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Forgot how to graph sin graphs(translation)

  1. Nov 22, 2008 #1
    Ok so i have a function
    cos4/3x between 0<= X <= 2π
    the amp is 1, max is 1 and min is -1.
    the period is 3π/2.

    normally in a cos graph, π/2=0, π=-1, 3π/2=0 and 2π=1.
    but since the period is 3π/2, i know the 1 is now at 3π/2. But Im getting confused as to where the 0s and the negative -1 would be at.
  2. jcsd
  3. Nov 22, 2008 #2


    Staff: Mentor

    No, π/2!=0, π != 1, and so on, but cos(π/2) = 0, and cos(π) = -1.

    The effect of the multiplier 4/3 in cos((4/3)x) is to compress the graph toward the y-axis so that each point on the compressed graph is 3/4 as far away from the y-axis as that of the normal cosine graph.

    As you said, the period of your function is 3π/2. cos((4/3)0) = cos(0) = 1, and cos((4/3)(3π/2)) = cos(2π) = 1, so those are the high points at each end of one period of the graph. The graph crosses the x-axis halfway between those points. You should be able to fill in the rest of one period of this graph, and from that, extend the graph for the interval you're given.
  4. Nov 23, 2008 #3
    Since the period is 3π/2, you should draw the cos graph in the interval (0,3π/2). So, the points should be:
    cos(3π/8)=0 (this is п/2 in y=cosx)
    cos(3π/4)=-1 (since you draw half of the period, and that is п in y=cosx)
    cos((3п/2+3п/4)/2)=cos(9п/8)=0 (3п/2 in y=cosx)
    cos(3π/2)=1 (2п in y=cosx)
    At the beginning you should point half of the period, and again one more half of the half.

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