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 10"

(→‎Solution)
(→‎Solution)
Line 28: Line 28:
 
As for <math>2(\text{sin}^6 x + \text{cos}^6 x)</math>, we may factor it as  
 
As for <math>2(\text{sin}^6 x + \text{cos}^6 x)</math>, we may factor it as  
 
<math>2(\text{sin}^2 x + \text{cos}^2 x)(\text{sin}^4 x + \text{cos}^4 x - \text{sin}^2 x \text{cos}^2 x)</math>, which can be rewritten as
 
<math>2(\text{sin}^2 x + \text{cos}^2 x)(\text{sin}^4 x + \text{cos}^4 x - \text{sin}^2 x \text{cos}^2 x)</math>, which can be rewritten as
−
<cmath>2(\text{sin}^4 x + \text{cos}^4 x - \text{sin}^2 x \text{cos}^2 x)</cmath>
+
<cmath>2(\text{sin}^4 x + \text{cos}^4 x - \text{sin}^2 x \text{cos}^2 x)</cmath> and then as
−
<cmath>= 2(\text{sin}^2 x + \text{cos}^2)^2) - 6\text{sin}^2 x \text{cos}^2 x</cmath>
+
<cmath>2(\text{sin}^2 x + \text{cos}^2)^2) - 6\text{sin}^2 x \text{cos}^2 x</cmath>, which is equivalent to
 +
<cmath>2 - 6\text{sin}^2 x \text{cos}^2 x</cmath>
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2002|ab=P|num-b=9|num-a=11}}
 
{{AMC12 box|year=2002|ab=P|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:17, 10 March 2024

Problem

Let $f_n (x) = \text{sin}^n x + \text{cos}^n x.$ For how many $x$ in $[0,\pi]$ is it true that

\[6f_{4}(x)-4f_{6}(x)=2f_{2}(x)?\]

$\text{(A) }2 \qquad \text{(B) }4 \qquad \text{(C) }6 \qquad \text{(D) }8 \qquad \text{(E) more than }8$

Solution

Divide by 2 on both sides to get \[3f_{4}(x)-2f_{6}(x)=f_{2}(x)\] Substituting the definitions of $f_{2}(x)$, $f_{4}(x)$, and $f_{6}(x)$, we may rewrite the expression as \[3(\text{sin}^4 x + \text{cos}^4 x) - 2(\text{sin}^6 x + \text{cos}^6 x) = 1\] We now simplify each term separately using some algebraic manipulation and the Pythagorean identity.

We can rewrite $3(\text{sin}^4 x + \text{cos}^4 x)$ as $3(\text{sin}^2 x + \text{cos}^2 x)^2 - 6\text{sin}^2 x \text{cos}^2 x$, which is equivalent to $3 - 6\text{sin}^2 x \text{cos}^2 x$.

As for $2(\text{sin}^6 x + \text{cos}^6 x)$, we may factor it as $2(\text{sin}^2 x + \text{cos}^2 x)(\text{sin}^4 x + \text{cos}^4 x - \text{sin}^2 x \text{cos}^2 x)$, which can be rewritten as \[2(\text{sin}^4 x + \text{cos}^4 x - \text{sin}^2 x \text{cos}^2 x)\] and then as \[2(\text{sin}^2 x + \text{cos}^2)^2) - 6\text{sin}^2 x \text{cos}^2 x\], which is equivalent to \[2 - 6\text{sin}^2 x \text{cos}^2 x\]

See also

2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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_10&oldid=216693"