Inverse Trigonometric Function Problems

[Data]

The x-coordinate is -1/2 because we're considering whatever is the argument to \cos^{-1} to be the cosine of some angle, so that the value of the expression is the angle itself.

Remember, the cosine is just the x-coordinate of a point on the circle! So, in this case, the cosine is -1/2, and that means that the corresponding point on the circle has an x-coordinate of -1/2. We then know that it's in the second quadrant, too, because like I said, \cos^{-1} only gives us values in quadrants 1 and 2, and no points in quadrant 1 have negative x-coordinates.

So, we have a point in quadrant 2 with an x-coordinate of -1/2. There is exactly one point on the circle satisfying these conditions. We now need to know what the angle is, though.

Notice that this point is the reflection over the y-axis of the point on the circle with x-coordinate 1/2 in quadrant 1 (ie. if we flip the point over the y-axis, we get the point with x-coordinate 1/2 in quadrant 1, also on the circle). But you know what angle corresponds to that point, it's just \pi / 3.

The angle for our real point, with x-coordinate -1/2, is then just \pi - \pi / 3 = 2\pi / 3 (which you should be able to see geometrically from your picture)

 

[Lucretius]

Hmm…interesting. I think I understand now. So cos^({-1} is different because of the range of values it can be in?

I guess I'll go memorize that chart, or better yet, I'll just write it down in my journal of notes. I can use those on tests…

Data, can't tell you how helpful you have been. Thanks for sticking it through with me !

 

[Data]

Let me state this a little bit more formally so that it might be a little clearer.

Okay. Let's say we want to find

\cos^{-1} x

for some given x. We know that

\cos^{-1} (\cos \gamma) = \gamma

(at least for \pi \geq \gamma \geq 0)

so we'll try to construct an expression something like that. Remember that \cos^{-1} always returns an angle, so we can let it be equal to some angle \theta:

\theta = \cos^{-1} x

but then, this means that we can interpret this to mean

x = \cos \theta

thus, if we can find an angle \theta such that

x = \cos \theta

then our question is solved. We also have to keep in mind that

0 \leq \theta \leq \pi

so we're only looking for angles in quadrants 1 and 2.

The hard part of the question, then, is just how to find \theta? You should do this by considering the situation geometrically, like I did in my last post to find the angle 2\pi / 3 (or by just taking \cos^{-1}(x) if you have your calculator handy ).

 

[Data]

good

if you have any more questions, just ask. I'm going to sleep for now, though! :zzz: