# Radians and the Unit Circle

#### eutas1

##### Junior Member
I am SO confused on how to do (d) - please refer to the attachments.

I asked my maths teacher today how to do all of question 15 because I did not understand why the domain was even needed to answer the question - I had just used the angle given in the question and then sketched the reference angle and got all of the answers correct, however I know how I did it is wrong because you must consider the 'x' domain angle in order to answer the question apparently.

The issue is, my teacher explained to me that to answer the questions, you draw the given angle, then sketch the reference angle into quadrant 1 (Q1), then reflect/translate the same angle into whatever quadrant 'x' is supposed to be in, in reference to the given domain. Attachment T shows what my teacher told me to do in order to answer the questions - I used (c) as a main example. This worked for (a), (b), and (c), however it does NOT work for (d). I have no idea why, I am just so confused and have been stuck on this question for like 2 weeks... I asked my teacher again later today about why it does not work for (d) but we did not have time to discuss it so I am still completely stuck...

Thank you!

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#### Dr.Peterson

##### Elite Member
It's definitely a confusing type of problem. Just to make sure I'm understanding it correctly, can you tell me what the "correct" answers are, as you understand them? You didn't show your actual answer to (c), and the answer as I understand it is very simple (without doing nearly so much work).

One trouble in your work is that what you label as x is not x. That, among other things, makes it hard to follow what you are doing.

#### pka

##### Elite Member
The issue is, my teacher explained to me that to answer the questions, you draw the given angle, then sketch the reference angle into quadrant 1 (Q1), then reflect/translate the same angle into whatever quadrant 'x' is supposed to be in, in reference to the given domain. Attachment T shows what my teacher told me to do in order to answer the questions - I used (c) as a main example. This worked for (a), (b), and (c), however it does NOT work for (d). I have no idea why, I am just so confused and have been stuck on this question for like 2 weeks... I asked my teacher again later today about why it does not work for (d) but we did not have time to discuss it so I am still completely stuck...
It is difficult for us to understand what you need.
$$\sin(\alpha-\beta)=sin(\alpha)\cos(\beta)-\cos(\alpha)\sin(\beta)$$
$$\sin\left(\dfrac{3\pi}{2}-x\right)=\sin\left(\dfrac{3\pi}{2}\right)\cos(x)-\cos\left(\dfrac{3\pi}{2}\right)\sin(x)$$
Now $$\sin\left(\dfrac{3\pi}{2}\right)=-1,~\cos\left(\dfrac{3\pi}{2}\right)=0$$
So $$\sin\left(\dfrac{3\pi}{2}-x\right)=-\cos(x)$$

#### skeeter

##### Elite Member
reference the diagram ...

$$\displaystyle \sin\left(\dfrac{3\pi}{2} - \theta \right) = -\cos{\theta}$$

For any of my colleagues who may be taken aback by this strange approach, the blue triangle in quadrant III is congruent to the red one in quadrant I.

Fot the OP ... so much easier if you were taught basic sum/difference identities.

#### Dr.Peterson

##### Elite Member
reference the diagram ...

$$\displaystyle \sin\left(\dfrac{3\pi}{2} - \theta \right) = -\cos{\theta}$$

For any of my colleagues who may be taken aback by this strange approach, the blue triangle in quadrant III is congruent to the red one in quadrant I.

Fot the OP ... so much easier if you were taught basic sum/difference identities.

View attachment 26624
The thing that is strange to me is not just the convoluted visual approach to solving the problems (which is perfectly sensible in simple cases like this one), but that the problems themselves don't need to refer to domains at all! Their answers are independent of all that.

So why bother teaching students to do all this in special cases, when they will soon learn that there is only one case, and it can be solved either in terms of whatever case is easiest (which, by the way, is what (d) is, being in quadrant I), or in terms of sum/difference identities?

I could demonstrate how I would solve these visually, taking the problem at face value and not playing games with the reference angle; but I'm not sure that would help with what the teacher wants them to do.

#### eutas1

##### Junior Member
It's definitely a confusing type of problem. Just to make sure I'm understanding it correctly, can you tell me what the "correct" answers are, as you understand them? You didn't show your actual answer to (c), and the answer as I understand it is very simple (without doing nearly so much work).

One trouble in your work is that what you label as x is not x. That, among other things, makes it hard to follow what you are doing.
Oh yeah, I forgot to include the answer for (d)...

(a) --> - cos(x)
(b) --> - sin(x)
(c) --> sin(x)
(d) --> - cos(x)

#### eutas1

##### Junior Member
It is difficult for us to understand what you need.
$$\sin(\alpha-\beta)=sin(\alpha)\cos(\beta)-\cos(\alpha)\sin(\beta)$$
$$\sin\left(\dfrac{3\pi}{2}-x\right)=\sin\left(\dfrac{3\pi}{2}\right)\cos(x)-\cos\left(\dfrac{3\pi}{2}\right)\sin(x)$$
Now $$\sin\left(\dfrac{3\pi}{2}\right)=-1,~\cos\left(\dfrac{3\pi}{2}\right)=0$$
So $$\sin\left(\dfrac{3\pi}{2}-x\right)=-\cos(x)$$
Ahh is this angle sum and difference identities? We have not learnt these so I was hoping I could find a method that does not involve that... (preferably visual)

#### skeeter

##### Elite Member
Ahh is this angle sum and difference identities? We have not learnt these so I was hoping I could find a method that does not involve that... (preferably visual)
see post #4

#### eutas1

##### Junior Member
reference the diagram ...

$$\displaystyle \sin\left(\dfrac{3\pi}{2} - \theta \right) = -\cos{\theta}$$

For any of my colleagues who may be taken aback by this strange approach, the blue triangle in quadrant III is congruent to the red one in quadrant I.

Fot the OP ... so much easier if you were taught basic sum/difference identities.

View attachment 26624
Oh, I see that I sketched my reference triangle wrong in Q1. However, I am still confused... Aren't you meant to compare the reference angle with the 'x' domain angle? The domain angle is in quadrant 1 so is it not just the exact same angle as the red triangle overlapping? So wouldn't that mean sin(3pi/2 - x) = cos(x) ???

#### eutas1

##### Junior Member
The thing that is strange to me is not just the convoluted visual approach to solving the problems (which is perfectly sensible in simple cases like this one), but that the problems themselves don't need to refer to domains at all! Their answers are independent of all that.

So why bother teaching students to do all this in special cases, when they will soon learn that there is only one case, and it can be solved either in terms of whatever case is easiest (which, by the way, is what (d) is, being in quadrant I), or in terms of sum/difference identities?

I could demonstrate how I would solve these visually, taking the problem at face value and not playing games with the reference angle; but I'm not sure that would help with what the teacher wants them to do.
the problems themselves don't need to refer to domains at all
^ Is this the same thing I was talking about regarding how I get all of the correct answers by ONLY referring to the given angle and drawing the reference angle, completely ignoring the domain?

#### skeeter

##### Elite Member
Oh, I see that I sketched my reference triangle wrong in Q1. However, I am still confused... Aren't you meant to compare the reference angle with the 'x' domain angle? The domain angle is in quadrant 1 so is it not just the exact same angle as the red triangle overlapping? So wouldn't that mean sin(3pi/2 - x) = cos(x) ???
no ... the y-coordinate in quad III is negative

#### Dr.Peterson

##### Elite Member
the problems themselves don't need to refer to domains at all
^ Is this the same thing I was talking about regarding how I get all of the correct answers by ONLY referring to the given angle and drawing the reference angle, completely ignoring the domain?
Yes, it explains why you will get the right answer doing that.

My concern is that your teacher clearly expects you to take domain into account in your drawing.

Here is how I would do (c):

#### nasi112

##### Junior Member
Yes, it explains why you will get the right answer doing that.

My concern is that your teacher clearly expects you to take domain into account in your drawing.

Here is how I would do (c):
View attachment 26629
Dr.Peterson, why do you care about quadrant IV while the question for (c) pointed that the domain is $$\displaystyle \frac{\pi}{2} < x < \pi$$?

#### Dr.Peterson

##### Elite Member
Dr.Peterson, why do you care about quadrant IV while the question for (c) pointed that the domain is $$\displaystyle \frac{\pi}{2} < x < \pi$$?
Where did I put the angle x??? It is not in quadrant IV! It's right where it belongs, in quadrant II.

But $$\frac{\pi}{2}-x$$ is in quadrant IV.

#### nasi112

##### Junior Member
Where did I put the angle x??? It is not in quadrant IV! It's right where it belongs, in quadrant II.

But $$\frac{\pi}{2}-x$$ is in quadrant IV.
I am confused. I understand that the original cosine was in quadrant IV, but does the question require us to use both quadrants, II and IV? Then, why it motioned the domain? We can draw them both without the need to know the domain of the simplified one.

#### Dr.Peterson

##### Elite Member
I am confused. I understand that the original cosine was in quadrant IV, but does the question require us to use both quadrants, II and IV? Then, why it motioned the domain? We can draw them both without the need to know the domain of the simplified one.
This is what I said makes the problem odd! The requirement stated in the problem that we pay attention to the (really irrelevant) quadrant is what makes it excessively complicated, as I said from the start.

But it says to simplify

So we are required to assume x is in quadrant II. And given that $$\frac{\pi}{2}<x<\pi$$, it follows that $$\frac{\pi}{2}-\frac{\pi}{2}>\frac{\pi}{2}-x>\frac{\pi}{2}-\pi$$, i.e. $$0>\frac{\pi}{2}-x>-\frac{\pi}{2}$$, which means $$\frac{\pi}{2}-x$$ is in quadrant IV.

I have no control over that. What you question is forced by the problem. Why do you want to do otherwise?

So I started by drawing x in the indicated quadrant, and its reference triangle (red); then I reflected the whole thing over the 45 degree line, which subtracts x from $$\frac{\pi}{2}$$, and looked at the cosine of that angle, which is clearly the same as the sine of the original angle, that is, $$\sin(x)$$.

What I really don't understand is how the solution to (c) we were shown, which is supposedly what was taught, has anything to do with the problem.

#### nasi112

##### Junior Member
This is what I said makes the problem odd! The requirement stated in the problem that we pay attention to the (really irrelevant) quadrant is what makes it excessively complicated, as I said from the start.

But it says to simplify
View attachment 26631

So we are required to assume x is in quadrant II. And given that $$\frac{\pi}{2}<x<\pi$$, it follows that $$\frac{\pi}{2}-\frac{\pi}{2}>\frac{\pi}{2}-x>\frac{\pi}{2}-\pi$$, i.e. $$0>\frac{\pi}{2}-x>-\frac{\pi}{2}$$, which means $$\frac{\pi}{2}-x$$ is in quadrant IV.

I have no control over that. What you question is forced by the problem. Why do you want to do otherwise?

So I started by drawing x in the indicated quadrant, and its reference triangle (red); then I reflected the whole thing over the 45 degree line, which subtracts x from $$\frac{\pi}{2}$$, and looked at the cosine of that angle, which is clearly the same as the sine of the original angle, that is, $$\sin(x)$$.

What I really don't understand is how the solution to (c) we were shown, which is supposedly what was taught, has anything to do with the problem.
Thanks for the explanation, Dr.Peterson. This is why, I prefer to see the answer of one similar problem in the book, then I proceed with the others in a similar way. But even in your case, I would never draw anything in Quadrant II since solving the angle would give me Quandrant IV! The question did not ask to draw, but if it has asked, I would draw only the one in Quadrant IV.

#### eutas1

##### Junior Member
Yes, it explains why you will get the right answer doing that.

My concern is that your teacher clearly expects you to take domain into account in your drawing.

Here is how I would do (c):
View attachment 26629
I don't understand your solution to (c)... Could you please explain step-by-step?

#### Dr.Peterson

##### Elite Member
I don't understand your solution to (c)... Could you please explain step-by-step?
I just did. Read post #16, take time to think about it, and tell me which parts you don't understand, and why.

Then please explain in words what you did in your solution to (c), because I have the same problem!

#### eutas1

##### Junior Member
I just did. Read post #16, take time to think about it, and tell me which parts you don't understand, and why.

Then please explain in words what you did in your solution to (c), because I have the same problem!
Wait am I explaining what I did in attachment T? Or the way I did it originally?