2002 AMC 12P Problems/Problem 10

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 $2 - 6\text{sin}^2 x \text{cos}^2 x$ .

Putting everything together, we have $(3 - 6\text{sin}^2 x \text{cos}^2 x) - (2 - 6\text{sin}^2 x \text{cos}^2 x) = 1$ or $1 = 1$ . Therefore, the given equation $3f_{4}(x)-2f_{6}(x)=f_{2}(x)$ is true for all real $x$ , meaning that there are more than 8 values of $x$ that satisfy the given equation and so the answer is.

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.

Page

Toolbox

Search

2002 AMC 12P Problems/Problem 10

Problem

Solution

See also