Jimmy started with \(\theta = \pi\). Playing around with the following double-angle cosine formula

\[\cos\left(2\theta\right) = 2\cos^2\left(\theta\right) - 1\]

He made the following list of all cosine values in the following descending angle order \[ \begin{array}{rl} \cos(\pi) &= -1\\ \cos\left(\dfrac{\pi}{2}\right) &= 0\\ \cos\left(\dfrac{\pi}{4}\right) &= \dfrac{\sqrt{2}}{2}\\ \cos\left(\dfrac{\pi}{8}\right) &= \dfrac{\sqrt{2 + \sqrt{2}}}{2}\\ \cos\left(\dfrac{\pi}{16}\right) &= \dfrac{\sqrt{2 + \sqrt{2 + \sqrt{2}}}}{2}\\ \cos\left(\dfrac{\pi}{32}\right) &= \dfrac{\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2}}}}}{2}\\ \vdots & \quad \vdots \end{array} \] Noticing the increasing number of \(\sqrt \ \)'s and 2's, he then wrote down \[\lim_{n \rightarrow \infty} \cos\left(\dfrac{\pi}{2^n}\right) = \dfrac{\sqrt{2 + \sqrt{2 + \sqrt{2 + \cdots}}}}{2} = \dfrac{2}{2} = 1\] If he were to follow these similar steps from \(\theta = \dfrac{\pi}{k}\), where \(k > 0\), can he also achieve the following limit to be 1? \[\lim_{n \rightarrow \infty} \cos\left(\dfrac{\pi}{2^n \cdot k}\right) = 1\]

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