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Albert Chan · (This post was last modified: 08-02-2022 10:35 AM by Albert Chan.)

Hi, Thomas Klemm

Thanks for clearing this up.

d(3,3) were used for 5+ digits accuracy, because there is no argument reduction.
Here is a version that use simpler d(2,2), but reduce asind() argument to 0 .. 1/2

a0=√(1-x²) , g0=1, a1=(g0+a0)/2, g1=√(a1*g0)

45 a_∞ ≈ a0 - 20*a1 + 64*a2 = a0 + 12*a1 + 32*g1

Code:

function asind(s,c) -- deg(asin(s)), 6+ digits accuracy

    c = c or sqrt(1-s*s)

    if s < 0   then return -asind(-s,c) end

    if s > c   then return 90 - asind(c,s) end

    if s > 0.5 then return 45 - asind(1-2*s*s)/2 end

    c = 6 + 7*c + sqrt(512*(1+c))

    return deg(45*s/c)

end

function acosd(x) return 90 - asind(x) end

function atand(x) return asind(x/hypot(1,x)) end

lua> for i=0,10 do x=i/10; printf('%.1f: %.6f (%.6f)\n', x, acosd(x), deg(acos(x))) end

0.0: 90.000000 (90.000000)
0.1: 84.260830 (84.260830)
0.2: 78.463041 (78.463041)
0.3: 72.542396 (72.542397)
0.4: 66.421818 (66.421822)
0.5: 59.999979 (60.000000)
0.6: 53.130102 (53.130102)
0.7: 45.572996 (45.572996)
0.8: 36.869898 (36.869898)
0.9: 25.841940 (25.841933)
1.0: 0.000000 (0.000000)

Update: Here is pade(asin(x),x,10,6) version

Code:

function asind(x) -- deg(asin(s)), 6+ digits accuracy

    if x < 0   then return -asind(-x) end

    local z = x*x

    if z > 0.5 then return 90 - asind(sqrt(1-z)) end

    if x > 0.5 then return 45 - asind(1-2*z)/2 end

    z = z*(5490-z*2717)/(32940-z*(31125-z*145125/28))

    return deg(x + x*z)

end

acosd(x) results, using Pade version asind(x)

0.0: 90.000000 (90.000000)
0.1: 84.260830 (84.260830)
0.2: 78.463041 (78.463041)
0.3: 72.542397 (72.542397)
0.4: 66.421824 (66.421822)
0.5: 60.000037 (60.000000)
0.6: 53.130102 (53.130102)
0.7: 45.572996 (45.572996)
0.8: 36.869898 (36.869898)
0.9: 25.841926 (25.841933)
1.0: 0.000000 (0.000000)