If $P = \sin\frac{2\pi}{7} + \sin\frac{4\pi}{7} + \sin\frac{8\pi}{7}$ and $Q = \cos\frac{2\pi}{7} + \cos\frac{4\pi}{7} + \cos\frac{8\pi}{7}$, then the point (P,Q) lies on the circle of radius
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Sums of sines and cosines of angles like $2\pi k/n$ are often solved using the properties of the $n^{th}$ roots of unity. The fact that the sum of all $n^{th}$ roots of unity is zero is a very powerful tool.