# Radians and the Unit Circle

#### nasi112

##### Junior Member
Wait am I explaining what I did in attachment T? Or the way I did it originally?
Can I ask you eutas1? Drawing is not required in the problem. Why do you draw?

#### pka

##### Elite Member
I don't understand your solution to (c)... Could you please explain step-by-step?
Assuming that by (c) you mean part (c) as posted in reply #1: 18(c) $$\cos\left(\dfrac{\pi}{2}-x\right)$$ given $$x\in \left(\dfrac{\pi}{2},\pi\right)$$.
In my opinion the given is totally extraneous. $$\cos(a-b)=sin(a) sin(b) + cos(a) cos(b)$$
We should know that $$\cos\left(\dfrac{\pi}{2}\right)=0~\&~\sin\left(\dfrac{\pi}{2}\right)=1$$.
Taken together we get simply $$\cos\left(\dfrac{\pi}{2}-x\right)=\sin(x)$$. SEE HERE

#### eutas1

##### Junior Member
Can I ask you eutas1? Drawing is not required in the problem. Why do you draw?
We have learnt to draw for simple questions, and it makes it easier to visually see what is going on.

#### eutas1

##### Junior Member
Assuming that by (c) you mean part (c) as posted in reply #1: 18(c) $$\cos\left(\dfrac{\pi}{2}-x\right)$$ given $$x\in \left(\dfrac{\pi}{2},\pi\right)$$.
In my opinion the given is totally extraneous. $$\cos(a-b)=sin(a) sin(b) + cos(a) cos(b)$$
We should know that $$\cos\left(\dfrac{\pi}{2}\right)=0~\&~\sin\left(\dfrac{\pi}{2}\right)=1$$.
Taken together we get simply $$\cos\left(\dfrac{\pi}{2}-x\right)=\sin(x)$$. SEE HERE
I have not learnt angle sum and difference identities...

#### Dr.Peterson

##### Elite Member
Wait am I explaining what I did in attachment T? Or the way I did it originally?
I want to know what you are being taught (and others want to know, too!). You said that what you showed in attachment T was what the teacher showed you for (c), so I want to know what was said about it. I don't care what you did earlier that was wrong; I want to understand what you were told is right.

My understanding is that you are early in the process of learning about trig functions in terms of the unit circle, which is why a visual approach is appropriate; there is a lot you haven't learned yet. I think this problem goes too far with that, but I have certainly taught such methods, using reference angles and a triangle, to explain basic ideas such as symmetries in the functions.

#### Dr.Peterson

##### Elite Member
Can I ask you eutas1? Drawing is not required in the problem. Why do you draw?
My understanding is that, in context, the problem expects this particular approach.

In fact, the reason for the specified domain would be to tell the student what quadrant to draw the angle in; sometimes you can draw an angle in the first quadrant and assume it works in all quadrants, but in fact it doesn't. When they learn more, they will learn that these particular identities are not restricted to certain quadrants, while some others are.

#### eutas1

##### Junior Member
I want to know what you are being taught (and others want to know, too!). You said that what you showed in attachment T was what the teacher showed you for (c), so I want to know what was said about it. I don't care what you did earlier that was wrong; I want to understand what you were told is right.

My understanding is that you are early in the process of learning about trig functions in terms of the unit circle, which is why a visual approach is appropriate; there is a lot you haven't learned yet. I think this problem goes too far with that, but I have certainly taught such methods, using reference angles and a triangle, to explain basic ideas such as symmetries in the functions.

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#### Dr.Peterson

##### Elite Member
The trouble is, this "x" is not in the right domain; I can only guess that it is not really meant to be the x of the problem. It appears to be the reference angle all the way through.

#### eutas1

##### Junior Member
The trouble is, this "x" is not in the right domain; I can only guess that it is not really meant to be the x of the problem. It appears to be the reference angle all the way through.
Oh yeah... does that mean it is wrong?
And why is x not in the right domain? Isn't the domain (pi/2, pi)?

#### Dr.Peterson

##### Elite Member
Oh yeah... does that mean it is wrong?
And why is x not in the right domain? Isn't the domain (pi/2, pi)?
Isn't it clear that the angle marked as x is acute?? The angle that is in the right domain is the unmarked angle from the positive x axis to the final yellow line. (A line is not an angle.)

My sense is that your teacher is not writing what they mean. What they mean, and perhaps even what they say to you in explaining it, may be right, but a math teacher really should teach how to write what you mean. The picture does not communicate truth.

#### eutas1

##### Junior Member
Isn't it clear that the angle marked as x is acute?? The angle that is in the right domain is the unmarked angle from the positive x axis to the final yellow line. (A line is not an angle.)

My sense is that your teacher is not writing what they mean. What they mean, and perhaps even what they say to you in explaining it, may be right, but a math teacher really should teach how to write what you mean. The picture does not communicate truth.
Ahhhhhhhhh I am not sure what to do............

#### JeffM

##### Elite Member
This problem is EASY if you use basic identities. Why are you even discussing reference angles?

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

Why is that relevant? Because

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

$$\displaystyle cos(\phi + k \pi) = - cos(\phi) \text { for any integer } k.$$

$$\displaystyle - 1 \text { is an integer.}$$

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

Do the signs make sense? Now we can use the unit circle to check for reasonableness. Is it not obvious that if x is in the first quadrant the cosine is non-negative and that therefore the additive inverse of that cosine is non-positive. Moreover

$$\displaystyle 0 \le x \le \dfrac{\pi}{2}\implies - \dfrac{\pi}{2} \le - x \le 0 \implies \pi \le \dfrac{3 \pi }{2} - x \le \dfrac{3 \pi}{2}.$$

Thus we are in the third quadrant, and the sine is non-positive.

#### JeffM

##### Elite Member
By the way, radians are not relevant to this logic.

$$\displaystyle 0 \le x \le 90 \implies - 90 \le - x \le 0 \implies 180 \le 270 - x \le 270.$$

So the sine is non-positive.

$$\displaystyle sin (\theta ) = cos(90 - \theta)$$

$$\displaystyle \therefore sin (270 - x) = cos\{90 - (270 - x)\} = cos(x - 180) = - cos(x).$$

The reason to learn radians is not to get different results in trig, but to make calculus less messy.

#### eutas1

##### Junior Member
This problem is EASY if you use basic identities. Why are you even discussing reference angles?

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

Why is that relevant? Because

$$\displaystyle sin \left ( \dfrac{3 \pi}{2} - x \right ) = cos \left \{ \dfrac{\pi}{2} - \left ( \dfrac{3 \pi}{2} - x \right ) \right \} = cos(x - \pi).$$
I don't get the part after 'Because' ....

#### JeffM

##### Elite Member
I don't get the part after 'Because' ....
Do you agree that

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

$$\displaystyle \text {Let } \theta = \dfrac{3 \pi}{2} - x \implies$$

$$\displaystyle \dfrac{\pi}{2} - \theta = \dfrac{\pi}{2} - \left ( \dfrac{3 \pi}{2} - x \right ) = \dfrac{\pi(1 - 3)}{2} + x = x - \pi.$$

Basic algebra.

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

What is there not to understand?

#### eutas1

##### Junior Member
Do you agree that

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

$$\displaystyle \text {Let } \theta = \dfrac{3 \pi}{2} - x \implies$$

$$\displaystyle \dfrac{\pi}{2} - \theta = \dfrac{\pi}{2} - \left ( \dfrac{3 \pi}{2} - x \right ) = \dfrac{\pi(1 - 3)}{2} + x = x - \pi.$$

Basic algebra.

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

What is there not to understand?
Ohh, I see now.
So with
sin(θ) = cos(π/2 − θ) , theta can be angle? It does not have to be acute?

#### JeffM

##### Elite Member
$$\displaystyle cos \left ( \dfrac{\pi}{2} - \theta \right ) = cos \left ( \dfrac{\pi}{2} \right )cos ( \theta ) + sin \left ( \dfrac{\pi}{2} \right ) sin ( \theta) = 0 * cos (\theta) + 1 * sin(\theta ) = sin (\theta).$$

Do you see anything in that referring to acute angles?