sin²(x) can be written in terms of an infinite sum over cos²(x) [and vice versa]pic.twitter.com/avj4yy6SPe
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Only when cos(x)^2 is inside the Taylor radius of 1/(1-x). For example, for x=0, it fails.
the taylor radius is precisely |cos^2(x)|<1 which is true for all x except x=0,π. However, testing these endpoints separately we get 1/(1 + 1 + 1 + 1 ...) --> 0=sin(0,π) so it works!
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Except for almost all complex values.
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well my original tweet was concerning real numbers.. but let's see the endpoints in the complex plane are the z such that |cos^2(z)|=1 and so would be a condition as a function of e^{ix} e^{y} that i don't feel like working out
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