# Inverse Trigonometric Function Problems

Data said:
Not at all.

Have you seen trigonometric functions defined using the unit circle, or just triangles?

In general, $\cos (-\theta) = \cos \theta$, though, so $\cos^{-1} (-x) \neq \cos^{-1} x$.
I guess just triangles. I was just wondering because $\cos$ seems to be the only one that will change if it's $\cos^{-1}$ to a different number.

I was just wondering what made it change like that, while $sin$ and $tan$ don't.

Take a look at this webpage. Read carefully:

http://encyclopedia.laborlawtalk.com/Unit_circle [Broken]

Here's my explanantion:

Draw the unit circle in the cartesian plane(circle of radius 1, centered at the origin). Now, rotate an angle $\theta$ counter-clockwise from the positive x-axis, and draw a line in this direction. This line intersects the circle at exactly one point. The $x$-coordinate of the intersection is defined to be $\cos \theta$ and the $y$-coordinate is defined to be $\sin \theta$. $\tan \theta$ is defined as $\sin \theta / \cos \theta$.

If you instead measure the angle clockwise from the positive $x$-axis, you put a negative sign in front of it.

Look at it carefully and see if you can figure out in which quadrants $\sin, \ \cos,$ and $\tan$ are positive and negative. Be sure to post with any questions!

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Data said:
Take a look at this webpage. Read carefully:

http://encyclopedia.laborlawtalk.com/Unit_circle [Broken]

Here's my explanantion:

Draw the unit circle in the cartesian plane(circle of radius 1, centered at the origin). Now, rotate an angle $\theta$ counter-clockwise from the positive x-axis, and draw a line in this direction. This line intersects the circle at exactly one point. The $x$-coordinate of the intersection is defined to be $\cos \theta$ and the $y$-coordinate is defined to be $\sin \theta$. $\tan \theta$ is defined as $\sin \theta / \cos \theta$.

If you instead measure the angle clockwise from the positive $x$-axis, you put a negative sign in front of it.

Look at it carefully and see if you can figure out in which quadrants $\sin, \ \cos,$ and $\tan$ are positive and negative. Be sure to post with any questions!
So $\sin \theta$ is positive in Quadrants: 1,2 and negative in 3,4
$\cos \theta$ is positive in Quadrants: 1,4 and negative in 2,3
and $\tan \theta$ is positive in Quadrant 1 and negative in 2,3, and 4?

So $$\cos^{-1}-\frac{1}{2} = \frac{2\pi}{3}$$ because $\cos$ is $$-\frac{1}{2}$$, which would put it in quadrant 2 which should make it negative? I'm confused. I thought the adjacent was between the right angle and $\theta$, which would place it on the $x$ axis? Since $x$ is negative in quadrant 2, why is the answer positive?

I'm sorry I'm not understanding, I hope you aren't getting frustrated :(

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Not at all!

Alright, first let's standardize terminology:

I assume you mean

quadrant 1 is where x and y are both positive

quadrant 2 is where x is negative and y is positive

quadrant 3 is where x is negative and y is negative

quadrant 4 is where x is positive and y is negative.

Assuming that that is right, your analysis of where each function is + and - is perfect, except for one quadrant with the $\tan$ function: it is positive in quadrant 3. See if you can tell why.

Now, let's look at your question. First, I want you to forget your old definitions for the functions!!! At least for now (when you're dealing with normal triangles again, you should remember it again of course. For the purposes of these questions, you want to use the one I've shown you).

You are perfectly right: $\cos \theta = -1/2$ implies that $\theta$ is in one of the quadrants where cos is negative, of course . Now, you might ask, how do we know which quadrant it is in (since cos is negative in two of the four!)?

Well, the answer is something that dextercioby hinted to earlier: the inverse trig functions only give you values in certain ranges. $\sin^{-1}$ and $\tan^{-1}$ only give you values between $-\pi /2$ and $\pi / 2$, and $\cos^{-1}$ only gives you values between $0$ and $\pi$.

Anyways, going back to your question. Based on what I just said, the angle that we're looking for, given by

$$\cos^{-1} \left(-\frac{1}{2}\right),$$

must be between $0$ and $\pi$. Also, it must be in the second quadrant. The $x$-coordinate is $-1/2$, so the angle is indeed $2\pi / 3$. The sign of $\cos \theta$ does not correspond to the sign on $\theta$.

Data said:
Assuming that that is right, your analysis of where each function is + and - is perfect, except for one quadrant with the $\tan$ function: it is positive in quadrant 3. See if you can tell why.
Oh yeah… $\frac{-x}{-y}=positive$

Data said:
Anyways, going back to your question. Based on what I just said, the angle that we're looking for, given by

$$\cos^{-1} \left(-\frac{1}{2}\right),$$

must be between $0$ and $\pi$. Also, it must be in the second quadrant. The $x$-coordinate is $-1/2$, so the angle is indeed $2\pi / 3$. The sign of $\cos \theta$ does not correspond to the sign on $\theta$.
You lost me here. I now get that $tan^{-1}$ ranges between $0$ and $\pi$. But why is the $x$-coordinate is $-1/2$? Shouldn't it just be $-1$ because $\cos$ is $\frac{adjacent}{hypotenuse}$? And from there I still don't understand how we realize it is $\frac{2\pi}{3}$.

I feel dumber as we go along lol. I bet the answer is real simple and I'm just not clicking for some reason.

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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)

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 !!

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 ).

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good

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