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

(→‎Solution 2 (Cofunction Identity))
(→‎Solution 2 (Cofunction Identity))
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<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
1 - \cos x &= \sin x + 4n \\
 
1 - \cos x &= \sin x + 4n \\
−
1 - 4n &= \sin x + \cos x. \\
+
1 - 4n &= \sin x + \cos x.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 +
By rough constraints, we know that <math>-2 < \sin x + \cos x < 2,</math> so that <math>-2 < 1 - 4n < 2.</math> The only possibility is <math>n=0.</math> From here, we get
 +
<cmath>\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*}</cmath>
 +
for some integer <math>k.</math>
 +
  
  

Revision as of 11:18, 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 (Cofunction Identity)

By the cofunction identity $\sin\theta=\cos\left(\frac{\pi}{2}-\theta\right)$ for all $\theta,$ we simplify the given equation: \begin{align*} \sin \left( \frac{\pi}2 \cos x\right) &= \cos \left( \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) \\ \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*} for some integer $n.$ We keep simplifying: \begin{align*} 1 - \cos x &= \sin x + 4n \\ 1 - 4n &= \sin x + \cos x. \end{align*} By rough constraints, we know that $-2 < \sin x + \cos x < 2,$ so that $-2 < 1 - 4n < 2.$ The only possibility is $n=0.$ From here, 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 $k.$


~MRENTHUSIASM

Solution 3 (Graphs and 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: \[\begin{array}{cccc} & x=0 & x=\frac{\pi}{2} & x=\pi \\ [1.5ex] \hline\hline & & & \\ [-1ex] \cos x & 1 & 0 & -1 \\ [1.5ex] \frac{\pi}{2}\cos x & \frac{\pi}{2} & 0 & -\frac{\pi}{2} \\ [1.5ex] f(x) & 1 & 0 & -1 \\ [1.5ex] \hline & & & \\ [-1ex] \sin x & 0 & 1 & 0 \\ [1.5ex] \frac{\pi}{2}\sin x & 0 & \frac{\pi}{2} & 0 \\ [1.5ex] g(x) & 1 & 0 & 1 \end{array}\] The graph of $f(x)$ in $[0,\pi]$ (from left to right) is the same as the graph of $\sin x$ 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 $g(x)$ in $[0,\pi]$ (from left to right) has two parts:

$(1) \ \cos x$ 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.

$(2) \ \cos x$ 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 $\left(\frac{\pi}{2},0\right)$), but we will prove/disprove that 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*}

Applying Solution 2's argument to deduce that $(0,1)$ and $\left(\frac{\pi}{2},0\right)$ are the only points of intersection. So, the answer is $\boxed{\textbf{(C) }2}.$

Graphs of $f(x)$ and $g(x)$ in Desmos: https://www.desmos.com/calculator/9ypgulyzov

~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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