Hmmm what am I missing ?



(yeah acos1 is acos, don't worry )

Wow, nice! How well does it work near the end points? I was just going to compute asin() and acos() from atan().

Cool, I got it to work and that is nice! If you check out this:
http://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Relationships_among_the_inverse_trigonometric_functions

You can get a representation of arccos with arctangent, and on [-1,1] it becomes 2tan^-1(sqrt((1-x)/(1+x)). You can probably modify it to work with the Axe atan() command.

[rambling]
Yeah, after analysing, the method I gave uses one less multiplication than yours, the same number of divisions and square roots. However, mine also uses arctangent, and arctangent is slower than the saved multiplication, so in all, my routine would be slower. In assembly, I could optimise the division in both cases, but your multiplication would be trivial and fast whereas my square root would be faster than yours (only 8 bits of precision would be needed).
[/rambling]

Anyways, after looking at the graph, there are actually a few values that are off by a little more than 1, but that is still very accurate and excellent. If you need speed, though, you probably already know that table methods are great. In fact, you can use the fact that it is basically linear from [-89,89] to make only a table from 90 to 128 and they only need to be 1 byte. Here is a routine that gets exact rounded values (111 bytes):
acos:
;Input:
;    L is the signed int from [-128,127] where -128 corresponds to -1
;Output:
;    HL
ld h,0
ld a,l
or a
jp p,acossub0
neg
call acossub0
ld a,128
sub l
ld l,a
ret
acossub0:
    cp 14
jr c,$+7
cp 57
jr nc,$+3
dec a
cp 68
jp nc,acossub1
    inc a
ld bc,0530h
cp c \ jr c,$+3 \ sub c
rla \ djnz $-5
    or $E0
add a,65
ld l,a
ret
acossub1:
;now
sub 68
;between 0 and 21, slope is -2/5
sub 22
jr nc,acossub2
ld c,5 \ ld l,h
inc l \ add a,c \ jr nc,$-2
sub 3 \ ld a,l
    adc a,a
add a,1Fh
ld l,a
ret
acossub2:
;original a>=90
;between 0 and 17, slope is -1/2
sub 18
jr nc,acossub3
rra
    cpl
add a,18h
ld l,a
ret
;original a>=107
acossub3:
ld hl,acosLUT
add a,l
ld l,a
jr nc,$+3
inc h
ld l,(hl)
ld h,0
ret
acosLUT:
.db $17
.db $16
.db $16
.db $15
.db $15
.db $14
.db $13
.db $13
.db $12
.db $11
.db $10
.db $F
.db $E
.db $E
.db $D
.db $B
.db $A
.db $9
.db $7
.db $5

And if you are okay with tiny error of only 1, this is 21 bytes smaller and around 300 t-states worst case:
acos:
;Input:
;    L is the signed int from [-128,127] where -128 corresponds to -1
;Output:
;    HL
ld h,0
ld a,l
or a
jp p,acossub0
neg
call acossub0
ld a,128
sub l
ld l,a
ret
acossub0:
cp 90
jp nc,acossub1
    inc a
ld bc,0530h
cp c \ jr c,$+3 \ sub c
rla \ djnz $-5
    or $E0
add a,65
ld l,a
ret
acossub1:
;now
sub 90
ld hl,acosLUT
add a,l
ld l,a
jr nc,$+3
inc h
ld l,(hl)
ld h,0
ret
acosLUT:
.db $20
.db $20
.db $1F
.db $1F
.db $1E
.db $1E
.db $1D
.db $1D
.db $1C
.db $1C
.db $1B
.db $1B
.db $1A
.db $1A
.db $19
.db $19
.db $18
.db $18
.db $17
.db $16
.db $16
.db $15
.db $15
.db $14
.db $13
.db $13
.db $12
.db $11
.db $10
.db $F
.db $E
.db $E
.db $D
.db $B
.db $A
.db $9
.db $7
.db $5


Plop,

Small optimisations of the cos^-1 routine (algorithm-related), and I figured a fast way to get sin^-1 from cos^-1 so now you have this :
:Lbl ArcCos
:~abs()→r₂
:
:If >>~89
:~r₂
:Else
:128-√(r₂+128*30)
:End
:→r₂
:r₁ee0??128-r₂,r₂
:Return
:
:Lbl ArcSin
:-ArcCos()+64
(:Return)