Turing Programming: Solve Cos Inverse Function

[eax]

Hi,

I'm programming a game and need a character to turn facing another character. I got far but the problem is the language(Turing) Iam provided with doesn't support a cos/sin inverse function but has sin/cos.

Anyone know the formula? I have cos(x) how can I get x? Searching on google is not help all I get is what cos (x) = to relative to sin/tan.

Thanks in advance!

 

[derekmohammed]

Hi
I am not sure about your question but I will try my best to anwser it...
Cosine(x)= adj/hyp.
cos(x)^-1= Ratio of the adjecent arm to the ratio of the hypotines.

Or in terms of the unit circle...
Cos(x) is equal to x/r
so...
Cos(x)^-1= Ratio of x/r. THis solves for the angle of x.

Ok and finally...
cot(x)=cos(x)/sin(x)
the inverse of Cos(x) is sec(x) ie. sec(x)=r/x
the inverse of Sin(x) is csc(x) ie. csc(x)=r/y
I hope that helped...
The ratio of sin(x)/

 

[eax]

This is what I have for mouse to character interaction, instead of character(enemy) to character(player)

set x/y to equal mouse position x/y
x := mousex
y := mousey
drawx/yoff are < 0 but we want to add them to x/y thus
subtracting adds them to x/y
x -= drawxoff
y -= drawyoff


bound.x/y are the coordinates for the character(player)
subtracting x/y by bound.x/y gives the x/y position reveloving
bound.x/y so (0,0) = (bound.x, bound.y)
y -= bound.y
x -= bound.x
get distance
dist := sqrt (x * x + y * y)
if dist < 1 then
    return
end if

I added 90 for a reason but it doesn't matter
pangle += 90
this next function just makes sure pangle is >= 0 and <= 360
rapanglei (pangle)

cos_t is a table of cos values because calculating cos everytime takes
lots of CPU time

calculating dotproduct (I think that's what it is called)
(and I like to reuse variables to save memory :))

We are treating these as vectors. bound.r = radius of character(player)
multiplying by cos(angle) gives us the "vector" its facing, then calculating
dot product 
x := (x * cos_t (pangle) * bound.r + y * sin_t (pangle) * bound.r) / (bound.r * dist)
puting pangle back to original value
pangle -= 90

and now x = cos(angle I want)
angle I want = cos inverse(x)

There is no cos/sin inverse function in this language. How can I get "angle I want"?
I hope my question is more clear now

Thanks in advance!

 

[nolachrymose]

Maybe you could make an arccos() look-up table? I don't know of any mathematical way to transfer cos(x) to x without arccos(). A look-up table would probably be faster anyways. You'd just need to make a function that finds the closest related value.

By the way, I've never seen the language Turing, but it kind of looks like Pascal...

 

[eax]

Im not even sure if http://www.holtsoft.com/turing/support/#currentversion is a language. Its more of an interpreter, I don't know why my school starts of with Turing then C++ is next year(or cemester . It would be better to start with C/ASM (in 16bit(DOS) mode)

Someone has to know. My math teacher said that before there were calculators people new the formula's for cos/sin and arccos/sin. If everyone forgot then everyone just copy's and pastes code into calculators and every where else :D.

 

[nolachrymose]

I'm pretty sure there's a Taylor series for all the trigonometric functions, but I can't seem to find the one for arccos(). Try looking up on Google the Taylor series for inverse cosine, and see if you can find it.

Learning an interpreted language before a compiled one? That's quite odd.

 

[Nylex]

Try looking up on Google the Taylor series for inverse cosine, and see if you can find it.d.

You could use the derivative of cos^-1 x and work out it's Taylor series, then integrate term by term.

 

[VietDao29]

Hi,
I am not an anvanced programmer, but anyway, I suggest this:
Since you know cos(x) then you know it's positive/negative/0.

If cos(x) = 0 then
x = 90
EndIf

If cos(x) < 0 then
For s = 91 to 180
...
(You can find the difference between cos(x) and cos(s). When the cos(s) - cos(x) < 0, stop the loop and you will get the s).
Next s
EndIf

If cos(x) > 0 then
For s = 1 to 90
...
(You can find the difference between cos(x) and cos(s). When the cos(s) - cos(x) > 0, stop the loop and you will get the s).
Next s
EndIf

And you will get cos(s) approximetely equal to cos(x)... But it's not the best way, though,...
Hope this help :-)
Bye bye,
Viet Dao,

 

[eax]

Thank you :)