SOLUTION 5: Homework pt. 2

What he did first was graph both graphs together to see what they both look like.

By looking at the graph he knows that the red graph, f(x), is a lot more compressed horizontally compared to the black graph. The vertical shift is off because f(x) winds around y= 1 while g(x) winds around y= 3.

The standard form for sin and cos are:

Where A is the amplitude, that determines the stretch of the graph vertically. B is not the period but determines the period, the larger the number the graph gets compressed horizontally and when the number is between 1 and 0 the graph stretches horizontally. C is the phase shift, it shifts the graph horizontally and D is the vertical shift.

Since both graphs have the same amplitude there's no need to change an of those numbers.
When you expand f(x):

1. f(x) = 3 [sin3(x + 1)] + 1
2. f(x) = 3 [sin3x + 3] + 1
3. f(x) = 3 sin 3x + 3 + 1
4. f(x) = 3 sin 3x + 4

Parameter D has to change and be zero for the graph of f(x) because the cos and sin graphs are similiar in many ways the only thing that makes them different is by a phase shift; sin(x) = cos(x - π/2) OR sin(x + π/2) = cos(x).

Since the parameter C of g(x) is -π/2 there is no need to change the parameter of C for the graph of g(x). If we get rid of parmeter D, once you expand it'll be:

f(x)2 = 3 sin 3x + 3

He has to change parameter B in f(x) because they have to have the same periods. To make it easier set parameter B, which is 3 to y, where y is a constant.

g(x) = f(x)2
3cos(x - π/2) + 3 = 3sin[y(x + 1)]

Dividing both sides by three we get:
cos(x - π/2) + 1 = sin[(yx) + 1]
cos(x - π/2) + 1 = sinyx + 1

Moving the one over to one side of the equations, the one's cancel leaving us with:
cos(x - π/2) = sinyx

Because cos(x - π/2) = sin(x), y must be equal to 1, therefore the two numbers that need to be changed are:

3 [ sin 1(x - 1)] + 0 = 3 cos (x - π/2) + 3