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"

(Incorporated the graph into Sol 3.)
m (→‎Solution 3 (Graphs and Analyses))
Line 65: Line 65:
 
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>
 
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>
  
−
By the Cofunction Identity <math>\cos\theta=\sin\left(\frac{\pi}{2}-\theta\right),</math> we have
+
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>
 
<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}\sin x,\frac{\pi}{2}\cos x\in\left[0,\frac{\pi}{2}\right],</math> we can apply the arcsine function to both sides, then rearrange and simplify:
 
Since <math>\frac{\pi}{2}\sin x,\frac{\pi}{2}\cos x\in\left[0,\frac{\pi}{2}\right],</math> we can apply the arcsine function to both sides, then rearrange and simplify:

Revision as of 06:37, 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 rewrite the given equation: \[\sin\left(\frac{\pi}{2}\cos x\right) = \sin\left(\frac{\pi}{2}-\frac{\pi}{2}\sin x\right).\] Since $\frac{\pi}{2}\sin x,\frac{\pi}{2}\cos x\in\left[0,\frac{\pi}{2}\right],$ 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 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, 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

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=162342"