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Algebra II Trigonometry Circular Functions

  1. Nov 20, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the exact values of the six trigonometric function of the given number. If a function is undefined for the number, say so.

    11[itex]\pi[/itex]/6


    2. Relevant equations



    3. The attempt at a solution

    The way I did it was to turn it into radians by simply doing the math...
    5.76 radians
    Then, I multiplied that by 57.3 to get degrees.
    330 degrees
    After that, I double checked it against -30 degrees so that I could use the table.

    I didn't have an issue with getting the correct answers. My issue is with the form.

    The answers weren't decimals.
    Example:
    sin = - 0.5000
    cos = 0.8660
    tan = - 0.5774

    They were left in fraction form.
    Example:
    sin = - 1/2
    cos = [itex]\sqrt{3}[/itex]/2
    tan = - [itex]\sqrt{3}[/itex]/3

    I have a feeling that my exams want me to answer in the fractional form. And, I don't have an idea of how to do that. Please forgive me if it is obvious.

    Also, sorry that my LaTex is rather sloppy. Apparently, the board has changed since I've been away.

    Thanks in advance for the help. :D
     
  2. jcsd
  3. Nov 20, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I can't imagine why you would do all that. The point of this exersize is to use what you know about the trig functions to get exact values, not to just to plug numbers into a calculator to get approximate answers.

    Just notice that [itex]11\pi/6= (12pi- \pi)/6= 2\pi- \pi/6[/itex] so, because sine and cosine have period [itex]2\pi[/itex], [itex]sin(11\pi/6)= sin(-\pi/6)[/itex] and [itex]cos(11\pi/6)= cos(-\pi/6)[/itex]. And since sine is an odd function and cosine is an even function those are equal to [itex]-sin(\pi/6)[/itex] and [itex]cos(\pi./6)[/itex].

    If you don't know, off hand, what those are, imagine an equilateral triangle with each side of length 2. Each angle has measure [itex]\pi/3[/itex] and if you draw a perpendicular from one vertex to the opposite side, you have a right triangle with angle [itex]\pi/6[/itex], hypotenuse of length 2 and "opposite side" of length 1. What is the length of the "near side" and what are [itex]sin(\pi/6)[/itex] and [itex]cos(\pi/6)[/itex]?
     
  4. Nov 24, 2012 #3
    @HallsofIvy: Thank you so much for your help. In particular, the equilateral triangle explanation really enabled my understanding. I drew everything out, figured that one of the legs is '1' because the perpendicular line bisects that side, and from the Pythagorean Theorem that the other leg of the triangle is [itex]\sqrt{3}[/itex]. Then, from there, I was able to just pay attention to sine and cosine. Now, I understand that in the example you gave me sine is 1/2 and cosine is [itex]\sqrt{3}/2[/itex].

    The visual of drawing out the triangle really was the key to me understanding and getting the correct answers in the correct form for this particular problem, but I also had a problem [itex]-5\pi/6[/itex] which I solved the same way as in my earlier post and got the correct answers but in the wrong form. How do I apply what you have taught me to all problems of this nature?
     
  5. Nov 24, 2012 #4
    Why don't you make use of the simple identities [itex]\cos(2\pi-x)=\cos(2\pi)\cos(x)+\sin(2\pi)\sin(x)=\cos(x)[/itex] and [itex]\sin(2\pi-x)=\sin(2\pi)\cos(x)-\cos(2\pi)\sin(x)=-\sin(x)[/itex] Also, [itex]\cos(-x)=\cos(x)[/itex] and [itex]\sin(-x)=-\sin(x)[/itex]. These are very well known properties.
     
  6. Nov 26, 2012 #5
    @Millennial: Hm, I haven't learned that yet or I somehow overlooked it. Today I'm going to try and go back to review in my textbook anyways, so I'll keep my eye out for that.

    Could you give me an example using one of the identities or point me in the direction of a website (or thread on on PF) that has examples of it? I normally need to see an identity in action to understand how to apply it correctly.

    Thank you for your help.
     
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