Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2009 AMC 12B Problems/Problem 24"

(→‎Better Solution)
m (→‎Better Solution)
Line 37: Line 37:
 
== Better Solution ==
 
== Better Solution ==
 
Like the previous solution, assume the inverse trig function properties and ranges and simplify the problem to <math>\sin^{ - 1}(\sin 6x) = x</math>.
 
Like the previous solution, assume the inverse trig function properties and ranges and simplify the problem to <math>\sin^{ - 1}(\sin 6x) = x</math>.
−
Since <math>\sin^{-1}</math> is between -pi/2 and pi/2, x is between that as well. Since x is between 0 and pi, combining the two gives x is between 0 and pi/2. Now, remove the inverse sine by taking the sin of both sides. Then, you get sin x = sin 6x.
+
Since <math>\sin^{-1}</math> is between <math>-\pi/2</math> and <math>\pi/2</math>, x is between that as well. Since x is between 0 and pi, combining the two gives x is between 0 and pi/2. Now, remove the inverse sine by taking the sin of both sides. Then, you get sin x = sin 6x.
  
 
From this, either x = 6x +2pik, or x+6x = pi +2pik. Solving, and remembering that 0<=x<=pi/2, we get 4 solutions.
 
From this, either x = 6x +2pik, or x+6x = pi +2pik. Solving, and remembering that 0<=x<=pi/2, we get 4 solutions.

Revision as of 01:05, 4 February 2019

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 $(2\pi/6,3\pi/6)$.

Better Solution

Like the previous solution, assume the inverse trig function properties and ranges and simplify the problem to $\sin^{ - 1}(\sin 6x) = x$. Since $\sin^{-1}$ is between $-\pi/2$ and $\pi/2$, x is between that as well. Since x is between 0 and pi, combining the two gives x is between 0 and pi/2. Now, remove the inverse sine by taking the sin of both sides. Then, you get sin x = sin 6x.

From this, either x = 6x +2pik, or x+6x = pi +2pik. Solving, and remembering that 0<=x<=pi/2, we get 4 solutions.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_24&oldid=101220"