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

See Also

2018 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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All AMC 12 Problems and Solutions

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