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 "2002 AMC 12P Problems/Problem 17"

m (→‎Solution 2 (Cheese))
m (→‎Solution 2 (Cheese))
Line 41: Line 41:
 
Therefore, our answer is <math>\boxed{\textbf{(E) } \cos {2x}}</math>.
 
Therefore, our answer is <math>\boxed{\textbf{(E) } \cos {2x}}</math>.
  
−
Comment: If you decide to cheese the problem, be very careful not to choose any <math>x</math> where <math>x</math> is a multiple of <math>\frac{7\pi}{4}</math>. It turns out that all of them except for <math>\frac{7 \pi}{4} + 2 \pik</math> result in equality between the correct answer and one or more wrong answers.
+
Comment: If you decide to cheese the problem, be very careful not to choose any <math>x</math> where <math>x</math> is a multiple of <math>\frac{\pi}{4}</math>. It turns out that all of them except for <math>\frac{7\pi}{4} + 2\pi k</math> result in equality between the correct answer and one or more wrong answers.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2002|ab=P|num-b=16|num-a=18}}
 
{{AMC12 box|year=2002|ab=P|num-b=16|num-a=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:15, 11 March 2024

Problem

Let $f(x) = \sqrt{\sin^4{x} + 4 \cos^2{x}} - \sqrt{\cos^4{x} + 4 \sin^2{x}}.$ An equivalent form of $f(x)$ is

$\text{(A) }1-\sqrt{2}\sin{x} \qquad \text{(B) }-1+\sqrt{2}\cos{x} \qquad \text{(C) }\cos{\frac{x}{2}} - \sin{\frac{x}{2}} \qquad \text{(D) }\cos{x} - \sin{x} \qquad \text{(E) }\cos{2x}$

Solution

Solution 1

By the Pythagorean identity we can rewrite the given expression as follows. \[\sqrt{\sin^4{x} + 4 \cos^2{x}} - \sqrt{\cos^4{x} + 4 \sin^2{x}} = \sqrt{\sin^4{x} + 4(1 - \sin^2{x})} - \sqrt{\cos^4{x} + 4(1 - \cos^2{x})}\]

Expanding each bracket gives \[\sqrt{\sin^4{x} - 4\sin^2{x} + 4} - \sqrt{\cos^4{x} - 4\cos^2{x} + 4}\]

The expressions under the square roots can be factored to get \[\sqrt{(\sin^2{x} - 2)^2} - \sqrt{(\cos^2{x} - 2)^2}\]

Since $\sin^2{x} < 2$ and $\cos^2{x} < 2$ for all real $x$, the expression must evaluate to $(2 - \sin^2{x}) - (2 - \cos^2{x})$, which simplifies to $\cos^2{x} - \sin^2{x} = \boxed {\text{(E) }\cos{2x}}$.

Solution 2 (Cheese)

We don't actually have to solve the question. Just let $x$ equal some easy value to calculate $\cos {x}, \cos {2x}, \sin {x}, \sin {\frac{x}{2}},$ and $\cos {\frac{x}{2}}.$ For this solution, let $x=60^\circ.$ This means that the expression in the problem will give $\sqrt{\sin^4{60^\circ} + 4 \cos^2{60^\circ}} - \sqrt{\cos^4{60^\circ} + 4 \sin^2{60^\circ}}=\sqrt{(\frac{\sqrt{3}}{2})^4 + 4(\frac{1}{2})^2} - \sqrt{(\frac{1}{2})^4 + 4(\frac{\sqrt{3}}{2})^2}=\sqrt{\frac{9}{16} +1} - \sqrt{\frac{1}{16} + 3} = \frac{-1}{2}.$ Plugging in $x=60^\circ$ for the rest of the expressions, we get

$\text{(A) }1-\sqrt{2}\sin{60^\circ}=1-\frac{\sqrt{6}}{2} \neq \frac{-1}{2}.$

$\text{(B) }1+\sqrt{2}\cos{60^\circ}=1+\frac{\sqrt{2}}{2} \neq \frac{-1}{2}.$

$\text{(C) }\cos{\frac{60^\circ}{2}} - \sin{\frac{60^\circ}{2}}=\frac{\sqrt{3}}{2}-1 \neq \frac{-1}{2}.$

$\text{(D) }\cos{60^\circ} - \sin{60^\circ}=\frac{1-\sqrt{3}}{2} \neq \frac{-1}{2}.$

$\text{(E) }\cos {120^\circ}=\frac{-1}{2}.$

Therefore, our answer is $\boxed{\textbf{(E) } \cos {2x}}$.

Comment: If you decide to cheese the problem, be very careful not to choose any $x$ where $x$ is a multiple of $\frac{\pi}{4}$. It turns out that all of them except for $\frac{7\pi}{4} + 2\pi k$ result in equality between the correct answer and one or more wrong answers.

See also

2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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=2002_AMC_12P_Problems/Problem_17&oldid=216896"