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

Revision as of 17:11, 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.

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

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

$\cos x = 1 - \sin x$

$\cos x + \sin x = 1$

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 C: 2

~Tucker

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.

@@ Line 17: / Line 17: @@
 <math>\cos x + \sin x = 1</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>.
+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 C: 2
 ~Tucker

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

Revision as of 17:11, 11 February 2021

Problem

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