Feel the Power of cos(2θ)\cos\left(2\theta\right)!

Calculus Level 3

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

cos(2θ)=2cos2(θ)1\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 cos(π)=1cos(π2)=0cos(π4)=22cos(π8)=2+22cos(π16)=2+2+22cos(π32)=2+2+2+22 \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 limncos(π2n)=2+2+2+2=22=1\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 θ=πk\theta = \dfrac{\pi}{k}, where k>0k > 0, can he also achieve the following limit to be 1? limncos(π2nk)=1\lim_{n \rightarrow \infty} \cos\left(\dfrac{\pi}{2^n \cdot k}\right) = 1

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