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Difference between revisions of "2021 AMC 12A Problems/Problem 19"
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==Solution 2 (Analysis)==
 
==Solution 2 (Analysis)==
−
Let <math>f(x)=\sin\left(\frac{\pi}{2}\cos x\right)</math>
+
Let <math>f(x)=\sin\left(\frac{\pi}{2}\cos x\right)</math> and <math>g(x)=\cos \left( \frac{\pi}2 \sin x\right).</math> This problem is equivalent to counting the intersections of the graphs of <math>f(x)</math> and <math>g(x)</math> in the closed interval <math>[0,\pi].</math> We make a table of values, as shown below:
 +
<cmath>\begin{array}{cccc}
 +
& x=0 & x=\frac{\pi}{2} & x=\pi \\ [1.5ex]
 +
\hline\hline
 +
\cos x & 1 & 0 & -1 \\ [1.5ex]
 +
\frac{\pi}{2}\cos x & \frac{\pi}{2} & 0 & -\frac{\pi}{2} \\ [1.5ex]
 +
f(x) & 1 & 0 & -1 \\ [1.5ex]
 +
\hline
 +
\sin x & 0 & 1 & 0 \\ [1.5ex]
 +
\frac{\pi}{2}\sin x & 0 & \frac{\pi}{2} & 0 \\ [1.5ex]
 +
g(x) & 1 & 0 & 1
 +
\end{array}</cmath>
 +
 
 +
~MRENTHUSIASM (credit given to TheAMCHub)
  
 
== Video Solution by OmegaLearn (Using Triangle Inequality & Trigonometry) ==
 
== Video Solution by OmegaLearn (Using Triangle Inequality & Trigonometry) ==

Revision as of 09:52, 15 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 1 (Inverse Trigonometric Functions)

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

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.

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

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

$\cos x = 1 - \sin x$

$\cos x + \sin x = 1$

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 $\boxed{C) 2}$

~Tucker

Solution 2 (Analysis)

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 $f(x)$ and $g(x)$ in the closed interval $[0,\pi].$ We make a table of values, as shown below: \[\begin{array}{cccc} & x=0 & x=\frac{\pi}{2} & x=\pi \\ [1.5ex] \hline\hline \cos x & 1 & 0 & -1 \\ [1.5ex] \frac{\pi}{2}\cos x & \frac{\pi}{2} & 0 & -\frac{\pi}{2} \\ [1.5ex] f(x) & 1 & 0 & -1 \\ [1.5ex] \hline \sin x & 0 & 1 & 0 \\ [1.5ex] \frac{\pi}{2}\sin x & 0 & \frac{\pi}{2} & 0 \\ [1.5ex] g(x) & 1 & 0 & 1 \end{array}\]

~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

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