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Math Help - Phase Shift

  1. #1
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    Phase Shift

    I don't understand how to put the coordinates for sin and cos. All I know is that c/b is the phase shift.
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  2. #2
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    Quote Originally Posted by thisisamazing View Post
    I don't understand how to put the coordinates for sin and cos. All I know is that c/b is the phase shift.
    Try posting a question you are trying to solve and then we will be able to help you...
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  3. #3
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    ok for example:

    Sin(x+pi/2)

    or

    -1/2cos(4x+pi)
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  4. #4
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    Hello thisisamazing
    Quote Originally Posted by thisisamazing View Post
    ok for example:

    Sin(x+pi/2)

    or

    -1/2cos(4x+pi)
    First, try thinking about the value of x that makes the expression in the brackets zero. This will tell you how far the 'basic' graph has been moved. So for example, in
    y = \sin(x+\pi/2)
    the expression x +\pi/2 is zero when x = -\pi/2. So, instead of 'starting' at (0,0) the sine graph has been moved to 'start' at (-\pi/2,0); i.e. it has been moved a distance of \pi/2 to the left.

    (You'll see that I've put the word 'start' in quotes. That's because, of course, the graph doesn't actually start there; it extends infinitely far in each direction. But the cycle of the graph that we usually look at will start there.)

    In the second example you give, y = -\tfrac12\cos(4x+\pi):
    4x + \pi =0
    when
    x=-\pi/4
    So the 'basic' cosine graph has been moved \pi/4 to the left to 'start' at x=-\pi/4.

    However, a couple of other things have happened here as well:

    • x is multiplied by 4, which means that things happen 4 times as fast along the x-axis as they will on the 'basic' graph. So instead of requiring values of x from 0 to 2\pi to make a complete cycle, you'll only need values from 0 to \pi/2.


    • The cosine expression has then been multiplied by -\tfrac12. This, in turn, does two things:


    • The minus sign flips the graph over, reflecting it in the x-axis.


    • The factor of \tfrac12 reduces the y-values to one-half of their original values, 'squashing' the graph down, so that, instead of varying between -1 and +1, it will now vary between -0.5 and +0.5.
    Does that help to make it clear?

    Grandad
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  5. #5
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    I may not be OP, but Grandad, that was so easy to understand and super helpful for me.

    I would like to double-check though. I recently found out about doubleangle formulae and different ways of writing sin(A+B)\:,\:cos(A+B)

    Would y=sin(x+\frac{\pi}{2}) be the same as

    y=sin\:x\:cos\:\frac{\pi}{2}+cos\:x\:sin\:\frac{\p  i}{2}

    and \frac{\pi}{2} being 90 degrees,

    cos90=0\:,\:sin90=1\:,\:y=cos\:x
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  6. #6
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    Hello davidman
    Quote Originally Posted by davidman View Post
    I may not be OP, but Grandad, that was so easy to understand and super helpful for me.

    I would like to double-check though. I recently found out about doubleangle formulae and different ways of writing sin(A+B)\:,\:cos(A+B)

    Would y=sin(x+\frac{\pi}{2}) be the same as

    y=sin\:x\:cos\:\frac{\pi}{2}+cos\:x\:sin\:\frac{\p  i}{2}

    and \frac{\pi}{2} being 90 degrees,

    cos90=0\:,\:sin90=1\:,\:y=cos\:x
    Yes it would; and the fact that \sin(x+\tfrac{\pi}{2}) = \cos x means that the cosine graph is just the same as the sine graph after it's been shifted \tfrac{\pi}{2} to the left.

    Using that formula and a variation of it (namely \sin(A-B) = \sin A \cos B - \cos A\sin B), you can also show that (for example):
    \sin(\tfrac{\pi}{2}-x) = \cos x

    \sin(\pi - x) = \sin x

    \sin(\pi + x) = -\sin x

    ... and so on.
    Grandad
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