Difference between revisions of "2002 AMC 12P Problems/Problem 17"

Revision as of 02:03, 31 December 2023

Problem

Let $f(x) = \sqrt{\sin^4{x} + 4 \cos^2{x}} - \sqrt{\cos^4{x} + 4 \sin^2{x}}.$ An equivalent form of $f(x)$ is

$\text{(A) }1-\sqrt{2}\sin{x} \qquad \text{(B) }-1+\sqrt{2}\cos{x} \qquad \text{(C) }\cos{\frac{x}{2}} - \sin{\frac{x}{2}} \qquad \text{(D) }\cos{x} - \sin{x} \qquad \text{(E) }\cos{2x}$

Solution

Solution 2 (Cheese)

We don't actually have to solve the question. Just let $x$ equal some easy value to calculate $\cos {x}, \cos {2x},$ and $\cos {\frac{x}{2}}.$ For this solution, let $x=60^\circ.$ This means that the expression in the problem will give

2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 16: / Line 16: @@
 [[2002 AMC 12P Problems/Problem 17|Solution]]
-== Solution ==
+== Solution 2 (Cheese)==
-If <math>\log_{b} 729 = n</math>, then <math>b^n = 729</math>. Since <math>729 = 3^6</math>, <math>b</math> must be <math>3</math> to some [[factor]] of 6. Thus, there are four (3, 9, 27, 729) possible values of <math>b \Longrightarrow \boxed{\mathrm{E}}</math>.
+We don't actually have to solve the question. Just let <math>x</math> equal some easy value to calculate <math>\cos {x}, \cos {2x},</math> and <math>\cos {\frac{x}{2}}.</math> For this solution, let <math>x=60^\circ.</math> This means that the expression in the problem will give
 == See also ==
 {{AMC12 box|year=2002|ab=P|num-b=16|num-a=18}}
 {{MAA Notice}}

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Difference between revisions of "2002 AMC 12P Problems/Problem 17"

Revision as of 02:03, 31 December 2023

Problem

Solution 2 (Cheese)

See also