2018 AMC 12A Problem 9

Attempt Problem 9 of the 2018 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 12A solutions, or check the answer key.

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9.

Which of the following describes the largest subset of values of yy within the closed interval [0,π][0, \pi] for which sin(x+y)sin(x)+sin(y) \sin(x + y) \le \sin(x) + \sin(y) for every xx between 00 and π,\pi, inclusive?

y=0y = 0

0yπ40 \le y \le \tfrac{\pi}{4}

0yπ20 \le y \le \tfrac{\pi}{2}

0y3π40 \le y \le \tfrac{3\pi}{4}

0yπ0 \le y \le \pi

Answer: E
Concepts:trigonometric identityinequality
Difficulty rating: 1500
Solution:

For 0xπ0 \le x \le \pi and 0yπ0 \le y \le \pi we have sinx0,\sin x \ge 0, siny0,\sin y \ge 0, cosx1,\cos x \le 1, and cosy1.\cos y \le 1. Hence sin(x+y)=sinxcosy+cosxsinysinx+siny. \begin{aligned} \sin(x+y) &= \sin x \cos y \\ &\quad {}+ \cos x \sin y \\ &\le \sin x + \sin y. \end{aligned} The inequality therefore holds for every yy with 0yπ.0 \le y \le \pi.

Thus, the correct answer is E.

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