Don't need to think of it using trig, just some simple vector mathematics (you get the same value but this one may be easier to understand).
1^2 = 1 = i^2 + j^2
Diagonal so i = j
1 = 2(i^2)
(1/2) = i^2
(1/2)^(1/2) = i
root(0.5) = 0.70710678118654752440084436210485, but you could just approximate that to 0.71 or 0.707 if you preferred.
Due to the use of floating points you'll get a tiny "rounding error" creep into things but it's negligible for these purposes. Obviously the more accurate a value the less rounding error but again, negligible.
I'll call whatever value you use for the root of 0.5 "n" for now. and the speed "s"
Up: x, y - s
Down: x, y + s
Left: x - s, y
Right: x + s, y
Up+Left: x - sn, y - sn
Up+Right: x + sn, y - sn
Down+Left: x - sn, y + sn
Down+Right: x + sn, y + sn
Editors Note: I have had something in the region of 5 vodka and red bulls, two tequila shots, half a bottle of white wine and some as-yet-unidentified licorice-flavoured drink a friend handed me tonight. Not a lot really, but on a close-to-empty stomach...well, as far as I can tell I've sobered up enough to be reasonably sure as to my maths here...maybe I've even hit the
Ballmer Peak, who knows?
