# Math Grade 11, Sinusoidal functions1) Describe how the graph of compares to the graph of y = sin(x) 2) How does the graph of y = -3cos(2θ + 45°) + 3 differ from the graph of y = cos(θ)?...

Math Grade 11, Sinusoidal functions

1) Describe how the graph of compares to the graph of y = sin(x)

2) How does the graph of y = -3cos(2θ + 45°) + 3 differ from the graph of y = cos(θ)?

Please answer both questions, please and thanks. if u are not able to answer both then please teach the concept so i know how to do the second one. thanks.

### 1 Answer | Add Yours

(2) Compare the graph of `y=-3cos(2theta+45^@)+3` to the graph of `y=cos(theta)` :

The general form is `y=Acos(B(theta-h))+k` (You can replace cos with sin )

A: A is the amplitude. If A<0 then the graph is reflected over the horizontal axis. (A is a vertical stretch/compression.)

B: The period is `p=(2pi)/B` . (B is a horizontal stretch/compression. If B<0 then the graph is reflected over a vertical axis.)

h: h is a horizontal translation, usually called a phase shift.

k: k is a vertical translation. The midline of the sinusoid is y=k.

In this case we can write as `y=-3cos(2(theta+22.5^@))+3`

Since k=3 the midline is y=3.

Since A=-3 the amplitude is 3; the maximum is 3+3=6 and the minimum is 3-3=0. Since A<0 the cos function has been reflected over the horizontal axis.

Since B=2 we have the period `p=360/2=180^@` (A period of `pi` when using radians)

With h=22.5 we have a phase shift of 22.5 left. (A phase shift of `pi/8~~.393` when using radians.)

The graph of `y=cos(theta)` and the transformed graph:

The units are in radians; `pi` radians is `180^@` .