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Confused on what should be negative when finding with half angle identities

  1. Apr 4, 2012 #1
    1. The problem statement, all variables and given/known data
    The question is to find [itex]sin 2x, cos 2x, tan 2x[/itex] from the given information: [itex]sin x = -\frac{3}{5}[/itex], x in Quadrant III


    2. Relevant equations
    Half Angle Identities
    [itex]cos2x = cos^{2}x - sin^{2}x[/itex]

    [itex]sin2x = 2sinxcosx[/itex]

    [itex]tan2x = \frac{2tanx}{2-tan^{2}x}[/itex]


    3. The attempt at a solution
    I can find the solution for the most part, the only thing I can't figure out are the signs. What I do is use the given sin[itex](-\frac{3}{5})[/itex] to make a right triangle and solve for the unknown side. I then use that triangle to set up the 3 half angle identities and just plug in the numbers. I can do all that fine, but I can't figure out what should be negative and positive. I thought that since it is in Quadrant 3 both sin2x and cos2x should end up negative. However the back of the book answers say that they are all positive. Why?
     
  2. jcsd
  3. Apr 4, 2012 #2

    rock.freak667

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    Homework Helper

    When you draw the triangle in quadrant 3, you should see that the opposite side is negative and the adjacent side (all to x) is negative as well. As the hypotenuse is always positive, you can understand why sin2x is positive.

    If you don't understand, draw the triangle within the 3rd quadrant, put in the appropriate signs and then write down what sinx and cosx are. Post back if you are still confused.
     
  4. Apr 4, 2012 #3
    I don't understand why the hypotenuse is always positive..or really why we can say any of these distances are negative. I must be missing something. I tried to convey my thoughts/procedure through this (ugly photoshop) image:
    In7vk.jpg

    Basically I'm just setting up a triangle. But why couldn't the hypotenuse be negative and the two sides positive?
     
  5. Apr 5, 2012 #4
    You don't have to use identities, there's and easier way. Sin is -3/5, so you know the opp side is 3 and the hyp is 5. Think of the numbers as lengths, and you cannot have negative length, but the SIN and COS can the negative if in the quadrants like you said. Use the Pythagorean theorem to solve for the unknown side, then just jot down the six trug functions.

    Sin is negative when it is moving down (below the x-axis, so quadrant 3 & 4).
    Cosine is negative when moving to the left (quadrant 2 & 3)

    Tangent = sin/cos
    So -sin/cos = -tan
    sin / -cos = -tan
    sin / cos = tan
    -sin / - cos = tan

    "The question is to find sin2x,cos2x,tan2x from the given information: sinx=−35, x in Quadrant III"

    Do you mean Sin2(x)? Because anything real squared is positive, in which case the book is right.
     
    Last edited: Apr 5, 2012
  6. Apr 5, 2012 #5
    If you double all the angles in 3rd. quadrant, the answers will be all in 1st. quadrant(all positive)
     
  7. Apr 5, 2012 #6
    By substituting you can see all are positive.
    If you double the angle, approximately 217° to 434°, all will be in first quadrant.
     
  8. Apr 5, 2012 #7
    I think I see what I messed up now. I left off a negative sign on the horizontal side's measurement. I'm going to practice some more of these today and hopefully will become more comfortable with them. (although most of it is just knowing the formulas) Thanks for all the help.
     
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