# Phase Shift

• January 25th 2010, 05:39 PM
thisisamazing
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.
• January 25th 2010, 06:50 PM
Prove It
Quote:

Originally Posted by thisisamazing
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...
• January 25th 2010, 07:52 PM
thisisamazing
ok for example:

Sin(x+pi/2)

or

-1/2cos(4x+pi)
• January 25th 2010, 10:40 PM
Hello thisisamazing
Quote:

Originally Posted by thisisamazing
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?

• January 26th 2010, 08:27 AM
davidman
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,

$cos90=0\:,\:sin90=1\:,\:y=cos\:x$
• January 26th 2010, 09:34 AM
Hello davidman
Quote:

Originally Posted by davidman
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.