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Difference between revisions of "2020 AMC 12A Problems/Problem 9"
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==Solution==
 
==Solution==
 +
We count the intersections of the graphs of <math>y=\tan(2x)</math> and <math>y=\cos\left(\frac x2\right):</math>
 +
<ol style="margin-left: 1.5em;">
 +
  <li>The graph of <math>y=\tan(2x)</math> has a period of <math>\frac{\pi}{2},</math> asymptotes at <math>x=\frac{\pi}{4}+\frac{k\pi}{2},</math> and zeros at <math>x=\frac{k\pi}{2}</math> for some integer <math>k.</math> <p>
 +
On the interval <math>[0,2\pi],</math> the graph has five branches: <cmath>\biggl[0,\frac{\pi}{4}\biggr),\left(\frac{\pi}{4},\frac{3\pi}{4}\right),\left(\frac{3\pi}{4},\frac{5\pi}{4}\right),\left(\frac{5\pi}{4},\frac{7\pi}{4}\right),\left(\frac{7\pi}{4},2\pi\right].</cmath>
 +
Note that <math>\tan(2x)\in[0,\infty)</math> for the first branch, <math>\tan(2x)\in(-\infty,\infty)</math> for the three middle branches, and <math>\tan(2x)\in(-\infty,0]</math> for the last branch. Moreover, all branches are strictly increasing.
 +
</li><p>
 +
  <li>The graph of <math>y=\cos\left(\frac x2\right)</math> has a period of <math>4\pi</math> and zeros at <math>x=\pi+2k\pi</math> for some integer <math>k.</math> <p>
 +
On the interval <math>[0,2\pi],</math> note that <math>\cos\left(\frac x2\right)\in[-1,1].</math> Moreover, the graph is strictly decreasing.</li><p>
 +
</ol>
 +
The graphs of <math>y=\tan(2x)</math> and <math>y=\cos\left(\frac x2\right)</math> intersect once on each of the five branches of <math>y=\tan(2x),</math> as shown below:
 +
<asy>
 +
/* Made by MRENTHUSIASM */
 +
size(800,200);
  
−
Draw a graph of tan<math>(2x)</math> and cos<math>(\frac{x}{2})</math>
+
real f(real x) { return tan(2*x); }
 +
real g(real x) { return cos(x/2); }
  
−
tan<math>(2x)</math> has a period of <math>\frac{\pi}{2},</math> asymptotes at <math>x = \frac{\pi}{4}+\frac{k\pi}{2},</math> and zeroes at <math>\frac{k\pi}{2}</math>. It is positive from <math>(0,\frac{\pi}{4}) \cup (\frac{\pi}{2},\frac{3\pi}{4}) \cup (\pi,\frac{5\pi}{4}) \cup (\frac{7\pi}{4},2\pi)</math> and negative elsewhere.
+
draw(graph(f,0,atan(3)/2),red,"$y=\tan(2x)$");
 +
draw(graph(f,-atan(3)/2+pi/2,atan(3)/2+pi/2),red);
 +
draw(graph(f,-atan(3)/2+2*pi/2,atan(3)/2+2*pi/2),red);
 +
draw(graph(f,-atan(3)/2+3*pi/2,atan(3)/2+3*pi/2),red);
 +
draw(graph(f,-atan(3)/2+4*pi/2,2*pi),red);
 +
draw(graph(g,0,2pi),blue,"$y=\cos\left(\frac x2\right)$");
  
−
cos<math>(\frac{x}{2})</math> has a period of <math>4\pi</math> and zeroes at <math>\pi</math>. It is positive from <math>[0,\pi)</math> and negative elsewhere.
+
real xMin = 0;
 +
real xMax = 9/4*pi;
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real yMin = -3;
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real yMax = 3;
  
−
Drawing such a graph would get <math>\boxed{\textbf{E) }5}</math> ~lopkiloinm
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//Draws the horizontal gridlines
 +
void horizontalLines()
 +
{
 +
  for (real i = yMin+1; i < yMax; ++i)
 +
  {
 +
    draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4));
 +
  }
 +
}
  
−
==Solution (Algebraically)==
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//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));
 +
  }
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}
  
−
<math>\tan(2x)=\frac{\sin(2x)}{\cos(2x)}</math>. Applying double angle identities for both, we have
+
//Draws the horizontal ticks
 +
void horizontalTicks()
 +
{
 +
  for (real i = yMin+1; i < yMax; ++i)
 +
  {
 +
    draw((-1/8,i)--(1/8,i), black+linewidth(1));
 +
  }
 +
}
  
−
<cmath>\tan(2x)=\frac{\sin(2x)}{\cos(2x)}=\frac{2\sin x \cos x}{2\cos^{2}x-1}</cmath>
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//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));
 +
  }
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}
  
−
Applying half angle identities on the RHS, we have <math>\cos\frac{x}{2}=\pm\sqrt{\frac{\cos x +1}{2}}</math>.
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horizontalLines();
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verticalLines();
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horizontalTicks();
 +
verticalTicks();
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draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
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draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
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label("$x$",(xMax,0),(2,0));
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label("$y$",(0,yMax),(0,2));
  
−
Setting both sides equal and squaring,  
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pair A[], B[];
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A[0] = (2pi,0);
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A[1] = (0,2);
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A[2] = (0,0);
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A[3] = (0,-2);
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B[0] = intersectionpoints(graph(f,0,atan(3)/2),graph(g,0,2pi))[0];
 +
B[1] = intersectionpoints(graph(f,-atan(3)/2+pi/2,atan(3)/2+pi/2),graph(g,0,2pi))[0];
 +
B[2] = intersectionpoints(graph(f,-atan(3)/2+2*pi/2,atan(3)/2+2*pi/2),graph(g,0,2pi))[0];
 +
B[3] = intersectionpoints(graph(f,-atan(3)/2+3*pi/2,atan(3)/2+3*pi/2),graph(g,0,2pi))[0];
 +
B[4] = intersectionpoints(graph(f,-atan(3)/2+4*pi/2,atan(3)/2+4*pi/2),graph(g,0,2pi))[0];
  
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<cmath>\frac{2\sin x \cos x}{2\cos^{2}x-1}=\pm\sqrt{\frac{\cos x + 1}{2}}</cmath>
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label("$2\pi$",A[0],(0,-2.5));
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label("$2$",A[1],(-2.5,0));
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label("$0$",A[2],(-2.5,0));
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label("$-2$",A[3],(-2.5,0));
  
−
<cmath>\frac{4\sin^2 x \cos^2 x}{4\cos^{4}x-4\cos^2 x+1}=\frac{\cos x + 1}{2}</cmath>
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for (int i = 0; i < 5; ++i)
 +
{
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dot(B[i],black+linewidth(5));
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}
  
−
Since <math>\sin^2 x + \cos^2 x = 1</math>, we can substitute <math>\sin^2 x = 1-\cos^2 x</math> to convert the whole equation into cosine.
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add(legend(),point(E),60E,UnFill);
 +
</asy>
 +
Therefore, the answer is <math>\boxed{\textbf{(E)}\ 5}.</math>
  
−
<cmath>\frac{4(1-\cos^2 x) (\cos^2 x)}{4\cos^{4}x-4\cos^2 x+1}=\frac{\cos x + 1}{2}</cmath>
+
~MRENTHUSIASM ~lopkiloinm ~hi13 ~annabelle0913 ~codecow
  
−
Cross multiplying, we get
+
==Video Solution==
−
 
+
https://youtu.be/ZdVut0V1O4g  with tips on graphing trig functions fast ~MathProblemSolvingSkills.com
−
<cmath>8(1-\cos^2 x) (\cos^2 x)=(4\cos^{4}x-4\cos^2 x+1)(1+\cos x)</cmath>
 
−
 
 
−
<cmath>0=(4\cos^{4}x-4\cos^2 x+1)(1+\cos x)-8(1-\cos^2 x) (\cos^2 x)</cmath>
 
−
 
 
−
Without expanding anything, we can see that the first two polynomials will expand into a polynomial with degree <math>5</math> and the <math>8(1-\cos^2 x) (\cos^2 x)</math> term will expand into a polynomial with degree <math>4</math>. This means that overall, the polynomial will have degree <math>5</math>. From this, we can see that there are <math>\boxed{\textbf{E) }5}</math> solutions. ~quacker88
 
  
 
==See Also==
 
==See Also==
  
 
{{AMC12 box|year=2020|ab=A|num-b=8|num-a=10}}
 
{{AMC12 box|year=2020|ab=A|num-b=8|num-a=10}}
 +
 +
[[Category:Introductory Trigonometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 03:07, 8 July 2022

Problem

How many solutions does the equation $\tan(2x)=\cos(\tfrac{x}{2})$ have on the interval $[0,2\pi]?$

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution

We count the intersections of the graphs of $y=\tan(2x)$ and $y=\cos\left(\frac x2\right):$

  1. The graph of $y=\tan(2x)$ has a period of $\frac{\pi}{2},$ asymptotes at $x=\frac{\pi}{4}+\frac{k\pi}{2},$ and zeros at $x=\frac{k\pi}{2}$ for some integer $k.$

    On the interval $[0,2\pi],$ the graph has five branches: \[\biggl[0,\frac{\pi}{4}\biggr),\left(\frac{\pi}{4},\frac{3\pi}{4}\right),\left(\frac{3\pi}{4},\frac{5\pi}{4}\right),\left(\frac{5\pi}{4},\frac{7\pi}{4}\right),\left(\frac{7\pi}{4},2\pi\right].\] Note that $\tan(2x)\in[0,\infty)$ for the first branch, $\tan(2x)\in(-\infty,\infty)$ for the three middle branches, and $\tan(2x)\in(-\infty,0]$ for the last branch. Moreover, all branches are strictly increasing.

  2. The graph of $y=\cos\left(\frac x2\right)$ has a period of $4\pi$ and zeros at $x=\pi+2k\pi$ for some integer $k.$

    On the interval $[0,2\pi],$ note that $\cos\left(\frac x2\right)\in[-1,1].$ Moreover, the graph is strictly decreasing.

The graphs of $y=\tan(2x)$ and $y=\cos\left(\frac x2\right)$ intersect once on each of the five branches of $y=\tan(2x),$ as shown below: [asy] /* Made by MRENTHUSIASM */ size(800,200); real f(real x) { return tan(2*x); } real g(real x) { return cos(x/2); } draw(graph(f,0,atan(3)/2),red,"$y=\tan(2x)$"); draw(graph(f,-atan(3)/2+pi/2,atan(3)/2+pi/2),red); draw(graph(f,-atan(3)/2+2*pi/2,atan(3)/2+2*pi/2),red); draw(graph(f,-atan(3)/2+3*pi/2,atan(3)/2+3*pi/2),red); draw(graph(f,-atan(3)/2+4*pi/2,2*pi),red); draw(graph(g,0,2pi),blue,"$y=\cos\left(\frac x2\right)$"); real xMin = 0; real xMax = 9/4*pi; real yMin = -3; real yMax = 3; //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[], B[]; A[0] = (2pi,0); A[1] = (0,2); A[2] = (0,0); A[3] = (0,-2); B[0] = intersectionpoints(graph(f,0,atan(3)/2),graph(g,0,2pi))[0]; B[1] = intersectionpoints(graph(f,-atan(3)/2+pi/2,atan(3)/2+pi/2),graph(g,0,2pi))[0]; B[2] = intersectionpoints(graph(f,-atan(3)/2+2*pi/2,atan(3)/2+2*pi/2),graph(g,0,2pi))[0]; B[3] = intersectionpoints(graph(f,-atan(3)/2+3*pi/2,atan(3)/2+3*pi/2),graph(g,0,2pi))[0]; B[4] = intersectionpoints(graph(f,-atan(3)/2+4*pi/2,atan(3)/2+4*pi/2),graph(g,0,2pi))[0]; label("$2\pi$",A[0],(0,-2.5)); label("$2$",A[1],(-2.5,0)); label("$0$",A[2],(-2.5,0)); label("$-2$",A[3],(-2.5,0)); for (int i = 0; i < 5; ++i) { dot(B[i],black+linewidth(5)); } add(legend(),point(E),60E,UnFill); [/asy] Therefore, the answer is $\boxed{\textbf{(E)}\ 5}.$

~MRENTHUSIASM ~lopkiloinm ~hi13 ~annabelle0913 ~codecow

Video Solution

https://youtu.be/ZdVut0V1O4g with tips on graphing trig functions fast ~MathProblemSolvingSkills.com

See Also

2020 AMC 12A (Problems • Answer Key • Resources)
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Problem 8 		Followed by
Problem 10
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