Difference between revisions of "2018 AMC 12A Problems/Problem 9"

Revision as of 20:20, 21 January 2020

Problem

Which of the following describes the largest subset of values of $y$ within the closed interval $[0,\pi]$ for which $\sin(x+y)\leq \sin(x)+\sin(y)$ for every $x$ between $0$ and $\pi$ , inclusive? $\textbf{(A) } y=0 \qquad \textbf{(B) } 0\leq y\leq \frac{\pi}{4} \qquad \textbf{(C) } 0\leq y\leq \frac{\pi}{2} \qquad \textbf{(D) } 0\leq y\leq \frac{3\pi}{4} \qquad \textbf{(E) } 0\leq y\leq \pi$

Solution 1

On the interval $[0, \pi]$ sine is nonnegative; thus $\sin(x + y) = \sin x \cos y + \sin y \cos x \le \sin x + \sin y$ for all $x, y \in [0, \pi]$ and equality only occurs when $\cos x = \cos y = 1$ , which is cosine's maximum value. The answer is $\boxed{\textbf{(E) } 0\le y\le \pi}$ . (CantonMathGuy)

Solution 2

Expanding, $\cos y \sin x + \cos x \sin y \le \sin x + \sin y$ Let $\sin x =a \ge 0$ , $\sin y = b \ge 0$ . We have that $(\cos y)a+(\cos x)b \le a+b$ Comparing coefficients of $a$ and $b$ gives a clear solution: both $\cos y$ and $\cos x$ are less than or equal to one, so the coefficients of $a$ and $b$ on the left are less than on the right. Since $a, b \ge 0$ , that means that this equality is always satisfied over this interval, or $\boxed{\textbf{(E) } 0\le y\le \pi}$ .

Solution 3

If we plug in $\pi$ , we can see that $\sin(x+\pi) \le \sin(x)$ . Note that since $\sin(x)$ is always nonnegative, $\sin(x+\pi)$ is always nonpositive. So, the inequality holds true when $y=\pi$ . The only interval that contains $\pi$ in the answer choices is $\boxed{\textbf{(E) } 0\le y\le \pi}$ .

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

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

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 == 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==
 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>.
+==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==
 {{AMC12 box|year=2018|ab=A|num-b=8|num-a=10}}
 {{MAA Notice}}

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