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Algebra Difficulty 4.9 AIME Find the answer

Given that x+siny=2008x+\sin y=2008 and x+2008cosy=2007x+2008 \cos y=2007, where 0yπ/20 \leq y \leq \pi / 2, find the value of x+yx+y.

A number or a short expression. Spacing and $ signs are ignored.

Solution

Subtracting the two equations gives siny2008cosy=1\sin y-2008 \cos y=1. But since 0yπ/20 \leq y \leq \pi / 2, the maximum of siny\sin y is 1 and the minimum of cosy\cos y is 0 , so we must have siny=1\sin y=1, so y=π/2y=\pi / 2 and x+y=2007+π2x+y=2007+\frac{\pi}{2}.

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