Phase Shift in Sine Function

Date: 02/19/99 at 22:37:08
From: Corine Clark
Subject: Phase shift in sine function

I have a trigonometry test on Monday, and I do not understand phase 
shift in functions. The problem I am stuck on is y = sin(4x+pi/3). I 
have figured out that the amplitude is 1, the period is 90 degrees, 
and the phase shift is -15 degrees. I know how to graph this function 
without the shift, and I know the graph shifts to the left. What I 
do not know is how to move the function over correctly. Could you 
please show me an example of a shifted graph? 

Thanks for your time. I appreciate it!

Date: 02/20/99 at 15:31:27
From: Doctor Rick
Subject: Re: Phase shift in sine function

 
            *                               *
         *  B  *                         *
       *         *                     *
      *           *                   *
     *             *                 *
----*---------------*---------------*---------
   *A               C*             *E
  *                   *           *
 *                     *         *
                         *     *
                            *
                            D

Okay, here is an attempt at drawing a sine curve. I labeled the points 
A, B, C, D, and E. These are points on the curve that you get when the 
argument of the sine function (what's inside the parentheses) is 0, 
pi/2, pi, 3pi/2, and 2pi.

Your function is y = sin(4x+pi/3). What will x be at point A? You want

  4x+pi/3 = 0

so that y = sin(0) = 0. You can solve for this and you will find 
x = -pi/12 (which is -15 degrees, but you should leave it as it is). 
What will x be at point C? You want

  4x+pi/3 = pi

so that y = sin(pi) = 0 again. You can solve this to get x = pi/6. 
Since this is twice pi/12, the y axis is 1/3 of the way between 
A and C.

         |  *                               *
         *  B  *                         *
       * |       *                     *
      *  |        *                   *
     *   |         *                 *
----*----+----------*---------------*---------
   *|    |          |*             *E
  *-pi   |         pi *           *
 *  --   |         --  *         *
    12   |          6    *     *
         |                  *
         |                  D

Another way of seeing this is to note that when x = 0 you are 
evaluating sin(pi/3), and pi/3 is 60 degrees, which falls where you see 
it on the sine curve: 2/3 of the way up the curve toward B.

You can work out the other points if you need to. This is a way to be 
sure you've got it right. You can see now that you had the information 
you needed: the sine curve is shifted left by pi/12. Perhaps the only 
thing that was confusing you was that you converted this shift to 
degrees.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/