Magical Trig Identity

Problem: Prove that \[\cos \frac{\pi}{7}-\cos \frac{2\pi}{7}+\cos \frac{3\pi}{7}=\frac{1}{2}\]

Solution: The RHSRHS is the same as cosπ7+cos3π7+cos5π7\cos \frac{\pi}{7}+\cos \frac{3\pi}{7}+\cos \frac{5\pi}{7}. Let S=cosπ7+cos3π7+cos5π7S=\cos \frac{\pi}{7}+\cos \frac{3\pi}{7}+\cos \frac{5\pi}{7}, then multiplying by sinπ7\sin \frac{\pi}{7}, we have sinπ7S=sinπ7cosπ7+sinπ7cos3π7+sinπ7cos5π7=sin2π72+sin4π7sin2π72+sin6π7sin4π72=sin6π72=sinπ72\sin \frac{\pi}{7}S=\sin \frac{\pi}{7} \cos \frac{\pi}{7}+\sin \frac{\pi}{7}\cos \frac{3\pi}{7}+\sin \frac{\pi}{7}\cos \frac{5\pi}{7}=\frac{\sin \frac{2\pi}{7}}{2}+\frac{\sin \frac{4\pi}{7}-\sin \frac{2\pi}{7}}{2}+\frac{\sin \frac{6\pi}{7}-\sin \frac{4\pi}{7}}{2}=\frac{\sin \frac{6 \pi}{7}}{2}=\frac{\sin \frac{\pi}{7}}{2} Thus, this implies that S=12S=\frac{1}{2}. And we are done!\blacksquare

#Geometry

Note by ChengYiin Ong
1 year, 1 month ago

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Comments

If we rewrite it as cos2π7+cos4π7+cos6π7=12\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2}, perhaps we can 'see' a general pattern: k=1ncos2kπ2n+1=12\sum_{k=1}^{n} \cos\frac{2k\pi}{2n+1}=-\frac{1}{2}

Chan Lye Lee - 1 year ago
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