some short and selected math problems of different levels in random order I try to keep the ans simple
Thursday, February 25, 2016
2016/019) If \theta = \frac{\pi}{2^{n+1}} then show that 2^n cos\theta \ cos 2 \theta \cos 2^2\theta \cdots cos 2^n \theta = 1
2^n cos\theta \cos 2 \theta \cos 2^2\theta \cdots cos 2^n \theta
= (2\cos\theta) (2\cos 2 \theta)(2 \cos 2^2\theta) \cdots (2cos 2^n \theta) = 1
= (\frac{\sin 2\theta}{\sin \theta}) (\frac{\sin 4\theta}{\sin 2\theta}) \cdots (\frac{\sin 2^n\theta}{\sin 2^{n-1}\theta}) = 1
=\frac{\sin 2^n \theta }{\sin \theta}
==\frac{\sin 2^n\frac{\pi}{2^n+1})}{\sin \frac{\pi}{2^n+1}}
==\frac{\sin (\frac{\pi}{2^n}-\frac{\pi}{2^n+1}) }{\sin \frac{\pi}{2^n+1}}
==\frac{\sin (\frac{\pi}{2^n+1}) }{\sin \frac{\pi}{2^n+1}}=1
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