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Thread: 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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    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 $\displaystyle x$ that makes the expression in the brackets zero. This will tell you how far the 'basic' graph has been moved. So for example, in
    $\displaystyle y = \sin(x+\pi/2)$
    the expression $\displaystyle x +\pi/2$ is zero when $\displaystyle x = -\pi/2$. So, instead of 'starting' at $\displaystyle (0,0)$ the sine graph has been moved to 'start' at $\displaystyle (-\pi/2,0)$; i.e. it has been moved a distance of $\displaystyle \pi/2$ 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, $\displaystyle y = -\tfrac12\cos(4x+\pi)$:
    $\displaystyle 4x + \pi =0$
    when
    $\displaystyle x=-\pi/4$
    So the 'basic' cosine graph has been moved $\displaystyle \pi/4$ to the left to 'start' at $\displaystyle x=-\pi/4$.

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

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


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


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


    • The factor of $\displaystyle \tfrac12$ reduces the $\displaystyle y$-values to one-half of their original values, 'squashing' the graph down, so that, instead of varying between $\displaystyle -1$ and $\displaystyle +1$, it will now vary between -0.5 and $\displaystyle +0.5$.
    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 $\displaystyle sin(A+B)\:,\:cos(A+B)$

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

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

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

    $\displaystyle cos90=0\:,\:sin90=1\:,\:y=cos\:x$
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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 $\displaystyle sin(A+B)\:,\:cos(A+B)$

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

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

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

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

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

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

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

    ... and so on.
    Grandad
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