# Sine graph with Phase Shift

The standard form of sine graph with phase shift is
y = a sin( bx + c) + d
Where, a= amplitude

$\frac{2\pi }{b}$ = Period

$\frac{-c}{b}$ = Horizontal shift

d = Vertical shift
Both b and c in these graphs affect the phase shift (or displacement).
The phase shift is the amount that the curve is moved in a horizontal direction from its normal position. If the phase shift is negative then the displacement will move to the left and if the phase shift is positive then the displacement will move to the right.
Phase shift is obtained by solving the expression
bx + c = 0
bx = - c
x = -c/b
'd' affects the vertical shift of the graph. If 'd' is positive then the graph move upward by d unit and if 'd' is negative then the graph will move down by d unit.
Note : The period for sine graph is $2\pi$.

How to plot the points on the x-axis ? Divide the period by 4
Let us name it as 'a'.
First point : Mark the phase shift on the X-axis. If the phase shift is negative then plot that on the left side of the zero and positive then at the right side of the zero.
Second point : Add phase shift and a.

## Example on sine graph with phase shift

Example 1: Graph y = sin ( x + $\frac{\pi }{6}$)
Solution : compare y = a sin ( bx + c ) + d and y = sin ( x + $\frac{\pi }{6}$)
a = amplitude = 1
b = 1
$\frac{2\pi }{b}$ = $\frac{2\pi }{1}$ = $2\pi$ = Period
For phase shift, solve x + $\frac{\pi }{6}$ = 0
x = $\frac{-\pi }{6}$ = phase shift
As the phase shift is negative so the graph will move to $\frac{\pi }{6}$ unit to left.
d = 0 = vertical shift , so there is no vertical shift.
Since the phase shift is negative so 1st point on the X-axis is $\frac{-\pi }{6}$ which we plot it on the left side of zero.

2nd point : $\frac{-\pi }{6}$ + $\frac{period}{4}$

= $\frac{-\pi }{6}$ + $\frac{2\pi}{4}$ = $\frac{2\pi }{6}$

3rd point : $\frac{2\pi }{6}$ + $\frac{2\pi}{4}$ = $\frac{5\pi}{6}$

4th point :$\frac{5\pi}{6}$ + $\frac{2\pi}{4}$ = $\frac{8\pi}{6}$

5th point : $\frac{5\pi}{6}$ + $\frac{2\pi}{4}$ = $\frac{11\pi}{6}$
Note : only one cycle of sine graph is shown below. Example 2: Graph y = -2sin ( x - $\frac{\pi }{3}$) -1
Solution : compare y = a sin ( bx + c ) + d and y = 2 sin ( x - $\frac{\pi }{3}$)-1
Negative sign at the front represents that the sine graph reflects at the X-axis.
a = amplitude = 2
b = 1
$\frac{2\pi }{b}$ = $\frac{2\pi }{1}$ = $2\pi$ = Period
For phase shift, solve x - $\frac{\pi }{3}$ = 0
x = $\frac{\pi }{3}$ = phase shift
As the phase shift is positive so the graph will move to $\frac{\pi }{3}$ unit to right.
d = -1 = vertical shift , so the graph will move 1 unit down.
So the new X-axis is y = -1 Since the phase shift is positive so 1st point on the X-axis is $\frac{\pi }{3}$ which we plot it on the right side of the zero.

2nd point : $\frac{\pi }{3}$ + $\frac{period}{4}$

= $\frac{\pi }{3}$ + $\frac{2\pi}{4}$ = $\frac{5\pi }{6}$

3rd point : $\frac{5\pi }{6}$ + $\frac{2\pi}{4}$ = $\frac{8\pi}{6}$

4th point :$\frac{8\pi}{6}$ + $\frac{2\pi}{4}$ = $\frac{11\pi}{6}$

5th point : $\frac{12\pi}{6}$ + $\frac{2\pi}{4}$ = $\frac{14\pi}{6}$
Note : only one cycle of sine graph is shown below. 11th grade math

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