A83: help with math
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A83: help with math
I've been having trouble figuring out how to do the
following 2 problems without a whole lot of code
a/tan(b degrees)
a*tan(b degrees)
a is -16 to 16 b is 0-90
(note: could someone explain this to me: i thought it
wasn't possible to have negative integers in the
registers but a bunch of stuff i've seen appears
otherwise, since there also is a "neg" opcode to find
the negative of a. How is the negative sign stored in
a 8 bit number???)
Would constructing a table be the only way? I don't
really want to use up that much space if there is an
approximate i can use, other than using floating point
numbers (would that be terribly slow???).
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