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Math Help - Help with tangent function

  1. #1
    Junior Member
    Joined
    Jun 2005
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    43

    Help with tangent function

    Hi, here's my example from my math book that I don't understand.

    Sketch the graph of y = -3 tan 2x.

    By solving the equations

    2x = - pi/2 and 2x = pi/2

    x = -pi/2 x = pi/4

    Here's the part that I have a problem understanding:

    when x = - pi/8
    - 3 tan 2x = 3

    when x = 0
    -3 tan 2x = 0

    when x = pi/8
    -3 tan 2x = -3


    Can I do this calculation without looking at a graph. I can perform a simple x and y cordinate using this technique but when it comes to tangent problems, I can't same to apply the same principle.

    Please help
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  2. #2
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Without looking at a graph?
    Sure.

    [Time will come, (when you are comfortable with Trigonometry already) when you don't have to look at graphs to analyze trig equations. Well, maybe seldom only.]

    y = -3 tan 2x

    ----------------
    when x = - pi/8,

    y = -3 tan 2(-pi/8)
    y = -3 tan -pi/4

    angle -pi/4 is in the 4th quadrant.
    Its reference angle is pi/4 (or 45 degrees). I assume you know by heart the tan(pi/4) or tan(45deg). It is 1.
    Since -pi/4 is in the 4th quadrant, where tangent is negative (you know that?), then,

    y = -3 *(-1)
    y = 3 ----answer

    ---------------------
    when x = 0

    y = -3 tan 2*0
    y = -3 tan 0

    tan(0) is 0, so,

    y = -3 * 0
    y = 0 -----answer.

    -------------------------
    when x = pi/8

    y = -3 tan 2(pi/8)
    y = -3 tan pi/4

    Angle pi/4 (or 45 deg) is in the 1st quadrant.
    You know that tan(pi/4) or tan(45deg) is 1.
    So,

    y = -3 *(1)
    y = -3 ----answer
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