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: