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Difference between revisions of "2009 AMC 12B Problems/Problem 24"
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Summary: The function <math>f</math> has <math>\boxed{4}</math> roots on <math>[0,\pi]</math>: the first one is <math>0</math>, the second one is in <math>(0,\pi/6)</math>, and the last two are in <math>(2\pi/6,3\pi/6)</math>.
 
Summary: The function <math>f</math> has <math>\boxed{4}</math> roots on <math>[0,\pi]</math>: the first one is <math>0</math>, the second one is in <math>(0,\pi/6)</math>, and the last two are in <math>(2\pi/6,3\pi/6)</math>.
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==Alternate solution==
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Notice that the graph of the second function is just a straight line that goes up to <math>\pi</math> at <math>\pi</math> and starts at <math>0</math>. So the second equation is really just the line <math> y = x</math> on the interval. In addition, notice that the first function has a range of <math>[-\frac{\pi}{2}, \frac{pi}{2}]</math>, so we only have to look at the first half of the domain -- <math>\cos^{-1}(\cos x) > \sin^{-1}(\sin x)</math> for <math>\frac {\pi}{2} < x < \pi</math>. So we now find values of x such that <math>\sin ^-1 (\sin 6x) = x</math>. Applying sin to both sides, this becomes <math>\sin 6x = \sin x</math>. Thus, we find values on the domain <math>[0, \frac{\pi}{2}]</math> such that <math>6x</math> is a reference angle of <math>x</math>.
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First, of course there is 0. Then, when <math>x</math> is in first quadrant, <math>6x</math> can be in second quadrant, first quadrant (with a revolution), and similarly second quadrant. These are modeled by <math>6x = \pi - x</math>, <math>6x = 2\pi + x</math>, and <math>6x = 3\pi - x</math>. So we find answers <math>\{\frac {\pi}{7}, \frac{2\pi}{5}, \frac{3\pi}{7}\}</math>. So there are four solutions, the answer is <math>\boxed{C}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 12:13, 28 November 2013

Problem

For how many values of $x$ in $[0,\pi]$ is $\sin^{ - 1}(\sin 6x) = \cos^{ - 1}(\cos x)$? Note: The functions $\sin^{ - 1} = \arcsin$ and $\cos^{ - 1} = \arccos$ denote inverse trigonometric functions.

$\textbf{(A)}\ 3\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ 7$

Solution

First of all, we have to agree on the range of $\sin^{-1}$ and $\cos^{-1}$. This should have been a part of the problem statement -- but as it is missing, we will assume the most common definition: $\forall x: -\pi/2 \leq \sin^{-1}(x) \leq \pi/2$ and $\forall x: 0\leq \cos^{-1}(x) \leq \pi$.

Hence we get that $\forall x\in[0,\pi]: \cos^{ - 1}(\cos x) = x$, thus our equation simplifies to $\sin^{ - 1}(\sin 6x) = x$.

Consider the function $f(x) = \sin^{ - 1}(\sin 6x) - x$. We are looking for roots of $f$ on $[0,\pi]$.

By analyzing properties of $\sin$ and $\sin^{-1}$ (or by computing the derivative of $f$) one can discover the following properties of $f$:

  • $f(0)=0$.
  • $f$ is increasing and then decreasing on $[0,\pi/6]$.
  • $f$ is decreasing and then increasing on $[\pi/6,2\pi/6]$.
  • $f$ is increasing and then decreasing on $[2\pi/6,3\pi/6]$.

For $x=\pi/6$ we have $f(x) = \sin^{-1} (\sin \pi) - \pi/6 = -\pi/6 < 0$. Hence $f$ has exactly one root on $(0,\pi/6)$.

For $x=2\pi/6$ we have $f(x) = \sin^{-1} (\sin 2\pi) - 2\pi/6 = -2\pi/6 < 0$. Hence $f$ is negative on the entire interval $[\pi/6,2\pi/6]$.

Now note that $\forall t: \sin^{-1}(t) \leq \pi/2$. Hence for $x > 3\pi/6$ we have $f(x) < 0$, and we can easily check that $f(3\pi/6)<0$ as well.

Thus the only unknown part of $f$ is the interval $(2\pi/6,3\pi/6)$. On this interval, $f$ is negative in both endpoints, and we know that it is first increasing and then decreasing. Hence there can be zero, one, or two roots on this interval.

To prove that there are two roots, it is enough to find any $x$ from this interval such that $f(x)>0$.

A good guess is its midpoint, $x=5\pi/12$, where the function $\sin^{-1}(\sin 6x)$ has its local maximum. We can evaluate: $f(x) = \sin^{-1}(\sin 5\pi/2) - 5\pi/12 = \pi/2 - 5\pi/12 = \pi/12 > 0$.

Summary: The function $f$ has $\boxed{4}$ roots on $[0,\pi]$: the first one is $0$, the second one is in $(0,\pi/6)$, and the last two are in $(2\pi/6,3\pi/6)$.

Alternate solution

Notice that the graph of the second function is just a straight line that goes up to $\pi$ at $\pi$ and starts at $0$. So the second equation is really just the line $y = x$ on the interval. In addition, notice that the first function has a range of $[-\frac{\pi}{2}, \frac{pi}{2}]$, so we only have to look at the first half of the domain -- $\cos^{-1}(\cos x) > \sin^{-1}(\sin x)$ for $\frac {\pi}{2} < x < \pi$. So we now find values of x such that $\sin ^-1 (\sin 6x) = x$. Applying sin to both sides, this becomes $\sin 6x = \sin x$. Thus, we find values on the domain $[0, \frac{\pi}{2}]$ such that $6x$ is a reference angle of $x$.

First, of course there is 0. Then, when $x$ is in first quadrant, $6x$ can be in second quadrant, first quadrant (with a revolution), and similarly second quadrant. These are modeled by $6x = \pi - x$, $6x = 2\pi + x$, and $6x = 3\pi - x$. So we find answers $\{\frac {\pi}{7}, \frac{2\pi}{5}, \frac{3\pi}{7}\}$. So there are four solutions, the answer is $\boxed{C}$.

See Also

2009 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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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