Difference between revisions of "2002 AMC 12P Problems/Problem 17"
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Therefore, our answer is <math>\boxed{\textbf{(E) } \cos {2x}}</math>. | Therefore, our answer is <math>\boxed{\textbf{(E) } \cos {2x}}</math>. | ||
− | Comment: If you decide to cheese the problem, be very careful not to choose any <math>x</math> where <math>x</math> is a multiple of <math>\frac{\pi}{4}</math>. It turns out that all of them except for <math>\frac{7 \pi}{4} + 2 \ | + | Comment: If you decide to cheese the problem, be very careful not to choose any <math>x</math> where <math>x</math> is a multiple of <math>\frac{\pi}{4}</math>. It turns out that all of them except for <math>\frac{7\pi}{4} + 2\pi k</math> result in equality between the correct answer and one or more wrong answers, which one could quickly verify by setting two choices equal and solving for <math>x</math>. |
== See also == | == See also == | ||
{{AMC12 box|year=2002|ab=P|num-b=16|num-a=18}} | {{AMC12 box|year=2002|ab=P|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 19:16, 11 March 2024
Problem
Let An equivalent form of
is
Solution 1
By the Pythagorean identity we can rewrite the given expression as follows.
Expanding each bracket gives
The expressions under the square roots can be factored to get
Since and
for all real
, the expression must evaluate to
, which simplifies to
.
Solution 2 (Cheese)
We don't actually have to solve the question. Just let equal some easy value to calculate
and
For this solution, let
This means that the expression in the problem will give
Plugging in
for the rest of the expressions, we get
Therefore, our answer is .
Comment: If you decide to cheese the problem, be very careful not to choose any where
is a multiple of
. It turns out that all of them except for
result in equality between the correct answer and one or more wrong answers, which one could quickly verify by setting two choices equal and solving for
.
See also
2002 AMC 12P (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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.