Difference between revisions of "2021 AMC 12A Problems/Problem 19"
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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 <math>\boxed{\textbf{(C) }2}.</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 <math>\boxed{\textbf{(C) }2}.</math> | ||
| − | ~Tucker | + | ~Tucker |
| − | |||
| − | |||
==Solution 2 (Cofunction Identity)== | ==Solution 2 (Cofunction Identity)== | ||
| Line 37: | Line 35: | ||
We get <math>x=0,\frac{\pi}{2}</math> for this case. Note that <math>x=\pi</math> is an extraneous solution by squaring <math>(*).</math></li><p> | We get <math>x=0,\frac{\pi}{2}</math> for this case. Note that <math>x=\pi</math> is an extraneous solution by squaring <math>(*).</math></li><p> | ||
<li><b><math>\boldsymbol{\frac{\pi}2 \cos x = \pi - \left(\frac{\pi}2 - \frac{\pi}2 \sin x\right) + 2n\pi}</math> for some integer <math>\boldsymbol{n}</math></b> <p> | <li><b><math>\boldsymbol{\frac{\pi}2 \cos x = \pi - \left(\frac{\pi}2 - \frac{\pi}2 \sin x\right) + 2n\pi}</math> for some integer <math>\boldsymbol{n}</math></b> <p> | ||
| − | + | Similar to Case 1, we conclude that <math>n=0,</math> so <cmath>\cos x - \sin x = 1.</cmath> | |
We get <math>x=0</math> for this case.</li><p> | We get <math>x=0</math> for this case.</li><p> | ||
</ol> | </ol> | ||
| Line 70: | Line 68: | ||
By the Cofunction Identity <math>\cos\theta=\sin\left(\frac{\pi}{2}-\theta\right),</math> we rewrite the given equation: | By the Cofunction Identity <math>\cos\theta=\sin\left(\frac{\pi}{2}-\theta\right),</math> we rewrite the given equation: | ||
<cmath>\sin\left(\frac{\pi}{2}\cos x\right) = \sin\left(\frac{\pi}{2}-\frac{\pi}{2}\sin x\right).</cmath> | <cmath>\sin\left(\frac{\pi}{2}\cos x\right) = \sin\left(\frac{\pi}{2}-\frac{\pi}{2}\sin x\right).</cmath> | ||
| − | Since <math>\frac{\pi}{2}\sin x,\frac{\pi}{2}\ | + | Since <math>\frac{\pi}{2}\cos x\in\left[0,\frac{\pi}{2}\right]</math> and <math>\frac{\pi}{2}\sin x\in\left[0,\frac{\pi}{2}\right],</math> it follows that <math>x\in\left[0,\frac{\pi}{2}\right]</math> and <math>\frac{\pi}{2}-\frac{\pi}{2}\sin x\in\left[0,\frac{\pi}{2}\right].</math> |
| + | |||
| + | We can apply the arcsine function to both sides, then rearrange and simplify: | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\frac{\pi}{2}\cos x &= \frac{\pi}{2}-\frac{\pi}{2}\sin x \\ | \frac{\pi}{2}\cos x &= \frac{\pi}{2}-\frac{\pi}{2}\sin x \\ | ||
| Line 162: | Line 162: | ||
~ pi_is_3.14 | ~ pi_is_3.14 | ||
| + | |||
| + | ==Video Solution (Quick and Easy)== | ||
| + | https://youtu.be/6AWb9cqFblU | ||
| + | |||
| + | ~Education, the Study of Everything | ||
==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}} | ||
Latest revision as of 19:10, 5 November 2022
Contents
Problem
How many solutions does the equation 
have in the closed interval ![$[0,\pi]$](http://latex.artofproblemsolving.com/e/c/e/ece27ba559bd93cdf2416aa0f88fc8d7d85e5791.png)
?
Solution 1 (Inverse Trigonometric Functions)
The ranges of 
and 
are both ![$\left[-\frac{\pi}2, \frac{\pi}2 \right],$](http://latex.artofproblemsolving.com/a/9/1/a9170cca00e4a24c4e6d2a174dc3152583601428.png)
which is included in the range of 
so we can use it with no issues.
This only happens at 
on the interval ![$[0,\pi],$](http://latex.artofproblemsolving.com/8/a/1/8a18c184a0b45c0a3996945b662b19e0f50328ad.png)
because one of 
and 
must be 
and the other 
Therefore, the answer is 
~Tucker
Solution 2 (Cofunction Identity)
By the Cofunction Identity 
we rewrite the given equation: ![\[\sin \left(\frac{\pi}2 \cos x\right) = \sin \left(\frac{\pi}2 - \frac{\pi}2 \sin x\right).\]](http://latex.artofproblemsolving.com/0/6/e/06e3540d357c76e8d482f1816252689b9d0e734f.png)
Recall that if 
then 
or 
for some integer 
Therefore, we have two cases:
for some integer 
We rearrange and simplify:
![\[\sin x + \cos x = 1 + 4n.\]](//latex.artofproblemsolving.com/5/d/7/5d77e5720cff52ea5b14fa3c610c4e3e27b05dc4.png)
By rough constraints, we know that 
from which 
The only possibility is 
so
for some integer 
We get
for this case. Note that 
is an extraneous solution by squaring 
for some integer Similar to Case 1, we conclude that
so ![\[\cos x - \sin x = 1.\]](//latex.artofproblemsolving.com/6/9/3/693d045cac17bba7e6edad4d9de233911c010433.png)
We get 
for this case.
for some integer 
![\[\sin x + \cos x = 1 + 4n.\]](http://latex.artofproblemsolving.com/5/d/7/5d77e5720cff52ea5b14fa3c610c4e3e27b05dc4.png)
By rough constraints, we know that 
from which 
The only possibility is 
so
for some integer 
for this case. Note that 
is an extraneous solution by squaring 
for some integer ![\[\cos x - \sin x = 1.\]](http://latex.artofproblemsolving.com/6/9/3/693d045cac17bba7e6edad4d9de233911c010433.png)
We get 
for this case.
Together, we obtain 
solutions: 
~MRENTHUSIASM
Solution 3 (Graphs and Analyses)
This problem is equivalent to counting the intersections of the graphs of 
and 
in the closed interval ![$[0,\pi].$](http://latex.artofproblemsolving.com/b/4/c/b4ccb3ab0522405552f4590e118eccf217cce9c4.png)
We construct a table of values, as shown below:
![\[\begin{array}{c|ccc} & & & \\ [-2ex] & \boldsymbol{x=0} & \boldsymbol{x=\frac{\pi}{2}} & \boldsymbol{x=\pi} \\ [1.5ex] \hline & & & \\ [-1ex] \boldsymbol{\cos x} & 1 & 0 & -1 \\ [1.5ex] \boldsymbol{\frac{\pi}{2}\cos x} & \frac{\pi}{2} & 0 & -\frac{\pi}{2} \\ [1.5ex] \boldsymbol{\sin\left(\frac{\pi}{2}\cos x\right)} & 1 & 0 & -1 \\ [1.5ex] \hline & & & \\ [-1ex] \boldsymbol{\sin x} & 0 & 1 & 0 \\ [1.5ex] \boldsymbol{\frac{\pi}{2}\sin x} & 0 & \frac{\pi}{2} & 0 \\ [1.5ex] \boldsymbol{\cos\left(\frac{\pi}{2}\sin x\right)} & 1 & 0 & 1 \\ [1ex] \end{array}\]](http://latex.artofproblemsolving.com/b/c/d/bcd742d378458e46b475f233696e281083331b63.png)
For ![$x\in[0,\pi],$](http://latex.artofproblemsolving.com/1/9/2/192702db67df074a645d5ad546b648098199dfef.png)
note that:
![$\frac{\pi}{2}\cos x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right],$](//latex.artofproblemsolving.com/a/5/f/a5fa15ed83f755f18132063b8f690810b0f0bb62.png)
so ![$\sin\left(\frac{\pi}{2}\cos x\right)\in[-1,1].$](//latex.artofproblemsolving.com/b/3/6/b367dc6c6b993f327240033cf658d12f29951b78.png)
![$\frac{\pi}{2}\cos x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right],$](http://latex.artofproblemsolving.com/a/5/f/a5fa15ed83f755f18132063b8f690810b0f0bb62.png)
so ![$\sin\left(\frac{\pi}{2}\cos x\right)\in[-1,1].$](http://latex.artofproblemsolving.com/b/3/6/b367dc6c6b993f327240033cf658d12f29951b78.png)
![$\frac{\pi}{2}\sin x\in\left[0,\frac{\pi}{2}\right],$](//latex.artofproblemsolving.com/b/4/a/b4a2bdb76ba5f3ea6721f944da15f1554f6fc1ba.png)
so ![$\cos\left(\frac{\pi}{2}\sin x\right)\in[0,1].$](//latex.artofproblemsolving.com/9/c/7/9c7d2a68cf4efebf6cd0fd0d71257fe1d475100b.png)
![$\cos\left(\frac{\pi}{2}\sin x\right)\in[0,1].$](http://latex.artofproblemsolving.com/9/c/7/9c7d2a68cf4efebf6cd0fd0d71257fe1d475100b.png)
For the graphs to intersect, we need ![$\sin\left(\frac{\pi}{2}\cos x\right)\in[0,1].$](http://latex.artofproblemsolving.com/4/1/8/418052bfbadf2ff056f256e0574a002ad15809d0.png)
This occurs when ![$\frac{\pi}{2}\cos x\in\left[0,\frac{\pi}{2}\right].$](http://latex.artofproblemsolving.com/f/8/f/f8ff205fc6994a190cc833ceff7e81a464f8298c.png)
By the Cofunction Identity we rewrite the given equation:
![\[\sin\left(\frac{\pi}{2}\cos x\right) = \sin\left(\frac{\pi}{2}-\frac{\pi}{2}\sin x\right).\]](http://latex.artofproblemsolving.com/e/e/6/ee6ed0735ea791a9591bb2ee0eda6a1e50923cd0.png)
Since ![$\frac{\pi}{2}\cos x\in\left[0,\frac{\pi}{2}\right]$](http://latex.artofproblemsolving.com/0/b/5/0b59b91069bf74a053ed6d23c2f2366575e64d00.png)
and ![$\frac{\pi}{2}\sin x\in\left[0,\frac{\pi}{2}\right],$](http://latex.artofproblemsolving.com/b/4/a/b4a2bdb76ba5f3ea6721f944da15f1554f6fc1ba.png)
it follows that ![$x\in\left[0,\frac{\pi}{2}\right]$](http://latex.artofproblemsolving.com/a/d/6/ad6d7672d2ef0d27254045d666d8a9cd52e94f59.png)
and ![$\frac{\pi}{2}-\frac{\pi}{2}\sin x\in\left[0,\frac{\pi}{2}\right].$](http://latex.artofproblemsolving.com/a/3/4/a34b74443d671784a6bd431e59c2453b5e0e1b72.png)
We can apply the arcsine function to both sides, then rearrange and simplify:
From Case 1 in Solution 2, we conclude that 
and 
are the only points of intersection, as shown below:
![[asy] /* Made by MRENTHUSIASM */ size(600,200); real f(real x) { return sin(pi/2*cos(x)); } real g(real x) { return cos(pi/2*sin(x)); } draw(graph(f,0,pi),red,"$y=\sin\left(\frac{\pi}{2}\cos x\right)$"); draw(graph(g,0,pi),blue,"$y=\cos\left(\frac{\pi}{2}\sin x\right)$"); real xMin = 0; real xMax = 5/4*pi; real yMin = -2; real yMax = 2; //Draws the horizontal gridlines void horizontalLines() { for (real i = yMin+1; i < yMax; ++i) { draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); } } //Draws the vertical gridlines void verticalLines() { for (real i = xMin+pi/2; i < xMax; i+=pi/2) { draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); } } //Draws the horizontal ticks void horizontalTicks() { for (real i = yMin+1; i < yMax; ++i) { draw((-1/8,i)--(1/8,i), black+linewidth(1)); } } //Draws the vertical ticks void verticalTicks() { for (real i = xMin+pi/2; i < xMax; i+=pi/2) { draw((i,-1/8)--(i,1/8), black+linewidth(1)); } } horizontalLines(); verticalLines(); horizontalTicks(); verticalTicks(); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2)); pair A[]; A[0] = (pi/2,0); A[1] = (pi,0); A[2] = (0,1); A[3] = (0,0); A[4] = (0,-1); label("$\frac{\pi}{2}$",A[0],(0,-2.5)); label("$\pi$",A[1],(0,-2.5)); label("$1$",A[2],(-2.5,0)); label("$0$",A[3],(-2.5,0)); label("$-1$",A[4],(-2.5,0)); dot((0,1),linewidth(5)); dot((pi/2,0),linewidth(5)); add(legend(),point(E),40E,UnFill); [/asy]](http://latex.artofproblemsolving.com/a/5/b/a5bb3fda3e60884de1b737e849a42cf5879ecf05.png)
Therefore, the answer is
~MRENTHUSIASM (credit given to TheAMCHub)
Video Solution by OmegaLearn (Using Triangle Inequality & Trigonometry)
~ pi_is_3.14
Video Solution (Quick and Easy)
~Education, the Study of Everything
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
