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ANSWERS: 1
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Here's one I prepared earlier... x + x^3/3 + 2x^5/15 + 17x^7/315 + ... I just looked up the formula and it's pretty ugly: sum {n=1..infinity, B[2n] ((-4)^n (1 - 4^n) / (2n)!) x^(2n-1) } "B[i]" are the Bernoulli numbers which themselves have no easy definition. I'd say, sin(x) is easy, cos(x) is easy, just divide them. The rational polynomial obviously requires less terms too.
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