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Homework Help: The Unit Circle

  1. Oct 22, 2008 #1
    1. The problem statement, all variables and given/known data

    This isnt exactly a problem. Also, it is only the first section of the chapter on trig, and this has nothing to do with trig, per se. I think it is more of like a primer for what is to come. So, whomever reads this, please try to imagine you know nothing about sin, cos, angles or anything like that. this is much more basic, which is why im having trouble finding it on the internet. plus, in my book, they give only one example and im more interested in the why, rather than memorising a problem. ok!

    It is about "the reference number." The book says let t be a real number. the reference number t associated with t is the shortest distance along the unit circle between the terminal point determined by t and the x axis.

    2. Relevant equations

    find the reference number for each value of t: (i thought there used to be a thing i could click on to put in math symbols? i cant find it now, so sorry about the annoying pi)

    1. t= 5pi/6
    2. t=7pi/4
    3. t= -2pi/3
    4. t= -3
    5. t= 6
    6. t= -7

    3. The attempt at a solution
    i seem to understand everything with pi.
    reference t for 5pi/6 is pi/6 because 5pi/6 is closest to the x axis at pi so i did pi - 5pi/6 = pi/6. so pi/6 is the reference t. is that the idea?
    then #2 is 2pi - 7pi/4 = pi/4.
    #3 is pi - 2pi/3 = pi/3.
    #4 started to confuse me, i think because of the negative. -3 is closest to the x axis at pi (going clockwise) so is it pi - 3? which is guess would be .14. would it ever be 3 - pi?
    #5 is closest to x at 2pi so 2pi - 6.
    #6 is closest to x at 2pi(going clockwise) so would it be 2pi - 7 ? at -.72?

    am i doing this correctly? i mean, i see my answers at least for the ones with pi are correct, but i want to make sure i am doing it right because i am thinking about it right.
  2. jcsd
  3. Oct 22, 2008 #2
    I must admit that I find it hard to make sense of your post :redface:

    However, maybe this will help your understanding of trig and on page two, there's a complete unit circle: Trig Cheat Sheet

    Hope it's a start.
  4. Oct 22, 2008 #3
    yes, when i did a search, a post with the same link came up. i bookmarked that site for future reference, but its still 2 sections ahead of what i am doing right now. i had a feeling the post might be confusing, as i couldnt find any information on it on the internet, so that must mean these types of questions are obscure. well, if they are obscure, when maybe i dont need to be spending so much time on them. ill try to make sense out of what my recitation teacher garbles when i ask him tomorrow. thanks.
  5. Nov 23, 2008 #4
  6. Nov 23, 2008 #5
    Ok, I know this is late as well, but I believe I understand what you are trying to ask. I remember them being called reference angles, but basically you are dividing your unit circle up into quarters to make it easier on you, sector I (0 to pi/2), II (pi/2 to pi), III (pi to 3pi/2), and IV (3pi/2 to 2pi), and finding the smallest angle to the x-axis. These reference angles become important later because angles 5pi/3, 4pi/3, 2pi/3 and pi/3 are different but they have the same reference angles (pi/3) and as you'll notice the absolute value of the trig functions will be the same for all four angles, [tex]sin = \frac{\sqrt{3}}{2}[/tex], [tex]cos = \frac{1}{2}[/tex], and [tex]tan = \sqrt{3}[/tex].

    The sum it up, I think all your answers were right except for the last one. That is in sector IV, so you were right in taking 2pi - 7, but you are not done because the angle was greater than 2pi. A reference angle will always be between 0 and pi/2. You need to go another step get 0.72.
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