\sin x\in\left[0,\frac{\pi}{2}\right],$ it follows that $x\in\left[0,\frac{\pi}{2}\right]$ and $\frac{\pi}{2}-\frac{\pi}{2}\sin x\in\left[0,\frac{\pi}{2}\right].$

We can apply the arcsine function to both sides, then rearrange and simplify: \begin{align*} \frac{\pi}{2}\cos x &= \frac{\pi}{2}-\frac{\pi}{2}\sin x \\ \sin x + \cos x &= 1. \end{align*} From Case 1 in Solution 2, we conclude that $(0,1)$ and $\left(\frac{\pi}{2},0\right)$ are the only points of intersection, as shown below: [asy] /* Made by MRENTHUSIASM */ size(600,200); real f(real x) { return sin(pi/2*cos(x)); } real g(real x) { return cos(pi/2*sin(x)); } draw(graph(f,0,pi),red,"$y=\sin\left(\frac{\pi}{2}\cos x\right)$"); draw(graph(g,0,pi),blue,"$y=\cos\left(\frac{\pi}{2}\sin x\right)$"); real xMin = 0; real xMax = 5/4*pi; real yMin = -2; real yMax = 2; //Draws the horizontal gridlines void horizontalLines() { for (real i = yMin+1; i < yMax; ++i) { draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); } } //Draws the vertical gridlines void verticalLines() { for (real i = xMin+pi/2; i < xMax; i+=pi/2) { draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); } } //Draws the horizontal ticks void horizontalTicks() { for (real i = yMin+1; i < yMax; ++i) { draw((-1/8,i)--(1/8,i), black+linewidth(1)); } } //Draws the vertical ticks void verticalTicks() { for (real i = xMin+pi/2; i < xMax; i+=pi/2) { draw((i,-1/8)--(i,1/8), black+linewidth(1)); } } horizontalLines(); verticalLines(); horizontalTicks(); verticalTicks(); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2)); pair A[]; A[0] = (pi/2,0); A[1] = (pi,0); A[2] = (0,1); A[3] = (0,0); A[4] = (0,-1); label("$\frac{\pi}{2}$",A[0],(0,-2.5)); label("$\pi$",A[1],(0,-2.5)); label("$1$",A[2],(-2.5,0)); label("$0$",A[3],(-2.5,0)); label("$-1$",A[4],(-2.5,0)); dot((0,1),linewidth(5)); dot((pi/2,0),linewidth(5)); add(legend(),point(E),40E,UnFill); [/asy] Therefore, the answer is $\boxed{\textbf{(C) }2}.$

~MRENTHUSIASM (credit given to TheAMCHub)

Video Solution by OmegaLearn (Using Triangle Inequality & Trigonometry)

https://youtu.be/wJxN1YPuyCg

~ pi_is_3.14

Video Solution (Quick and Easy)

https://youtu.be/6AWb9cqFblU

~Education, the Study of Everything

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

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