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

(Added a solution)
(Rephrasing of statement on arc functions)
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<math>\sin \left( \frac{\pi}2 \cos x\right)=\cos \left( \frac{\pi}2 \sin x\right)</math>
 
<math>\sin \left( \frac{\pi}2 \cos x\right)=\cos \left( \frac{\pi}2 \sin x\right)</math>
  
−
The interval is <math>[0,\pi]</math>, so both <math>\arccos</math> and <math>\arcsin</math> will lie within the range.
+
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.
  
 
<math>\frac{\pi}2 \cos x=\arcsin \left( \cos \left( \frac{\pi}2 \sin x\right)\right)</math>
 
<math>\frac{\pi}2 \cos x=\arcsin \left( \cos \left( \frac{\pi}2 \sin x\right)\right)</math>
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~Tucker
 
~Tucker
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==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}}

Revision as of 16:58, 11 February 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

$\sin \left( \frac{\pi}2 \cos x\right)=\cos \left( \frac{\pi}2 \sin x\right)$

The interval is $[0,\pi]$, which is included in the range of both $\arccos$ and $\arcsin$, so we can use them with no issues.

$\frac{\pi}2 \cos x=\arcsin \left( \cos \left( \frac{\pi}2 \sin x\right)\right)$

$\frac{\pi}2 \cos x=\frac{\pi}2 - \frac{\pi}2 \sin x$

$\cos x = 1 - \sin x$

This only happens at $x = 0, \frac{\pi}2$ on the interval $[0,\pi]$, because either $\cos x = 0$ and $\sin x = 1$ or $\cos x = 1$ and $\sin x = 0$.

~Tucker

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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