Difference between revisions of "2021 AMC 12A Problems/Problem 19"
(→Solution: fixed problem with range of arcsin; only cos has the range) |
m (→Solution 4 (Risky Interpolations)) |
||
(67 intermediate revisions by 6 users not shown) | |||
Line 4: | Line 4: | ||
<math>\textbf{(A) }0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3\qquad \textbf{(E) }4</math> | <math>\textbf{(A) }0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3\qquad \textbf{(E) }4</math> | ||
− | ==Solution== | + | ==Solution 1 (Inverse Trigonometric Functions)== |
− | <math>\sin \left( \frac{\pi}2 \ | + | The ranges of <math>\frac{\pi}2 \sin x</math> and <math>\frac{\pi}2 \cos x</math> are both <math>\left[-\frac{\pi}2, \frac{\pi}2 \right],</math> which is included in the range of <math>\arcsin,</math> so we can use it with no issues. |
+ | <cmath>\begin{align*} | ||
+ | \frac{\pi}2 \cos x &= \arcsin \left( \cos \left( \frac{\pi}2 \sin x\right)\right) \\ | ||
+ | \frac{\pi}2 \cos x &= \arcsin \left( \sin \left( \frac{\pi}2 - \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. | ||
+ | \end{align*}</cmath> | ||
+ | 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> | ||
− | The interval | + | ~Tucker |
+ | |||
+ | ==Solution 2 (Cofunction Identity)== | ||
+ | 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> | ||
+ | Recall that if <math>\sin\theta=\sin\phi,</math> then <math>\theta=\phi+2n\pi</math> or <math>\theta=\pi-\phi+2n\pi</math> for some integer <math>n.</math> Therefore, we have two cases: | ||
+ | <ol style="margin-left: 1.5em;"> | ||
+ | <li><b><math>\boldsymbol{\frac{\pi}2 \cos x = \left(\frac{\pi}2 - \frac{\pi}2 \sin x\right) + 2n\pi}</math> for some integer <math>\boldsymbol{n}</math></b> <p> | ||
+ | We rearrange and simplify: <cmath>\sin x + \cos x = 1 + 4n.</cmath> | ||
+ | By rough constraints, we know that <math>-2 < \sin x + \cos x < 2,</math> from which <math>-2 < 1 - 4n < 2.</math> The only possibility is <math>n=0,</math> so | ||
+ | <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> <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> | ||
+ | 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> | ||
+ | </ol> | ||
+ | Together, we obtain <math>\boxed{\textbf{(C) }2}</math> solutions: <math>x=0,\frac{\pi}{2}.</math> | ||
+ | |||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | ==Solution 3 (Graphs and Analyses)== | ||
+ | This problem is equivalent to counting the intersections of the graphs of <math>y=\sin\left(\frac{\pi}{2}\cos x\right)</math> and <math>y=\cos\left(\frac{\pi}{2}\sin x\right)</math> in the closed interval <math>[0,\pi].</math> We construct a table of values, as shown below: | ||
+ | <cmath>\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}</cmath> | ||
+ | For <math>x\in[0,\pi],</math> note that: | ||
+ | |||
+ | * <math>\frac{\pi}{2}\cos x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right],</math> so <math>\sin\left(\frac{\pi}{2}\cos x\right)\in[-1,1].</math> | ||
+ | |||
+ | * <math>\frac{\pi}{2}\sin x\in\left[0,\frac{\pi}{2}\right],</math> so <math>\cos\left(\frac{\pi}{2}\sin x\right)\in[0,1].</math> | ||
− | <math>\frac{\pi}2 \ | + | For the graphs to intersect, we need <math>\sin\left(\frac{\pi}{2}\cos x\right)\in[0,1].</math> This occurs when <math>\frac{\pi}{2}\cos x\in\left[0,\frac{\pi}{2}\right].</math> |
− | <math>\frac{\pi}2 \sin x=\frac{\pi}2 - \frac{\pi}2 \cos x</math> | + | 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> | ||
+ | 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*} | ||
+ | \frac{\pi}{2}\cos x &= \frac{\pi}{2}-\frac{\pi}{2}\sin x \\ | ||
+ | \sin x + \cos x &= 1. | ||
+ | \end{align*}</cmath> | ||
+ | From Case 1 in Solution 2, we conclude that <math>(0,1)</math> and <math>\left(\frac{\pi}{2},0\right)</math> 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> | ||
+ | Therefore, the answer is <math>\boxed{\textbf{(C) }2}.</math> | ||
+ | |||
+ | ~MRENTHUSIASM (credit given to TheAMCHub) | ||
== Video Solution by OmegaLearn (Using Triangle Inequality & Trigonometry) == | == Video Solution by OmegaLearn (Using Triangle Inequality & Trigonometry) == | ||
Line 25: | 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 ?
Solution 1 (Inverse Trigonometric Functions)
The ranges of and are both which is included in the range of so we can use it with no issues. This only happens at on the interval 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: Recall that if then or for some integer Therefore, we have two cases:
- for some integer
We rearrange and simplify: 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 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 We construct a table of values, as shown below: For note that:
- so
- so
For the graphs to intersect, we need This occurs when
By the Cofunction Identity we rewrite the given equation: Since and it follows that and
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: 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.