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

m (→‎Solution 3 (Graphs and Analyses))
(Condensed Sol 2 a bit.)
Line 18: Line 18:
  
 
==Solution 2 (Cofunction Identity)==
 
==Solution 2 (Cofunction Identity)==
−
By the cofunction identity <math>\sin\theta=\cos\left(\frac{\pi}{2}-\theta\right)</math> for all <math>\theta,</math> we simplify the given equation:
+
By the Cofunction Identity <math>\cos\theta=\sin\left(\frac{\pi}{2}-\theta\right),</math> we simplify the given equation:
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
\sin \left( \frac{\pi}2 \cos x\right) &= \cos \left( \frac{\pi}2 \sin x\right) \\
+
\sin \left( \frac{\pi}2 \cos x\right) &= \sin \left(\frac{\pi}2 - \frac{\pi}2 \sin x\right) \\
−
\cos \left(\frac{\pi}2-\frac{\pi}2 \cos x\right) &= \cos \left( \frac{\pi}2 \sin x\right) \\
+
\frac{\pi}2 \cos x &= \frac{\pi}2 - \frac{\pi}2 \sin x + 2n\pi
−
\cos \left(\frac{\pi}2 \left(1 - \cos x \right)\right) &= \cos \left( \frac{\pi}2 \sin x\right) \\
 
−
\frac{\pi}2 \left(1 - \cos x \right) &= \frac{\pi}2 \sin x + 2n\pi,
 
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
for some integer <math>n.</math> We keep simplifying:
+
for some integer <math>n.</math>
−
<cmath>\begin{align*}
+
 
−
1 - \cos x &= \sin x + 4n \\
+
We rearrange and simplify: <cmath>\sin x + \cos x = 1 + 4n.</cmath>
−
1 - 4n &= \sin x + \cos x.
+
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 we get
−
\end{align*}</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> From here, we get
 
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
\sin x + \cos x &= 1 \ \ \ \ \ (*) \\
+
\sin x + \cos x &= 1 && (*) \\
 
\sin^2 x + \cos^2 x + 2\sin x \cos x &= 1 \\
 
\sin^2 x + \cos^2 x + 2\sin x \cos x &= 1 \\
 
2\sin x \cos x &= 0 \\
 
2\sin x \cos x &= 0 \\
 
\sin(2x) &= 0 \\
 
\sin(2x) &= 0 \\
 
2x &= k\pi \\
 
2x &= k\pi \\
−
x &= \frac{k\pi}{2},
+
x &= \frac{k\pi}{2}
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 
for some integer <math>k.</math>
 
for some integer <math>k.</math>
  
−
The <i><b>possible</b></i> solutions in <math>[0,\pi]</math> are <math>x=0,\frac{\pi}{2},\pi,</math> but only <math>x=0,\frac{\pi}{2}</math> check the original equation (Note that <math>x=\pi</math> is an extraneous solution formed by squaring <math>(*)</math> above.). Therefore, the answer is <math>\boxed{\textbf{(C) }2}.</math>
+
The <i><b>possible</b></i> solutions in <math>[0,\pi]</math> are <math>x=0,\frac{\pi}{2},\pi.</math> However, <math>x=\pi</math> is an extraneous solution by squaring <math>(*).</math> Therefore, the answer is <math>\boxed{\textbf{(C) }2}.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM

Revision as of 21:57, 16 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)

Let $f(x)=\sin\left(\frac{\pi}{2}\cos x\right)$ and $g(x)=\cos \left( \frac{\pi}2 \sin x\right).$ This problem is equivalent to counting the intersections of the graphs of $y=f(x)$ and $y=g(x)$ in the closed interval $[0,\pi].$ We make a table of values, as shown below: \[\begin{array}{c|ccc} & & & \\ [-2ex] & \boldsymbol{x=0} & \boldsymbol{x=\frac{\pi}{2}} & \boldsymbol{x=\pi} \\ [1.5ex] \hline & & & \\ [-1ex] \boldsymbol{\cos x} & 1 & 0 & -1 \\ [1.5ex] \boldsymbol{\frac{\pi}{2}\cos x} & \frac{\pi}{2} & 0 & -\frac{\pi}{2} \\ [1.5ex] \boldsymbol{f(x)} & 1 & 0 & -1 \\ [1.5ex] \hline & & & \\ [-1ex] \boldsymbol{\sin x} & 0 & 1 & 0 \\ [1.5ex] \boldsymbol{\frac{\pi}{2}\sin x} & 0 & \frac{\pi}{2} & 0 \\ [1.5ex] \boldsymbol{g(x)} & 1 & 0 & 1 \\ [1ex] \end{array}\] For the graphs of $y=f(x)$ and $y=g(x),$ we will analyze their increasing/decreasing behaviors in $[0,\pi]:$

  • The graph of $y=f(x)$ in $[0,\pi]$ (from left to right) has the same behavior as the graph of $y=\sin x$ in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ (from right to left): The output is from $1$ to $-1$ (from left to right), inclusive, and strictly decreasing.
  • The graph of $y=g(x)$ in $[0,\pi]$ (from left to right) has two parts:
  1. The graph of $y=g(x)$ in $\left[0,\frac{\pi}{2}\right]$ has the same behavior as the graph of $y=\cos x$ in $\left[0,\frac{\pi}{2}\right]$ (from left to right): The output is from $1$ to $0$ (from left to right), inclusive, and strictly decreasing.

  2. The graph of $y=g(x)$ in $\left[\frac{\pi}{2},\pi\right]$ has the same behavior as the graph of $y=\cos x$ in $\left[0,\frac{\pi}{2}\right]$ (from right to left): The output is from $0$ to $1$ (from left to right), inclusive, and strictly increasing.

If $x\in\left(\frac{\pi}{2},\pi\right],$ then $f(x)<0$ and $g(x)>0.$ So, their graphs do not intersect.

If $x\in\left[0,\frac{\pi}{2}\right],$ then $0\leq f(x),g(x)\leq1.$ Clearly, the graphs intersect at $x=0$ and $x=\frac{\pi}{2}$ (at points $(0,1)$ and $\left(\frac{\pi}{2},0\right),$ respectively), but we will determine whether they are the only points of intersection:

Let $A=\frac{\pi}{2}\cos x$ and $B=\frac{\pi}{2}\sin x.$ It follows that $A,B\in\left[0,\frac{\pi}{2}\right].$ Since $\sin A = \cos B,$ we know that $A+B=\frac{\pi}{2}$ by the cofunction identity: \begin{align*} \frac{\pi}{2}\cos x + \frac{\pi}{2}\sin x &= \frac{\pi}{2} \\ \cos x + \sin 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

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