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

Revision as of 00:33, 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 $\frac{\pi}{2}\sin x,\frac{\pi}{2}\cos x\in\left[0,\frac{\pi}{2}\right],$ 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. 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

@@ Line 67: / Line 67: @@
 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:
+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:
 <cmath>\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*}</cmath>

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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