Difference between revisions of "2002 AMC 12P Problems/Problem 10"
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We now simplify each term separately using some algebraic manipulation and the Pythagorean identity. | We now simplify each term separately using some algebraic manipulation and the Pythagorean identity. | ||
We can rewrite <cmath>3(\text{sin}^4 x + \text{cos}^4 x)</cmath> as | We can rewrite <cmath>3(\text{sin}^4 x + \text{cos}^4 x)</cmath> as | ||
| − | <cmath>3(\text{sin}^2 x + \text{cos}^2 x)^2 - 6\text{sin} x | + | <cmath>3(\text{sin}^2 x + \text{cos}^2 x)^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 12:52, 10 March 2024
Problem
Let 
For how many 
in ![$[0,\pi]$](http://latex.artofproblemsolving.com/e/c/e/ece27ba559bd93cdf2416aa0f88fc8d7d85e5791.png)
is it true that
![\[6f_{4}(x)-4f_{6}(x)=2f_{2}(x)?\]](http://latex.artofproblemsolving.com/c/6/7/c67025172dfa980ef0254c8fcf756bb8b99952d4.png)
Solution
Divide by 2 on both sides to get ![\[3f_{4}(x)-2f_{6}(x)=f_{2}(x)?\]](http://latex.artofproblemsolving.com/f/b/1/fb1570dbc82659dffc949a1c44bff680cf6b5325.png)
Substituting the definitions of 
, 
, and 
, we may rewrite the expression as
![\[3(\text{sin}^4 x + \text{cos}^4 x) - 2(\text{sin}^6 x + \text{cos}^6 x) = 1\]](http://latex.artofproblemsolving.com/6/1/0/610d996271e12623e5740b1a9fb40b874c3d9c7f.png)
We now simplify each term separately using some algebraic manipulation and the Pythagorean identity.
We can rewrite ![\[3(\text{sin}^4 x + \text{cos}^4 x)\]](http://latex.artofproblemsolving.com/3/d/4/3d4e8c9d22a15dd455e08dabdef020caa85934ad.png)
as
![\[3(\text{sin}^2 x + \text{cos}^2 x)^2 - 6\text{sin}^2 x \text{cos}^2 x\]](http://latex.artofproblemsolving.com/d/5/e/d5eea9fb348d78022f86b193cfe3a8a1f44441ab.png)
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. 
