Question

which cosine function has maximum of .5 a minimum of -.5 and a period of (2pi/3)?

Answer 1

@hermom1967 @e.mccormick @Hero

Answer 2

@Frostbitten @ForeverNever

Answer 3

@Frostbitten you there?

Answer 4

Sorry, this is beyond me, I don't know the answer.

Answer 5

It's ok

Answer 6

@hero can you help me with this one?

Answer 7

OK. Normally, cosine has a range of -10 to 1. What would change the range to -.5 TO .5?

Answer 8

Um... would it be... ah I don't know. Can you help explain the procedure?

Answer 9

*by explaining

Answer 10

@e.mccormick ? Are you there?

Answer 11

Yah. Just typing something up and my bos is asking me a question.

Answer 12

oh wow, okay thank you. I didn't know you where at work :) I can wait

Answer 13

Sorry, this is wwwwaaaaayyyyyyy beyond me. I can ask my friend though.

Answer 14

For pretty much any function, trig or otherwise, the following is true:

If the normal version is \(f(x)\) then transformations can be done with \(Af(Bx-h)+k\)

A is the vertical stretching.
B is the horizontal stretching
h is the horizontal shift
k is the vertical shift

Answer 15

Okay, that makes sense.

and okay @ForeverNever

Answer 16

So, if you look at the unit circle, cosine has a range of -1 to 1. If A is the vertical stretching, and cosine is normally \(\cos (x)\) what would the A in \(A\cos (x)\) be to change the range to -.5 to .5?

Answer 17

Wouldn't it be .5?

Answer 18

Yes! So that is part of the answer:

\(.5 \cos(x)\) is where we are at and that is half the work.

The period is the horizontal domain. The normal period is \(2\pi\). Now, changes in period are inverted. That means that to do a period that is twice as long, you multiply by \(\frac{1}{2}\). To do a period that is half as long, you multiply by 2.

So, what would change \(2\pi\) to \(\dfrac{2\pi}{3}\) and then what is the inverse of that?

Answer 19

Sorry, he doesn't know. all he said was his brain exploded when he read it

Answer 20

Umm, dividing by three and the inverse is... I don't know, isn't it just the reverse?

It's okay @ForeverNever

Answer 21

If I have \(\dfrac{5}{a}\) then the inverse is \(\dfrac{a}{5}\). So you flip the fraction.

Answer 22

Ok so yeah, it's just the reverse. So 3/2pi?

Answer 23

Well, you only want what would change it. So what would change it is \(\dfrac{1}{3}\) and the inverse is \(\dfrac{3}{1}\) or just 3.

Answer 24

Okay. Yeah I get that...

Answer 25

So, where would the 3 need to go. Just look at my list of rules and what I said it is changing and see what you think.

Answer 26

Well, If we multiply 2*3.14, we get around 6. So 3/6=1/3?

Answer 27

Well, I mean where in the \(.5cos(x)\) does the 3 need to go? That is all that is left. You already changed the range, so all you need it to finish changing the period and it is done.

Answer 28

Well the three goes in the x right?

Answer 29

By in the x you mean in place of or next to?

Answer 30

In the place of it.

Answer 31

No, the x still needs to be there or it is no longer a function, just an equation.

Answer 32

AH, if you want to sue \(\theta\) instead of x, then yes.

Answer 33

Use... not sue. Ignore the dyslexia.

You just need a variable, and \(\theta\) works.

Answer 34

Okay, that makes sense. Thank you for helping me :)

Answer 35

No problem. Have fun!

Oh, and those rules, with a little extra info on what -A and -B do, are valid for most anything. So the things that shift a parabola or a circle or just about anything will be the same or very similar.

Answer 36

Okay, I will keep that in mind :)

Answer 37

-A flips over the x axis. -B flips over the y axis. That is the only other thing to know.

Here are some examples that are from non-trig functions and you can see the numbers in the same places doing the same sort of changes:
http://tutorial.math.lamar.edu/Classes/Alg/Transformations.aspx

Answer 38

Okay I will check the link out :) Thank you!

Answer 39

=) Welcome!