Question

The cosine equation for a function that has a period of 4 straight pi and an amplitude of 8

Answer 1

Answer:

y=4sin[2(x−π2)]−6

Step-by-step explanation:

The standard form of a sine function is

y=asin[b(x−h)]+k

where

•a : is the amplitude,

•2π/b : is the period,

•h : is the phase shift, and

•k : is the vertical displacement.

We start with classic

y=sinx :

graph{(y-sin(x))(x^2+y^2-0.075)=0 [-15, 15, -11, 5]}

(The circle at (0,0) is for a point of reference.)

The amplitude of this function is

a=1 .To make the amplitude 4, we need

a to be 4 times as large, so we set

a=4

.

Our function is now

y=4sinx ,and looks like:

graph{(y-4sin(x))(x^2+y^2-0.075)=0 [-15, 15, -11, 5]}

The period of this function—the distance between repetitions—right now is

2π , with b=1

.To make the period π , we need to make the repetitions twice as frequent, so we need

b=normal period/desired period

=2π/π=2

.

Our function is now

y=4sin(2x), and looks like: graph

{(y-4sin(2x))(x^2+y^2-0.075)=0 [-15, 15, -11, 5]}

This function currently has no phase shift, since

h=0

. To induce a phace shift, we need to offset

xby the desired amount, which in this case is

π2 to the right. A phase shift right means a positive

h, so we set

h = π2

.

Our function is now

y=4sin[2(x−π2)] , and looks like:graph

{(y-4sin(2(x-pi/2)))((x-pi/2)^2+y^2-0.075)=0 [-15, 15, -11, 5]}

Finally, the function currently has no vertical displacement, since

k=0

.To displace the graph 6 units down, we set

k=−6

.

Our function is now

y=4sin[2(x−π2)]−6, and looks like:graph {(y-4sin(2(x-pi/2))+6)((x-pi/2)^2+(y+6)^2-0.075)=0 [-15, 15, -11, 5]}