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.
(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^@` .