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
0≤y≤π40 \le y \le \tfrac{\pi}{4}
0≤y≤π20 \le y \le \tfrac{\pi}{2}
0≤y≤3π40 \le y \le \tfrac{3\pi}{4}
0≤y≤π0 \le y \le \pi
Answer: E
Concepts:trigonometric identityinequality
Small hint:
Expand sin⁡(x+y)\sin(x+y) =sin⁡xcos⁡y+cos⁡xsin⁡y= \sin x \cos y + \cos x \sin y
Big hint:
For x,y∈[0,π]x, y \in [0, \pi] both sin⁡x,sin⁡y≥0\sin x, \sin y \ge 0 and cos⁡x,cos⁡y≤1;\cos x, \cos y \le 1; compare term by term
Solution:
For 0≤x≤π0 \le x \le \pi and 0≤y≤π0 \le y \le \pi we have sin⁡x≥0,\sin x \ge 0, sin⁡y≥0,\sin y \ge 0, cos⁡x≤1,\cos x \le 1, and cos⁡y≤1.\cos y \le 1. Hence sin⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡y≤sin⁡x+sin⁡y. \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 0≤y≤π.0 \le y \le \pi.
Thus, the correct answer is E.
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