A345644 Decimal expansion of the radius of the circle tangent to the curves y=cos(x), y=-cos(x) and to the y-axis for x in [0,Pi/2].

6, 4, 2, 7, 0, 7, 8, 7, 2, 5, 4, 6, 5, 3, 2, 4, 4, 5, 7, 7, 9, 2, 1, 1, 7, 7, 8, 4, 6, 8, 6, 0, 7, 9, 1, 8, 2, 8, 5, 0, 4, 7, 8, 2, 4, 0, 8, 1, 4, 6, 3, 0, 3, 9, 8, 5, 3, 3, 1, 5, 0, 7, 9, 4, 6, 4, 4, 9, 0, 0, 0, 9, 9, 3, 4, 6, 5, 2, 5, 4, 5, 3, 1, 3, 3, 8, 2, 4, 4, 2, 8, 0, 9, 7, 2, 7, 3, 7, 8

Offset: 0

Author

Gleb Koloskov, Jun 21 2021

Keywords

• nonn
• cons

Comments

Let r and (x,y) denote the radius of the circle and the point of tangency in the first quadrant, respectively.

Then r in [0,1] is the root of equation cos(r+sqrt(r^2-1+sqrt(1-r^2)))^2 = 1-sqrt(1-r^2),

r = 0.642707872546532445779211778468607918285047824...,

x = r+sqrt(r^2-1+sqrt(1-r^2)) = 1.066010072972971718857583783392083793389510385...,

y = sqrt(1-sqrt(1-r^2)) = 0.483620364074368181073730094271148302685427120...

Examples

			0.642707872546532445779211778468607918285047824...
		

Crossrefs

Cf. A197016, A197017, A197018, A197019, A197020.

Programs

Mathematica

r = r /. FindRoot[Cos[r + Sqrt[-1 + r^2 + Sqrt[1 - r^2]]]^2 == 1 - Sqrt[1 - r^2], {r, 1/2}]; Show[Plot[Cos[x], {x, 0, Pi/2}], Plot[-Cos[x], {x, 0, Pi/2}], Graphics[Circle[{r, 0}, r]], PlotRange -> All, AspectRatio -> Automatic] (* Vaclav Kotesovec, Jul 01 2021 *)

PARI

solve(r=0,1,cos(r+sqrt(r^2-1+sqrt(1-r^2)))^2-1+sqrt(1-r^2))