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Another trig problem

  1. Oct 18, 2013 #1
    1. The problem statement, all variables and given/known data

    Evaluate the following exactly.

    cos 300°
    sin(-120°)
    2. Relevant equations
    Unit circle


    3. The attempt at a solution
    I seem to have a problem with the remembering if it is going to be the position on the unit circle such as in these cases I thought it was
    cos 300°= 5∏/3 but it is actually I think 1/2 so I got the problem wrong
    then you have
    sin(-120°) = 4∏/3 but this is wrong and I think it is -√3 / 2 so yeah then I got this one right
    csc120°= 2√3 / 3
    I just never know when I am supposed to use which one for the exact value. Could someone help out please?
     
  2. jcsd
  3. Oct 18, 2013 #2

    Mark44

    Staff: Mentor

    You're really garbling up a bunch of different stuff. cos 300° ≠ 5∏/3, but 300° = 5∏/3. The reference angle for 5∏/3 is -∏/3, and this angle has the same cosine as ∏/3.

    Writing cos 300°= 5∏/3 is incorrect, but cos(300°)= cos(5∏/3) would be correct. Be mindful that the cosine of something is not the same as the something.
     
  4. Oct 18, 2013 #3
    So then why is arcsin(-1/2) = -Pi/6 Is the inverse just going back to the reference angle then?
     
  5. Oct 19, 2013 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, because the trig functions are periodic, they are not "one to one" and so do not have true "inverses". In order to talk about inverse for the trig functions, we have to restrict their domains. The standard convention for "sine" is to restrict to between [itex]-\pi/2[/itex] (-90 degrees) and [itex]\pi/2[/itex] (90 degrees). If x is positive, [itex]sin^{-1}(x)[/itex] is between 0 and [itex]\pi/2[/itex]. If x is negative, [itex]sin^{-1}(x)[/itex] is between [itex]-\pi/2[/itex] and 0.

    Personally, for "sin(300)" I would note that 300= 360- 60 and use "sin(a-b)= sin(a)cos(b)- cos(a)sin(b)" so that sin(300)= sin(360)cos(60)- cos(360)sin(60)=0- sin(60).
     
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