Question

Am very Confused

Answer 1

how would you transform f(x) = 2 sin(2x - π) + 3 into a cosine function in the form f(x) = a cos(bx - c) + d

Answer 2

it is 2 sin(2x - pi) + 3

Answer 3

I already know the answer, its just that I do not know why it is that.
The asnwer is g(x) = 2 cos(2x + 0.5π) + 3

Answer 4

replace pi with pi/2

Answer 5

the question marks are pi

Answer 6

hint: cos(pi/2 - x) = sin(x)

http://www.mathwords.com/c/cofunction_identities.htm

Answer 7

How would I apply that equation?

Answer 8

well the easiest thing to do is use a phase shift

Answer 9

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

f(x) = 2sin(z) + 3 ... let z = 2x-pi

f(x) = 2*cos(pi/2 - z) + 3 ... use that cofunction identity

f(x) = 2*cos(pi/2 - (2x-pi)) + 3

I'll let you simplify

Answer 10

f(x) = 2* cos(-2x + 3pi/2) + 3

Answer 11

When I graph this function on a graphing software, it shows that it is different than 2 sin(2x - pi) + 3

Answer 12

how are you typing it in? I'm getting the same graph

Answer 13

I type them in exactly as they are here

Answer 14

Like this:
f(x) = 2cos(-2x + 3pi / 2) + 3
h(x) = 2cos(2x - 2pi) + 3

Answer 15

idk how you got 2cos(2x - 2pi) + 3

Answer 16

There was a mistake in what I inserted into the graphing software. Now That I insert the correct equations, I find that they are the same:
Here are the equations that I believe are the true:
f(x) = 2sin(2x - π) + 3
h(x) = 2cos(2x - 3pi / 2) + 3

Answer 17

replace the question marks with pi

Answer 18

yes they are the same graph

Answer 19

since 3pi/2 and -pi/2 are coterminal angles, you can replace 3pi/2 with -pi/2 to get

2cos(2x - 3pi / 2) + 3

2cos(2x - (-pi/2)) + 3

2cos(2x+pi/2) + 3

Answer 20

I noticed that when I simplified the hint you gave me earlier, I ended up with f(x) = 2* cos(-2x + 3pi/2) + 3.
Does this equal h(x) = 2cos(2x - 3pi / 2) + 3

Answer 21

those two graphs are also the same, yes

Answer 22

2* cos(-2x + 3pi/2) + 3

2* cos(-(2x - 3pi/2)) + 3

2* cos(2x - 3pi/2) + 3 .... using the rule cos(-x) = cos(x)

Answer 23

That makes sense, as it does not matter if you rotate clockwise or counterclockwise in the unit circle because you would still end up with the same x coordinate both ways.
The rule that I did not understand where it derived from was cos(pi/2 - x) = sin(x)

Answer 24

this page

http://www.mathwords.com/c/cofunction_identities.htm

explains how the cofunctions are related

Answer 25

if f(x) is a cofunction of g(x), then

f(pi/2 - x) = g(x)

Answer 26

notice how we have COsine and sine

cosine is a COfunction of sine

Answer 27

it might be easier to see in degree mode

cos(90 - x) = sin(x)

Answer 28

and using a drawing like a 30-60-90 triangle