Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.0 AIME, harder Find the answer

9. Real numbers $x, y$ satisfy $\left\{\begin{array}{l}x+\sin y=2008, \\ x+2008 \cos y=2007\end{array}\right.$ $\left(0 \leqslant y \leqslant \frac{\pi}{2}\right)$ . Then $x+y=$ $\qquad$

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

9. $2007+\frac{\pi}{2}$ .

Subtracting the two equations gives $\sin y=1+2008 \cos y$ . Given $0 \leqslant y \leqslant \frac{\pi}{2}$ , we know $1+2008 \cos y \geqslant 1$ . Therefore, it can only be that $\sin y=1, \cos y=0$ . Hence, $y=\frac{\pi}{2}$ . Consequently, $x=2007$ .

Therefore, $x+y=2007+\frac{\pi}{2}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.