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Homework Help: How to find θ by knoving sin/cos

  1. Dec 9, 2008 #1
    1. The problem statement, all variables and given/known data

    Given that sin θ =1/2, cos θ =–((√3)/2) and 0° < θ < 360°,
    (a) find the value of θ;
    (b) write down the exact value of tan θ.

    2. Relevant equations

    A; sin(180-θ)=sin(θ) and cos(180-θ)=-cos(θ)

    3. The attempt at a solution
    Using an calculator left me with that θ=30° but i asked a friend of mine wich got it to be 150. Now she has b-level math while i have a-level math so she siad she used some short of tabel to find the answer and that high level where supposed some furmula to find it... now i suspect that furmula A (wich i found in my course book)has something to do with this but i'm unnable (tired maybe:smile:) to wrap my head around it.

    as to proplem b i think the sullution is
    cos(θ)/sin(θ)=tan(θ) or opposit but i dont remember wich way it is...

    Please help and thanks in advance:smile:

    Youre my heroes :-p

    Edit:just uploaded the original homework document just to be sure :-)

    Attached Files:

    Last edited: Dec 9, 2008
  2. jcsd
  3. Dec 9, 2008 #2
    I think you've got issues with your math (at least, they're not showing up on Firefox for me). Can you retype your sine and cosine values?
  4. Dec 9, 2008 #3


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    Hi hostergaard! :smile:

    I'm not sure what you're doing (not all the question is visible),

    but that equation should be cos(180-θ) = -cos(θ)

    (from the formula cos(180-θ) = cos180cosθ + sin180sinθ :wink:)
  5. Dec 9, 2008 #4
    ups... sorry all done! :-) (with editing that is, not my homework)
    if you can help me with some of the other questions in my assigment that would be apreciatet, but i will post those wich i dont understand, So it's not to neccesary
    Last edited: Dec 9, 2008
  6. Dec 9, 2008 #5
    Okay, so let's consider your problem now. First, look at the sine values. This gives a few possibilities for theta, namely pi/6 and 5pi/6. Which of these values of theta give the appropriate cosine value?

    Your second part looks good!

    EDIT: Whoops! Sorry, forgot you were dealing with degrees. pi/6 radians is 30 degrees, and 5pi/6 radians is 150 degrees.
  7. Dec 9, 2008 #6
    pingpong, i'm not really sure if i understod that (english is not my main language, sorry) arent there a formula i can use?
  8. Dec 9, 2008 #7
    To reformulate that; exactly what should i write as the answer?
    i think i can understand it then :-p

    thanks for the help!
    apreciate it!
  9. Dec 9, 2008 #8


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    Hi hostergaard!

    (btw, i never open .doc files :wink:)

    ah, i can see it now … sin θ =1/2, cos θ =–((√3)/2) :wink:

    ok, this is a quadrants thing … you'd better learn which quadrants sin and cos are positive or negative in …

    sin is +ve in the 1st and 2nd quadrants; cos is +ve in the 1st and 4th quadrants …

    does that help? :smile:
  10. Dec 9, 2008 #9
    Okay, let's try to go slow. If you don't understand what I'm saying, let me know what doesn't make sense. There's not really just one formula you can use, it just takes some reasoning out.

    If [itex]\sin(\theta)=1/2[/itex], then what does [itex]\theta[/itex] have to be? Either [itex]\theta=30[/itex] or [itex]\theta=150[/itex] (remember, [itex]\sin(150)=\sin(180-30)=\sin(30)[/itex].

    Which of these values gives [itex]\cos(\theta)=-\sqrt{3}/2[/itex]? Well, we know that [itex]\cos(30)=\sqrt{3}/2[/itex], so it's not going to be 30 degrees. On the other hand, [itex]\cos(150)=\cos(180-30)=-\cos(30)=-\sqrt{3}/2[/itex], which is what we want.

    Hopefully that makes sense.
  11. Dec 9, 2008 #10
    oohh, now i understand the logic! (i can be so dense when i'm tired :-p)
    ok, so i find that θ is either 30 or 150 by saying sin^-1(1/2)
    Ohh, i just rember when i use calculator i only get one answer theres suposed to two.
  12. Dec 9, 2008 #11
    Really many thanks for the help! you guys really saved my skin :-)
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