Only when cos(x)^2 is inside the Taylor radius of 1/(1-x). For example, for x=0, it fails.
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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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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1-x² = 1/(1+x²+x^4+...) not necessarily using sin and cos
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indeed, that’s how I came up with it! But it’s always need to see something relating sine and cosine as it hints as something geometrical
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sin^2(x)=1-cos^2(x)= 1-p^2= 1/(1+p^2+p^4+p^6...), where |p|<=1, p=cos(x) ;-)
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The sin^2 and cos^2 of two vectors with rational components is also itself a rational number.
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This is actually just the geometric series formula combined with 1 - cos^2(x) = sin^2(x) 1/(1+x^2+x^4+...) = 1/(1/1-x^2)=1-x^2 and if x=cos(x) you're done
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s^2=1/(1+c^2/s^2) =1/(1+c^2(1+c^2/s^2)) =1/(1+c^2(1+c^2(1+c^2/s^2))) =...
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Geometric series...hmm...
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