Question

So I'm studying for my wonderful trig final, (yay me!) and I'm going over the whole transformations of a basic trig equation.

I don't understand what effect changing the period has on a graph? I'm looking at my textbook and the pictures for 4pi and pi look exactly the same.

I know that the higher the frequency (pi) the more waves you get, but it looks the same in my book. No extra lines curves, the two pictures look exactly the same. I emailed my teacher, and she said "they're different, look closely".

Answer 1

that is why they call it trickonometry

Answer 2

not sure what "pictures for \(4\pi\) and \(\pi\) look exactly the same" means

Answer 3

probably just the scale on the x axis is my guess

Answer 4

f(x) = asin(kx - c) + h

That's the basic equation, right?

Answer 5

yeah i guess so
now sure why the minus, but if that is what the book has, fine

Answer 6

From what I understand "k" should be the change in period, right? So like 4π or π. But when I look at the graphs

sin(4π)x

&

sin(π)x

They look the same to me.

Answer 7

yeah they do, but look at the scale on the x axis

Answer 8

It actually has + and - but not sure how to make that symbol on the keyboard. same thing for h

Answer 9

I agree with satellite73 that it's the x-axis scale that differs from graph to graph. The shape of the sine function is always the same, but the period, amplitude, starting point, etc., can all vary.

What specific help are you looking for in regard to y = a*sin (kx-c) + h?

Answer 10

so for example \[\sin(4\pi x)\] has period \[\frac{2\pi}{4\pi}=\frac{1}{2}\]

Answer 11

I'm look at the x-axis, and they both say 180 and 360

Answer 12

whoa nelly !!
really?

Answer 13

they are graphing with degrees??!!

Answer 14

Yes...?

Answer 15

holy cow, make them stop

Answer 16

Alright, I'll just go right ahead and email whatever publishing company made my text book. I'll do that right after I get an A on this final.

Answer 17

sine and cosine are functions, functions of numbers, i.e put in a number, get out a number
they only correspond to functions of angles if the angles are measured in radian, not degrees

Answer 18

@sloppycanada can you show a picture of the graph you're referring to?

Answer 19

a is the amplitude
k is associated with stretching the graph of the sinusoid longer or compressing it so that it is shorter.

c has to do with shifting of the graph left or right.
h has to do with shifting of the graph up or down.

Answer 20

yeah what @jim_thompson5910 said, a screen shot would help

Answer 21

Isn't that against the rules? I mean this is a practice worksheet? It's technically cheating, even if I'm not being graded. and I suppose the worksheet is optional.

Answer 22

As long as it's not a quiz or test or anything, it's fine

Answer 23

just post it i won't tell
btw you did write \(\sin(\pi x)\) right?

Answer 24

a clear indication that we are working with numbers (or radians if you prefer)

Answer 25

http://prntscr.com/99cg2h

Answer 26

That's the equation I'm working with, and it says to chose a graph, the practice question after it say the same thing but instead of 4π it says π

Answer 27

I've got it narrowed to these two graphs (which thankfully are right next to each other)

http://prntscr.com/99cgia

Answer 28

It gives me the same choices for the two questions

Answer 29

I'm pretty sure the changed equation of the 4π is the following -

sin(4πx)

Answer 30

period of \(4\pi\) means it does all it is going to do on an interval of length \(4\pi\)
start at 0, go up to one , then back to 0, down to -1, and finally back up to zero in that length

Answer 31

holy moly you are right!!!!

Answer 32

it is the first one , but they confused degrees with radians how weird

Answer 33

In other words, a period of 4Pi involves one complete cycle of either the sine or the cosine function.

Answer 34

if you call \(-360\) \(-2\pi\) and rename \(360\) as \(2\pi\) then the first graph has period \(4\pi\)

Answer 35

Wait so, the first one is sin(πx)? And then the second one is sin(4πx)?

Answer 36

although in actuality it is \(720\)

Answer 37

no, i think there is some confusion here
you are not being told the function, just the period

Answer 38

the one up top, the "longer" one, has period 720

Answer 39

the one below it, the more compact one, has period 180

Answer 40

Oh so the longer one is the period of 720?

Answer 41

yes

Answer 42

which means that it'd period is 4π

Answer 43

Okay, see, this is why I like vectors much more.

Answer 44

they want you to mentally convert \(720^\circ\) to \(4\pi\) because

well because they are morons and have no business teaching math

Answer 45

can you please tell me where this question comes from? what the source is? i

Answer 46

What do you mean?

Answer 47

is it in a book? by whom
is it a worksheet? what is the source?

Answer 48

Ohhh Keystone.

Answer 49

keystone an on line system? in PA?

Answer 50

Yeah?

Answer 51

Not sure what the relevance is.

Answer 52

got a reference number for the question ?

Answer 53

Ladies and gentlemen: Please stick to solving this problem.

Answer 54

problem is solved
unsolved problem is the author of this question, who needs to be contacted

Answer 55

Not sure who made the problem, but I can get the teacher name?

Answer 56

that would be great, they are usually happy to correct errors, once they are pointed out

Answer 57

I don't find that the problem posted here is clearly stated. What specificallyl do you want to learn, sloppycanada? Let's leave blame out of this game.

Answer 58

question was which one had a period of \(4\pi\)
it was the "longer " one

Answer 59

problem was that the graph had the x axis measured in degrees, so saying the period is \(4\pi\) makes no sense

Answer 60

My understanding is that you're working with y = a*sin(bx + c) + h and want to know how varying each quantity a, b, c and h will affect the shape of the graph. We should be focusing on that.

Answer 61

@mathmale I 've learned what I wanted to do, and solved one of those practice problems on this worksheet.

Answer 62

If y = a* sin (bx + c), the period is given by the algebraic expression 2Pi/b.

Answer 63

Good! Further questions?

Answer 64

I wanted to be able to tell the difference between π and 4π or something similar and how that changed the graph. I'm good for now. My brain hurts. I think I'll move to my happy vectors.

Answer 65

@sloppycanada make sure you do not confuse having a period of \(4\pi\) with the function \[\sin(4\pi x)\] they are not the same thing at all

Answer 66

If you begin with y=sin x, the period is 2Pi/1, or 2Pi;
if you begin with y=sin 4x, the period is 2Pi/(4), or Pi/2.

Answer 67

They're not the same? Then why the heck do I have this stupid "k" in this equation? I thought I had to include that so it made it clear that it was not 4*x

Answer 68

sloppycanada: Could you possibly share a diagram or drawing? I still find your "question" to be very vague and imprecise.
Also, I explained earlier that the quantity k affects the stretching or the compression of the graph of the sine function.

the period of sin 2x is half the period of sin x.

the period of sin x/2 is twice the period of sin x.

Getting to the point where y ou can ask precise questions using the correct math vocabulary will inevitably improve your understanding of trig.

Answer 69

Okay so I suppose it all boils down to the question of - say I have to write a trig equation where they gave me an amplitude of 2, a vertical shift of -4, and a period of 3. How would I write this? Where would I put the period of 3?

Answer 70

You are GIVEN the period, 3.

Answer 71

So I don't write it the equation?

Answer 72

start with the graph of y = sin x (or cos x, if that's the case). You need to name the function.

y = sin x is the most basic form of this function.

If the period is 3, recall that period is given by 2Pi/b. Thus, equate 3 to 2Pi/b and solve for b.

If theres's a vertical shift of -4, move the whole graph downward 4 units.

Answer 73

If the amplitude is to be 2, then draw your sine curve so that it begins at (0,0) and traces a curve as high as 2 units above the x-axis and as low as 2 units beneath the x-axis.

Answer 74

b = 2π/3

Answer 75

very good. That's the correct b (or k) value when the period is 3. Good!

Answer 76

Yes, but if I had to write the equation of it,

2sin((2π/3)x)) -4?

Answer 77

Start with y=sin x.

If the amplitude is 2, modify that equation to read y=2sin x

Answer 78

Yes. Good. Must label your expression with "y="
Nice work!

y = 2 sin ( (2Pi/3)x ) - 4.

Answer 79

Ohhhhhh so I have to actually change the period of 3 into the 2pi/b thing. got it. Thanks so much for your help.

Answer 80

Not precisely. Better: If I'm given the fact that the period is 3, I must use the formula for period, 2Pi/b, equating it to the given period (3), and finally solving for b.

if b is 2Pi/3, then I write the equation as y = sin (2Pi/3)x.

Answer 81

If the period is 4, let 4 = 2pi/b and solve for b.

And so on.

You're welcome!

Good luck!!

Bye.

Answer 82

So no matter what, I can't just put the period that they give me in, I have to equate it to 2pi/b. I think that's why I've been having problems.

Answer 83

I'm glad you've made this discovery.