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Math Help - How to do: Phase Shift of tan

  1. #1
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    How to do: Phase Shift of tan

    I was wondering how do i do the phase shift for




    all i know is:


    ?? 3x ??
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  2. #2
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    Hello S2Krazy
    Quote Originally Posted by S2Krazy View Post
    I was wondering how do i do the phase shift for




    all i know is:


    ?? 3x ??
    The graph of y = \tan(\theta+ \tfrac{\pi}{6}) is the same as the graph of y =\tan(\theta), shifted a distance \tfrac{\pi}{6} to the left.

    If you want to see why, just look at the zero values.
    When \theta = 0, y=\tan(\theta)=\tan(0) = 0. So the graph 'starts' at (0,0).
    But when \theta=0, y=\tan(\theta+\tfrac{\pi}{6})=\tan(\tfrac{\pi}{6}). So this graph 'starts' at (0,\tfrac{\pi}{6}), and it's when \theta = -\tfrac{\pi}{6} that we get the value y = 0.
    Now if we multiply the independent variable by 3, and replace \theta with 3x, we get the graphs of y = \tan(3x) and y = \tan(3x+\tfrac{\pi}{6}). Now, in the second graph, we shall need to take 3x = -\tfrac{\pi}{6} to get our zero value of y. So that's x = -\tfrac{\pi}{18}. So the graph has now been shifted \tfrac{\pi}{18} to the left.

    Finally, note that multiplying by 5, to give y = 5\tan(3x+\tfrac{\pi}{6}), makes no difference to the phase-shift: it merely 'stretches' the graph in a vertical direction either side of the x-axis by a factor of 5.

    Grandad
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  3. #3
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    ah icic cool thxs.

    so for



    the phase shift would be -4pi/5 correct?
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  4. #4
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    Quote Originally Posted by S2Krazy View Post
    ah icic cool thxs.

    so for



    the phase shift would be -4pi/5 correct?
    Correct!

    Grandad
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