# Difference between revisions of "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 $\boxed{\textbf{(E) } 0\le y\le \pi}$.