Jimmy started with θ=π. Playing around with the following double-angle cosine formula
cos(2θ)=2cos2(θ)−1
He made the following list of all cosine values in the following descending angle order
cos(π)cos(2π)cos(4π)cos(8π)cos(16π)cos(32π)⋮=−1=0=22=22+2=22+2+2=22+2+2+2⋮
Noticing the increasing number of 's and 2's, he then wrote down
n→∞limcos(2nπ)=22+2+2+⋯=22=1
If he were to follow these similar steps from θ=kπ, where k>0, can he also achieve the following limit to be 1?
n→∞limcos(2n⋅kπ)=1
Your answer seems reasonable.
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