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Sines, Cosines, and Powers of 2... Oh My!

Source: 2013 AIME I Problem 14

March 15, 2013
trigonometryquadraticsrationumber theoryrelatively primetrig identitiesgeometric series

Problem Statement

For π≤θ<2π\pi\leq\theta<2\pi, let
P=12cos⁡θ−14sin⁡2θ−18cos⁡3θ+116sin⁡4θ+132cos⁡5θ−164sin⁡6θ−1128cos⁡7θ+… P=\dfrac12\cos\theta-\dfrac14\sin2\theta-\dfrac18\cos3\theta+\dfrac1{16}\sin4\theta+\dfrac1{32}\cos5\theta-\dfrac1{64}\sin6\theta-\dfrac1{128}\cos7\theta+\ldots and Q=1−12sin⁡θ−14cos⁡2θ+18sin⁡3θ+116cos⁡4θ−132sin⁡5θ−164cos⁡6θ+1128sin⁡7θ+… Q=1-\dfrac12\sin\theta-\dfrac14\cos2\theta+\dfrac1{8}\sin3\theta+\dfrac1{16}\cos4\theta-\dfrac1{32}\sin5\theta-\dfrac1{64}\cos6\theta+\dfrac1{128}\sin7\theta +\ldots so that PQ=227\tfrac PQ = \tfrac{2\sqrt2}7. Then sin⁡θ=−mn\sin\theta = -\tfrac mn where mm and nn are relatively prime positive integers. Find m+nm+n.
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