Difference between revisions of "2018 AMC 12A Problems/Problem 9"
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== Solution 1 == | == Solution 1 == | ||
− | On the interval <math>[0, \pi]</math> sine is nonnegative; thus <math>\sin(x + y) = \sin x \cos y + \sin y \cos x \le \sin x + \sin y</math> for all <math>x, y \in [0, \pi]</math>. The answer is <math>\boxed{\textbf{(E) } 0\le y\le \pi}</math>. (CantonMathGuy) | + | On the interval <math>[0, \pi]</math> sine is nonnegative; thus <math>\sin(x + y) = \sin x \cos y + \sin y \cos x \le \sin x + \sin y</math> for all <math>x, y \in [0, \pi]</math> and equality only occurs when <math>\cos x = \cos y = 1</math>, which is cosine's maximum value. The answer is <math>\boxed{\textbf{(E) } 0\le y\le \pi}</math>. (CantonMathGuy) |
==Solution 2== | ==Solution 2== | ||
Expanding, <cmath>\cos y \sin x + \cos x \sin y \le \sin x + \sin y</cmath> Let <math>\sin x =a \ge 0</math>, <math>\sin y = b \ge 0</math>. We have that <cmath>(\cos y)a+(\cos x)b \le a+b</cmath> Comparing coefficients of <math>a</math> and <math>b</math> gives a clear solution: both <math>\cos y</math> and <math>\cos x</math> are less than or equal to one, so the coefficients of <math>a</math> and <math>b</math> on the left are less than on the right. Since <math>a, b \ge 0</math>, that means that this equality is always satisfied over this interval, or <math>\boxed{\textbf{(E) } 0\le y\le \pi}</math>. | Expanding, <cmath>\cos y \sin x + \cos x \sin y \le \sin x + \sin y</cmath> Let <math>\sin x =a \ge 0</math>, <math>\sin y = b \ge 0</math>. We have that <cmath>(\cos y)a+(\cos x)b \le a+b</cmath> Comparing coefficients of <math>a</math> and <math>b</math> gives a clear solution: both <math>\cos y</math> and <math>\cos x</math> are less than or equal to one, so the coefficients of <math>a</math> and <math>b</math> on the left are less than on the right. Since <math>a, b \ge 0</math>, that means that this equality is always satisfied over this interval, or <math>\boxed{\textbf{(E) } 0\le y\le \pi}</math>. | ||
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+ | ==Solution 3== | ||
+ | If we plug in <math>\pi</math>, we can see that <math>\sin(x+\pi) \le \sin(x)</math>. Note that since <math>\sin(x)</math> is always nonnegative, <math>\sin(x+\pi)</math> is always nonpositive. So, the inequality holds true when <math>y=\pi</math>. The only interval that contains <math>\pi</math> in the answer choices is <math>\boxed{\textbf{(E) } 0\le y\le \pi}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2018|ab=A|num-b=8|num-a=10}} | {{AMC12 box|year=2018|ab=A|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:20, 21 January 2020
Problem
Which of the following describes the largest subset of values of within the closed interval for which for every between and , inclusive?
Solution 1
On the interval sine is nonnegative; thus for all and equality only occurs when , which is cosine's maximum value. The answer is . (CantonMathGuy)
Solution 2
Expanding, Let , . We have that Comparing coefficients of and gives a clear solution: both and are less than or equal to one, so the coefficients of and on the left are less than on the right. Since , that means that this equality is always satisfied over this interval, or .
Solution 3
If we plug in , we can see that . Note that since is always nonnegative, is always nonpositive. So, the inequality holds true when . The only interval that contains in the answer choices is .
See Also
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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All AMC 12 Problems and Solutions |
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