Chris,
Here's your Lingo solution followed by a trig lesson as to how it was derived...
----------------------
-- The Lingo Lesson --
----------------------
--
on arcCos (thisValue)
-- compute appropriate theta value using
-- the arctan function
numerator = sqrt(1 - (thisValue * thisValue))
theta = arctan( numerator/thisValue )
-- return the angle value
return theta
end arcCos
--
--
on arcSin (thisValue)
-- compute appropriate theta value using
-- the arctan function
denominator = sqrt(1 - (thisValue * thisValue))
theta = arctan( thisValue/denominator )
-- return the angle value
return theta
end arcSin
--
---------------------
-- The Trig Lesson --
---------------------
There is no inverse cosine, or inverse sine functions, we only have access to the
inverse tangent function, arctan. But can can in fact get the desired information
(inverse cosine/sine) using arctan and the following:
Whave a variable someValue and we want to find the inverse sine of that value. Let's
start by declaring what we "know":
(1) sin(theta) = someValue
What we now want is the value of theta. What else do we know with respect to trig
relationships (sine/cosine/tangent) and the mysterious theta angle:
(2) tan(theta) = sin(theta)/cos(theta)
(3) sin(theta)^2 + cos(theta)^2 = 1
Let's rewrite equation (3) to read:
(4) cos(theta) = sqrt(1 - sin(theta)^2)
and put that into equation (2) to get:
(5) tan(theta) = sin(theta)/(sqrt(1 - sin(theta)^2))
Ahhhh, but we know the numerical value of sin(theta), it is someValue (as stated above
in equation (1)). Therefore we can rewrite (5) as:
(6) tan(theta) = someValue/(sqrt(1 - someValue^2))
And now, being able to calculate the value on the right, we can use the inverse
tangent function to get the value of theta itself (the desired goal):
(7) theta = arctan( someValue/(sqrt(1 - someValue^2))
Now, you asked about cosine in particular whereas I discussed the sine option above, I
guess that's just cause I wanted to teach a lil' general trig so that you can add this
to your skillset in the future while mucking about in Lingo. ;) For the case of
wanting to get the value of the inverse cosine, we'd need to change lines (1) and then
(4) through (7) to be:
(1b) cos(theta) = someValue
(4b) sin(theta) = sqrt(1 - cos(theta)^2)
(5b) tan(theta) = (sqrt(1 - cos(theta)^2))/cos(theta)
(6b) tan(theta) = (sqrt(1 - someValue^2))/someValue
(7b) theta = arctan( (sqrt(1 - someValue^2))/someValue )
Rock on,
Tom
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