Difference between revisions of "2002 AMC 12P Problems/Problem 10"
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We can rewrite <math>3(\text{sin}^4 x + \text{cos}^4 x)</math> as | We can rewrite <math>3(\text{sin}^4 x + \text{cos}^4 x)</math> as | ||
<math>3(\text{sin}^2 x + \text{cos}^2 x)^2 - 6\text{sin}^2 x \text{cos}^2 x</math>, which is equivalent to | <math>3(\text{sin}^2 x + \text{cos}^2 x)^2 - 6\text{sin}^2 x \text{cos}^2 x</math>, which is equivalent to | ||
− | <math>3 - 6\text{sin}^2 x \text{cos}^2 x</math> | + | <math>3 - 6\text{sin}^2 x \text{cos}^2 x</math>. |
+ | As for <math>2(\text{sin}^6 x + \text{cos}^6 x)</math>, we may factor it as | ||
+ | <cmath>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)</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 12:10, 10 March 2024
Problem
Let For how many
in
is it true that
Solution
Divide by 2 on both sides to get
Substituting the definitions of
,
, and
, we may rewrite the expression as
We now simplify each term separately using some algebraic manipulation and the Pythagorean identity.
We can rewrite
as
, which is equivalent to
.
As for
, we may factor it as
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.