Is there something I'm missing about this unit circle thing..?

  • #1
JR Sauerland
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ImageUploadedByPhysics Forums1438184151.524020.jpg


I get the dynamics of what the question is asking. What I don't get is how to know what pi/8 is. What I've been doing this entire time is plugging it into the calculator as a fraction, and getting the decimal value of it and comparing it to other values I know.

I know that pi radians = 180 degrees, if you divide both sides by two, pi/2 radians = 90 degrees. Then, I know that if I divide pi by 2 in a calculator and get the decimal value and compare it to the decimal value of pi/8, I know if it falls within the first quadrant or not.

But is there an easier or simpler way to do this? Again, I'm just focusing on pi over 8, and how they know precisely where it falls.
 

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  • #2
phinds
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But is there an easier or simpler way to do this? Again, I'm just focusing on pi over 8, and how they know precisely where it falls.
Can you figure out where 180/8 falls?
 
  • #3
gleem
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If there are 2π rad in a circle, then what fraction of a circle is π/8?
 
  • #4
Mentallic
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pi radians are 180 degrees, so pi/2 is 90 degrees, which is a right angle, hence anything smaller is an acute angle. pi/8 is 1/4 of pi/2 since (pi/2)/4 = pi/8, and since pi/2 is a right angle, then pi/8 is 1/4 of a right angle, which is pretty much what's been shown in that picture.
 
  • #5
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You're on the right track. I think the piece you're missing is that the y value of any point on the unit circle is the sin(angle).

Hence D is correct because of that. However C is also correct do you see why?
 
  • #6
JR Sauerland
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Wow. I never even thought about it that way! See, I've been rushing trigonometry, and studying it in such high volumes that I miss out on such small facets of mathematics that are totally valuable. 1/4th of pi/2 is way easier to understand so thank you.

On another question however, I was thrown for a real loop.
ImageUploadedByPhysics Forums1438186110.008911.jpg

This is the statement of the problem, and my answers I have selected. Given the problem, I know sine on the unit circle refers to the y-values. I ruled out sin(3) because it would have brought it to rest on -1, while the other two lead right back to the exact same spot. However the question threw me for a loop in the way that it expected me to know somehow that cos(4pi) is I guess the same?
ImageUploadedByPhysics Forums1438186287.313240.jpg

Here is the explanation of why I got the problem wrong. I still don't really know what it's talking about or what the explanation means. It states "B's y coordinate is different from A's, but it's x-coordinate is the same, 1. Therefore cos(4pi) is also a correct answer."

For these types of problems I've trained my mind to understand that when the problem asks for the y-coordinate of point-A, I assess that sine is associated with the y, so I look for y. As such, I fail to understand why the cosine option even works.
 
  • #7
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The cos(angle) on a unit circle is related to the x-values of the point. cos(0) = cos(2pi) = cos(2*2pi) = 1

I think in your problem, they want to know the value only so they show graphically a value of one in the y direction and so they are asking for what trig functions of the four listed have a value of 1.

sin( 2 *(1/2)pi ) = sin(pi) =0 and similarly for sin(2 pi) = 0 so they are the right answers but what about B and C?
 
  • #8
JR Sauerland
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The cos(angle) on a unit circle is related to the x-values of the point. cos(0) = cos(2pi) = cos(2*2pi) = 1

I think in your problem, they want to know the value only so they show graphically a value of one in the y direction and so they are asking for what trig functions of the four listed have a value of 1.

sin( 2 *(1/2)pi ) = sin(pi) =0 and similarly for sin(2 pi) = 0 so they are the right answers but what about B and C?
So they are literally asking which options return a value of 0? I know cos(4pi) brings it back to the exact same spot but 3 (1/2pi) brings it to the bottom Ray. Perhaps I can understand it as they want the values that bring it back to the exact same spot it was at?
 
  • #9
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JR,

It might be more helpful to look at the graphs of the sine and cosine so that comparing the domain and the codomain are easier to see graphically.


Remember the inputs (the x-axis) are mapped to their respective trig values. I find it helpful to see things in terms of functions, sequences, and ordered pairs:

##{F(x)_{sin}}_0 = (..., (0, 0),(2\pi, 0), (4\pi, 0), ... ) ##
##{F(x)_{cos}}_1 =(..., (0, 1),(2\pi, 1), (4\pi, 1)... \} ##

(input which is an angle measure in radians, output which is the height of the wave)
 

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  • #10
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So they are literally asking which options return a value of 0? I know cos(4pi) brings it back to the exact same spot but 3 (1/2pi) brings it to the bottom Ray. Perhaps I can understand it as they want the values that bring it back to the exact same spot it was at?
In this case, I think they are referring to the value only but fooling you by referring the the unit circle diagram.

Its like showing you two vectors pointing in different directions and then saying which two have the same magnitude.
 
  • #11
JR Sauerland
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In this case, I think they are referring to the value only but fooling you by referring the the unit circle diagram.

Its like showing you two vectors pointing in different directions and then saying which two have the same magnitude.
But this was on Khan Academy. I don't think a traditional course would ask a question like this anyway. Edit: you hit me with a physics example! :P magnitude is just the value of the vector quantity, and the vector is just speed in a direction?
 
  • #12
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Actually the vector example is just a math example.

Its the fact that you first hear about them in physics that leads you to think its a physics problem.

And you're right magnitude is the length of the vector and a vector encodes a magnitude and a direction.
 
  • #13
JR Sauerland
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As far as trigonometry goes though, should I be worried about this type of problem?
 
  • #14
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Add it to your list of difficult problems, move to the next ones and when done review your list of difficult problems. In that way, you'll cover the material and then can go fill in the gaps.
 

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