Trigonometry Functions with a Negative Domain

[June]

Hello,

I'm having some difficulty understanding a negative domain and how that works with calculating a reference angle and rotational angle. Attached are screenshots of practice questions Im working on and got incorrect but I don't know why they're incorrect. Could somebody walk me through this please? I full understand positive domains, just lost with the negative ones. Maybe I'm missing something really obvious? I feel silly asking this but I spent some time watching youtube videos and still don't get it.

For the picture with the 45-90-46 triangle in the top right - why do we subtract 60 from 0?


Thanks in advance.

 

[Dr.Peterson]

Hello,

I'm having some difficulty understanding a negative domain and how that works with calculating a reference angle and rotational angle. Attached are screenshots of practice questions Im working on and got incorrect but I don't know why they're incorrect. Could somebody walk me through this please? I full understand positive domains, just lost with the negative ones. Maybe I'm missing something really obvious? I feel silly asking this but I spent some time watching youtube videos and still don't get it.

For the picture with the 45-90-46 triangle in the top right - why do we subtract 60 from 0?

I'm not quite sure what you mean by "negative domain". The domain of a function is a set, and sets can't be negative; no function here has a domain consisting only of negative numbers. My guess is that you mean "negative argument" -- the argument of a function is the input value itself.

In addition, your pictures are quite fuzzy, and the question about a 45-90-45 triangle does not involve 60°! So I have to work around a lot of little problems to try to help.

But let's take the one involving 60°, which I can read better. The question appears to be,

Without using technology, determine the exact value(s) of θ, where cosθ = 1/2, -270° <= θ <= 90°.

The picture shows the two possible angles, in quadrants I and IV. Are you okay with that?

The cosine being 1/2 tells you that the reference angle is 60°. Are you okay with that?

Now, one name for the angle in the first quadrant is 60°; since that is in the indicated interval, we can use it. If we couldn't, we could add or subtract 360° until we found an angle that was acceptable. For example, if they had wanted -360° <= θ <= 0°, you might use 60 - 360 = -300°.

Similarly, one name for the angle in the fourth quadrant is -60°. We are starting at the positive x-axis and moving clockwise by 60°, which makes it negative. And that, too, is in the acceptable interval.

The way they show both angles is that you are starting at 0° (the positive x axis), and either moving counterclockwise (0 + 60) or clockwise (0 - 60), because counterclockwise angles are considered positive. I didn't quite say it that way, but it is reasonable.

If this isn't enough, please tell us what you are thinking. What would you do that is different from what I've described? What specific points are unclear to you?

 

[Dr.Peterson]

I looked again at your work to see where you might be having trouble.

On the 60° problem, your work suggests that you okay with the answer of -60°, even though that seemed to be what you were asking about; your error was on the other. What you seem to have done is to find a positive angle coterminal with the first, by adding 360 to -60°. That is not how you find the other angle; and it is not in the required interval. Perhaps you need to explain in words what you were thinking there.

On the other problem, it's harder to read, but the problem appears to be solving sinθ = -√2/2, with -180° <= θ <= 180°, and their final step appears to be

-180° + 45° = -135°, or
0 - 45° = -45°.
​

What they are doing there is, for quadrant III, starting at the negative x axis (-180°) and rotating 45° counterclockwise, which means adding 45°; and, for quadrant IV, starting at 0° and rotating clockwise, which means subtracting 45°. Both of these are in the required interval, so there is no additional work needed.

What you seem to have done is to get quadrant III correct, but for some reason add 45° to 0°, despite the fact that you are rotating clockwise and the angle in the picture clearly is not +45°. Your picture looks exactly right. I wonder if you are just letting your mind wander and forgetting that clockwise angles are negative?

Going back to the other problem, with this understanding: In the picture, you show the second angle as clockwise, which would be the -300° that I mentioned; that would be correct if the required interval were different. But you calculated it as 360° - 60° = 300°, suggesting that you see it as starting at the positive x axis (360°) and moving clockwise. Did you again forget that counterclockwise is positive, so that you would get 360° + 60° = 420° if you go in that direction? Of course, you didn't need a coterminal angle; starting at 0° puts it in the right interval.

 

[June]

I looked again at your work to see where you might be having trouble.

On the 60° problem, your work suggests that you okay with the answer of -60°, even though that seemed to be what you were asking about; your error was on the other. What you seem to have done is to find a positive angle coterminal with the first, by adding 360 to -60°. That is not how you find the other angle; and it is not in the required interval. Perhaps you need to explain in words what you were thinking there.

On the other problem, it's harder to read, but the problem appears to be solving sinθ = -√2/2, with -180° <= θ <= 180°, and their final step appears to be

-180° + 45° = -135°, or
0 - 45° = -45°.
​

What they are doing there is, for quadrant III, starting at the negative x axis (-180°) and rotating 45° counterclockwise, which means adding 45°; and, for quadrant IV, starting at 0° and rotating clockwise, which means subtracting 45°. Both of these are in the required interval, so there is no additional work needed.

What you seem to have done is to get quadrant III correct, but for some reason add 45° to 0°, despite the fact that you are rotating clockwise and the angle in the picture clearly is not +45°. Your picture looks exactly right. I wonder if you are just letting your mind wander and forgetting that clockwise angles are negative?

Going back to the other problem, with this understanding: In the picture, you show the second angle as clockwise, which would be the -300° that I mentioned; that would be correct if the required interval were different. But you calculated it as 360° - 60° = 300°, suggesting that you see it as starting at the positive x axis (360°) and moving clockwise. Did you again forget that counterclockwise is positive, so that you would get 360° + 60° = 420° if you go in that direction? Of course, you didn't need a coterminal angle; starting at 0° puts it in the right interval.

Good evening,

Firstly thank you for taking the time out to actually look over my work and write such a through explanation. So the point I was missing with this question was that when you are moving in the clockwise direction you subtract the reference angle and when you are moving in the counter clockwise direction you add the reference angle. I've attached a clearer version of the question with the answer mentioning 0-60+-45. I'm going to go ahead and assume thats wrong and thats what threw me off completely.


As for the other question, again I was missing the point of adding/subtracting based on clockwise or counterclockwise direction so that makes sense to me now. Just incase, to verify I'm not missing the point could I ask you to take a look at question #4 (bottom right) and see if I calculated the values correctly?



One last thing, I'm lost as to what I do in question C because when I input ϴ=cos^-1(-5) I get an error in the calculator. Am I taking the wrong approach answering this question?




Thank you once again. I look forward to hearing from you. :grin:

 

[Dr.Peterson]

Firstly thank you for taking the time out to actually look over my work and write such a through explanation. So the point I was missing with this question was that when you are moving in the clockwise direction you subtract the reference angle and when you are moving in the counter clockwise direction you add the reference angle. I've attached a clearer version of the question with the answer mentioning 0-60+-45. I'm going to go ahead and assume thats wrong and thats what threw me off completely.
View attachment 9713

Okay, now that I can read it, it's just a typo. They didn't mean to say 60 at all here; it's supposed to be 0° - 45° = -45°.

As for the other question, again I was missing the point of adding/subtracting based on clockwise or counterclockwise direction so that makes sense to me now. Just in case, to verify I'm not missing the point could I ask you to take a look at question #4 (bottom right) and see if I calculated the values correctly?

View attachment 9714

Looks perfect.

One last thing, I'm lost as to what I do in question C because when I input ϴ=cos^-1(-5) I get an error in the calculator. Am I taking the wrong approach answering this question?

View attachment 9715

You didn't show how you got that -5! You should have multiplied both sides by the reciprocal of 5/2, namely 2/5, so you'd get 2/5 * -2 = -4/5. That is within the range of the cosine, so things should go right from here.

You apparently multiplied by 5/2 on the right. This is why it's a good idea to write down more of your work than many students do.

If you really had to find cos^-1(-5), then you would say there is no solution, because no angle can have cosine -5.

 

[June]

Okay, now that I can read it, it's just a typo. They didn't mean to say 60 at all here; it's supposed to be 0° - 45° = -45°.


Looks perfect.



You didn't show how you got that -5! You should have multiplied both sides by the reciprocal of 5/2, namely 2/5, so you'd get 2/5 * -2 = -4/5. That is within the range of the cosine, so things should go right from here.

You apparently multiplied by 5/2 on the right. This is why it's a good idea to write down more of your work than many students do.

If you really had to find cos^-1(-5), then you would say there is no solution, because no angle can have cosine -5.

Thank you for all your guidance Dr. Peterson. I really appreciate all the help I can get as I'm doing this course online. Sometimes its difficult understanding a concept just reading about it but you made it simple and clear.