Why don't you simplify those fractions?

Geometry Level 2

k=1145[sin(2πk8)i cos(2πk14)]=1aicos(πb). \displaystyle \sum_{k=1}^{145} \left [ \sin \left ( \frac {2\pi k}{8} \right ) - i \ \cos \left ( \frac {2\pi k}{14} \right ) \right ] = \frac {1}{\sqrt{a}} - i \cos \left ( \frac {\pi}{b} \right ). Find the sum of the positive integers aa and b.b.

Note: i=1.i = \sqrt{-1}.

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