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same story, If we have $s$ circle with the diameter $AB$ (with length $1$) and the center $O$, then we can approximate $\operatorname{chord} AC$ where $x$ represents the value of the $\angle AOC$ in degrees, and $t=90-\frac{x}{2}$.So formula is $1-\frac{2\left(\frac{{\pi}}{360}t\right)^{2}}{\left(\dfrac{\left(\frac{\pi}{360}\right)}{\sin\left(\frac{\pi}{360}\right)} +3.222\cdot 10^{-11}\cdot \left(t^{2}+\frac{239t^4}{83.9^4}+\frac{10t^6}{83.08^6}-1\right)\right)^{\,2t^{2}}}$

For example, if $x=60(t=60)$, we will get $0.499999996\ldots$,or $x=120,(t=30)$ we will get $0.86602540377\dots$.So if $x<360$ then the value of the error cannot be greater than $4.55\cdot 10^{-9}$

Is there formula with a more precise approximate value for $\sin\frac{{\pi}}{360}x$ than mine?

and another question whether it can be simplified?

Srbin
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    This is very similar to your previous question. – Peter Phipps Sep 14 '21 at 08:50
  • Perhaps cite your previous question https://math.stackexchange.com/q/4233359 where some of us indicated other approximations. – Gary Sep 14 '21 at 08:51
  • “Is there formula with a more precise approximate value” – I would assume that you can take the approach from https://math.stackexchange.com/a/4233460/42969 (the answer to your previous question) and add another term. – It would also be interesting to know how you came up with that formula, and what kind of improved formula you are looking for. – Martin R Sep 14 '21 at 08:51
  • Do you reallze the cost of the evaluation of such an expression ? – Claude Leibovici Sep 16 '21 at 06:30
  • the idea is to capture the accuracy of a few dozen conditions from the Taylor series with a few more changes (conditions), I think that's possible – Srbin Sep 16 '21 at 07:14

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If you use the procedure used here, writing $$\sin(x)=\sum_{n=1}^p \frac{a_n}{\pi^{2n-1}} \big[x(\pi-x)\big]^n=f_p(x)$$ the first coefficients are $$\left( \begin{array}{cc} n & a_n \\ 1 & 1 \\ 2 & 1 \\ 3 & 2-\frac{\pi ^2}{6} \\ 4 & 5-\frac{\pi ^2}{2} \\ 5 & 14-\frac{3 \pi ^2}{2}+\frac{\pi ^4}{120} \\ 6 & 42-\frac{14 \pi ^2}{3}+\frac{\pi ^4}{24} \\ 7 & 132-15 \pi ^2+\frac{\pi ^4}{6}-\frac{\pi ^6}{5040} \\ 8 & 429-\frac{99 \pi ^2}{2}+\frac{5 \pi ^4}{8}-\frac{\pi ^6}{720} \\ 9 & 1430-\frac{1001 \pi ^2}{6}+\frac{55 \pi ^4}{24}-\frac{\pi ^6}{144}+\frac{\pi ^8}{362880} \\ 10 & 4862-572 \pi ^2+\frac{1001 \pi ^4}{120}-\frac{11 \pi ^6}{360}+\frac{\pi ^8}{40320} \end{array} \right)$$

For $p=8$, the maximum error is $4.995\times 10^{-13}$.

Claude Leibovici
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