Re: cos theta = 1

On Aug 21, 10:57 am, hobbes <sardonicStuffedTi...@xxxxxxxxx> wrote:
On Aug 21, 10:32 am, Randy Poe <poespam-t...@xxxxxxxxx> wrote:



On Aug 21, 10:19 am, hobbes <sardonicStuffedTi...@xxxxxxxxx> wrote:

hi yes cos = 1 was just an example, i was seeking the general
solution. i wasn't sure if you could just divide both sides of the
equation by cos. i guess that's what's happening when you use arccos?

No, that's not what's happening. "Divide by cos" doesn't
mean anything, since "cos" is not a number.

Imagine you have a table with two columns. In column 1 is
a bunch of values of theta. In column 2 is the corresponding
values of cos(theta). The operation "arccosine" is like
starting with a value in column 2, and seeing which value
in column 1 corresponds to that. It has nothing to do
with dividing. Division never comes into it.

Here is a plot of acos(x). An acos function starts with
a value from -1 to 1 on the x-axis of this plot, and finds
the corresponding value on the y-axis.http://www.mathworks.com/access/helpdesk/help/techdoc/ref/acos.html

It has nothing to do with division.

so, in the general case given cos( ) = x, = acos(x) ?

Yes, but -theta is also a solution, and there are
infinitely many others. I'm not sure what you're doing
but I keep reminding you of this because it may turn
out to be important. In some cases -acos(x) might be
the answer that makes more sense for your application.

- Randy

"I'm not sure what you're doing but I keep reminding you of this
because it may turn out to be important."

i'm not doing anything useful now except reminding myself how much
math i forgot (and how much i probably never really understood in the
first place) but i have some function that claims to calculate the
angle between two vectors in degrees. using the fact that the angle
between two vectors can be determined by using the dot product: cos
&theta; = (<b>u</b> &bull; <b>v</b>) / (||u|| ||v||), the logic of the
function does this:

return Math.acos((u.dotproduct(v) / u.magnitude() * v.magnitude()));

do you think this method would give someone a reasonable number for
the angle between two 2D vectors?

by the way, how do you display the html math symbols in this place?

besides the fact that the paranethesis are probably wrong:

return Math.acos(u.dot(v) / (u.magnitude() * v.magnitude()));

.

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