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

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==Solution 3==
 
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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}}

Revision as of 02:31, 18 April 2018

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 $x, y \in [0, \pi]$. 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 $\boxed{\textbf{(E) } 0\le y\le \pi}$.

See Also

2018 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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

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