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Math Help - Moving a Trigonometric function

  1. #1
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    Moving a Trigonometric function

    y=d+sin(bx+c)

    So if I have the equation:
    sin(2x)=y; I know that the period will be pi. So the whole graph will occur between (0,0) and (pi,0)
    Start at (0,0), Maximum (pi/4, 1); intercept at (pi/2, 0); Minimum (3Pi/4, -1); Intercept (pi,0)

    Now if I have:
    y=sin(x-(pi/4)) the graph gets shifted to the right pi/4 radians. So key points are
    Start: (pi/4, 0); Maximum: (3pi/4, 1); Intercept: (5pi/4,0); minimum (7pi/4, -1); Complete (9pi/4,0)

    If I put them together as Sin(2x-(pi/4)
    Why does my horizontal shift get cut in half? Why does it not start at (pi/4)? Why does it start at (pi/8)? In my graphing calculator I see everything gets moved my (pi/8) when I have this equation.
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  2. #2
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    Quote Originally Posted by IDontunderstand View Post
    y=d+sin(bx+c)

    So if I have the equation:
    sin(2x)=y; I know that the period will be pi. So the whole graph will occur between (0,0) and (pi,0)
    Start at (0,0), Maximum (pi/4, 1); intercept at (pi/2, 0); Minimum (3Pi/4, -1); Intercept (pi,0)

    Now if I have:
    y=sin(x-(pi/4)) the graph gets shifted to the right pi/4 radians. So key points are
    Start: (pi/4, 0); Maximum: (3pi/4, 1); Intercept: (5pi/4,0); minimum (7pi/4, -1); Complete (9pi/4,0)

    If I put them together as Sin(2x-(pi/4)
    Why does my horizontal shift get cut in half? Why does it not start at (pi/4)? Why does it start at (pi/8)? In my graphing calculator I see everything gets moved my (pi/8) when I have this equation.
    because ... y = \sin\left[2\left(x - \frac{\pi}{8}\right)\right]
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  3. #3
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    Thanks, that makes perfect sence. I did not see it that way. Thank you so much.
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