Question

How do i change y = a sin(bx+c)+d to a cosine graph y = e cos (fx + g ) + h. what transformation , so I am given, a,b,c,d. find e,f,g,h

Answer 1

Do you mean to say both the graphs should look the same?

Answer 2

yes

Answer 3

i think i got it, use sin x = cos ( pi/2 - x )

Answer 4

but then we have the d and stuff on the end

Answer 5

given a,b,c,d for a y = asinb(bx+c)+d , find e,f,g,h equivalent cos graph

Answer 6

OK!
We'll fix e=a and h=d

Then,
sin(bx+c)=cos(fx+g), is what we want right?

But we know sin(pi/2 - x)=cos(x)
Therefore, sin(bx+c)=cos(pi/2 - (bx+c))

Therefore,
cos(pi/2 - bx - c) = cos(fx+g)

This has infinite sols, but the most trivial one would be to equate all the coefficients of x and constants,
ie. f=-b
& g = pi/2 - c

Answer 7

Yes you are right to that point!

Answer 8

If i interpreted the question correctly and its to make both the graphs same, not transforming one into another!

Answer 9

yes both graphs the same , hmmm
so you can set e= a and h = d ?

Answer 10

Yea no restrictions on that!

Answer 11

yes it works,

Answer 12

yes, that makes sense, so by setting e= a and h = d, you get a simpler problem of equating
sin(bx+c) and cos (fx+g)

Answer 13

this corresponds to a flip about the y axis , and a shift , then you can apply the amplification factor (amplitude) and then the vertical shift

Answer 14

First the vertical shift (d), then the amplification(a), then the expansion (b) and then the horizontal shift (c).

Answer 15

yes but you have to be careful, these transformations are not always commutative. for instance flipping about the x axis after you have shifted up vertically

Answer 16

in our case all we have are vertical, horizontal shift, and flip about the y axis, so the order does not count (commutativity)

Answer 17

so im trying to show that sin (bx+c) = cos [-bx + (pi/2-c)] using tranformations

Answer 18

I know, the order I've given is commutative. And horizontal shift and flip about the y axis is also non-commutative!

Answer 19

no it is commutative

Answer 20

horizontal shift and flip about y axis is commutiative

Answer 21

-f(x) + d is not equal to - ( f(x) + d ) , essentially

Answer 22

f(-(x+c)) , is that what you mean ?

Answer 23

I'm saying f((-x)-c) <> f(-(x-c)). Yea, that's what you said right?

Answer 24

Which means that they are non-commutative

Answer 25

no a shift is f(x) + d

Answer 26

h(x) = f(x) + d

Answer 27

a vertical shift i mean, a horizontal shift is h(x) = f(x-c)

Answer 28

you would never write f( (-x) - c ) , since that doesnt make sense,
f ((-x) - c ) = f ( -x - c ) = f ( -(x+c))

Answer 29

you always have to think of it as f(u), where u is some function of x (possibly x by itself )

Answer 30

so you have f(ku), f(u-c) , and f(u) + c

Answer 31

those are the vertical, horiz. and reflections. for rotations you can throw in a matrix

Answer 32

ok for your example it might be easier to use a square root function, since its more clear
sqrt(x) --> sqrt ( x - 4) ---> sqrt ( - ( x - 4))
versus
sqrt(x) ---> sqrt ( -x) ---> sqrt (

Answer 33

sqrt(x) ---> sqrt ( -x) ---> sqrt ((-x)-4). oh then it is not commutative.

Answer 34

Yup, you can write it as f((-x)+c) . Hope that helped!