# Sketch this sine graph

• Jul 25th 2009, 11:55 PM
smmmc
Sketch this sine graph
Hello,
y=2sin3(theta + pi/4)

amplitude: 2
Period: 2pi/3

Thanks
• Jul 26th 2009, 12:22 AM
songoku
Set 3(theta + pi/4) = certain special angles, such as 0, pi/6, pi/4, pi/3, pi/2, -pi/6, -pi/4, etc to find theta.

Example :
3(theta + pi/4) = 0
theta = -pi/4

Then subs the value of theta back to the equation to obtain y.

Plot the graph

EDIT :
it will be much easier if you use transformation of graph
1.start by drawing y = sin theta ---------------------------------> you get y = sin theta
2.multiply all the values of theta by 1/3 --------------------------> you get y = sin (3 theta)
3.translate the graph by 3pi/4 unit in the negative x-direction -----> you get y = sin (3 theta + 3pi/4)
4.multiply all the value of y by 2 ---------------------------------> you get y = 2 sin (3 theta + 3pi/4)
• Jul 26th 2009, 02:53 AM
HallsofIvy
Quote:

Originally Posted by smmmc
Hello,
y=2sin3(theta + pi/4)

amplitude: 2
Period: 2pi/3

Thanks

Do you mean draw a "free-hand" graph?

I would do it like this: 3(theta+ pi/4)= 0 when theta= -\pi/4. 3(theta+ pi/4)= 2pi when theta+ pi/4= (3/2)pi or theta= (3/2)pi- pi/4= (5/4)pi.

Mark the points (-pi/4, 0) and ((5/4)pi, 0) as the starting and ending points of one "cycle". Mark the midpoint, (pi/2, 0), and the midpoints of the two intervals between, (pi/8, 0) and (7pi/8, 0). Draw the horizontal line y= 2 and y= -2. Now, use those to draw a smooth curve starting at (-pi/4, 0), rising to (pi/8, 2), down to (pi/2, 0), down to (7pi/8, -2), and up to (5pi/4, 0). That is one cycle. The rest of the graph consists of copies of that.