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In this post just another $\pi$ formula, I gave a kind of Riemann sum to compute the area of a quarter of circle based on a very simple geometric trick, and same reasoning can be used to compute any inverse trigonometric formulas based on the angle area.

But I think that because of the simplicity of the trick, they could have been found since the Ancient Greeks (at least), who already managed technics like exhaustion...

Here are the formulas for $\text{acos}(x)$, $\text{asin}(x)$ and $\text{atan}(x)$:

Let's $u=\sqrt{1-x}$ and $v=\sqrt{1+x}$, then: $$ \text{arccos}(x)=x\sqrt{1-x^2}+4\left(\sqrt{1-x^2}\right)^3\sum _{k=1}^n \frac{n^3(2 k-1)^2\left((k-1) k u^2+n^2 v^2\right)}{\left((k-1)^2 u^2+n^2 v^2\right)^2 \left(k^2 u^2+n^2 v^2\right)^2} $$ Let's $u=1-\sqrt{1-x^2}$ and $v=x$, then: $$ \text{arcsin}(x)=x\sqrt{1-x^2}+4x^3\left(1-\sqrt{1-x^2}\right)^3\sum _{k=1}^n \frac{n^3(2 k-1)^2\left((k-1) k u^2+n^2 v^2\right)}{\left((k-1)^2 u^2+n^2 v^2\right)^2 \left(k^2 u^2+n^2 v^2\right)^2} $$ Let's $u=\sqrt{1+x^2}-1$ and $v=x$, then: $$ \text{arctan}(x)=\frac{x}{1+x^2}+4x^3\left(\sqrt{1+x^2}-1\right)^3\sum _{k=1}^n \frac{n^3(2 k-1)^2\left((k-1) k u^2+n^2 v^2\right)}{\left((k-1)^2 u^2+n^2 v^2\right)^2 \left(k^2 u^2+n^2 v^2\right)^2} $$

Here the Mathematica code if someone is interested:

(* acos(x) *)
Clear[x,k,n,u,v];
x1=.7;
u[x_]:=Sqrt[1-x];
v[x_]:=Sqrt[1+x];
Ak[x_,k_,n_]:=n^3*(2 k-1)^2((k-1)k*u[x]^2+n^2*v[x]^2)/((k-1)^2*u[x]^2+n^2*v[x]^2)^2/(k^2*u[x]^2+n^2*v[x]^2)^2
SumAk[x_,n_]:=x*Sqrt[1-x^2]+4*(Sqrt[1-x^2])^3*Sum[Ak[x,k,n],{k,1,n}];
ArcCos[x1]
SumAk[x1,100]

(* asin(x) *)
Clear[x,k,n,u,v];
x1=.7;
u[x_]:=1-Sqrt[1-x^2];
v[x_]:=x;
Ak[x_,k_,n_]:=n^3*(2 k-1)^2((k-1)k*u[x]^2+n^2*v[x]^2)/((k-1)^2*u[x]^2+n^2*v[x]^2)^2/(k^2*u[x]^2+n^2*v[x]^2)^2
SumAk[x_,n_]:=x*Sqrt[1-x^2]+4*x^3*(1-Sqrt[1-x^2])^3*Sum[Ak[x,k,n],{k,1,n}];
ArcSin[x1]
SumAk[x1,100]

(* atan(x) *)
Clear[x,k,n,u,v];
x1=.7;
u[x_]:=Sqrt[1+x^2]-1;
v[x_]:=x;
Ak[x_,k_,n_]:=n^3*(2 k-1)^2((k-1)k*u[x]^2+n^2*v[x]^2)/((k-1)^2*u[x]^2+n^2*v[x]^2)^2/(k^2*u[x]^2+n^2*v[x]^2)^2
SumAk[x_,n_]:=x/(1+x^2)+4*x^3*(Sqrt[1+x^2]-1)^3*Sum[Ak[x,k,n],{k,1,n}];
ArcTan[x1]
SumAk[x1,100]

For small $x$, the formulas are really slow compared to classic inverse trig series, and better than the series as $x$ tends to $1$. In fact, it still works for higher $x$ values counter to the series.

What is interesting is that it works for any complex $z$ without the limitation $|z|<1$, thus without any limitations at all...

Anyone knows about?

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