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Difference between revisions of "2002 AMC 12P Problems/Problem 10"

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== Solution ==
 
== Solution ==
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Divide by 2 on both sides to get <cmath>3f_{4}(x)-2f_{6}(x)=f_{2}(x)?</cmath>
+
Divide by 2 on both sides to get <cmath>3f_{4}(x)-2f_{6}(x)=f_{2}(x)</cmath>
 
Substituting the definitions of <math>f_{2}(x)</math>, <math>f_{4}(x)</math>, and <math>f_{6}(x)</math>, we may rewrite the expression as
 
Substituting the definitions of <math>f_{2}(x)</math>, <math>f_{4}(x)</math>, and <math>f_{6}(x)</math>, we may rewrite the expression as
 
<cmath>3(\text{sin}^4 x + \text{cos}^4 x) - 2(\text{sin}^6 x + \text{cos}^6 x) = 1</cmath>
 
<cmath>3(\text{sin}^4 x + \text{cos}^4 x) - 2(\text{sin}^6 x + \text{cos}^6 x) = 1</cmath>

Revision as of 12:54, 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 $3 - 6\text{sin}^2 x \text{cos}^2 x$

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

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