2018 AMC 12A Problems/Problem 9

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 $\boxed{\textbf{(E) } 0\le y\le \pi}$ .

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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2018 AMC 12A Problems/Problem 9

Contents

Problem

Solution 1

Solution 2

Solution 3

See Also