Question

Consider the following.
A = π/3; B = 4

(a) Find a cosine function whose graph matches the given curve.
(b) Find a sine function whose graph matches the given curve.

Answer 1

The graph: http://i.imgur.com/INydrmi.gif

Answer 2

I have y=5cosBx SO FAR.

What do I do to find B?

Answer 3

Ah, trig...
that image is a sine graph btw.

Answer 4

but it could be both

Answer 5

for a sine graph it would have a phase shift of (+A)

Answer 6

B is your cycles from 0 to A

Answer 7

hmmm wonder where "5" came from

Answer 8

y = A sin (B(x-c)) + D

Answer 9

That's a typo. I meant to put 4.

Answer 10

hmmm B = 4 = A

Answer 11

A is your amplitude, B is your cycles, x is your period (2pi/b) c is your phase shift, and d is your vertical shift

Answer 12

well., I should have said B = 4 = Amplitude

Answer 13

wel... I was referring to the (0, B) coordinate =)

Answer 14

I'm probably not much help, I've forgotten some of the trig I've learned so I apologize for that.

Answer 15

the graph has 4 A regions, each one is pi/3 so the full period for that function will be \(\bf \cfrac{\pi}{3}+\cfrac{\pi}{3}+\cfrac{\pi}{3}+\cfrac{\pi}{3}\implies \cfrac{4\pi}{3}\)

since after that rise back up to 0, it'll just loop

Answer 16

\(\bf \textit{full regular period for sine and cosine is } 2\pi\\ \quad \\
A(Bx+C)+D\implies \textit{new period}= \cfrac{2\pi}{B}\qquad thus\\ \quad \\
\cfrac{4\pi}{3} = \cfrac{2\pi}{B}\)

Answer 17

so, solving for "B" there from the template, you can get what's multiplying "x"

Answer 18

it has a phase shift to the left, that is a positive pi/3, and \(\bf \cfrac{\pi}{3} = \cfrac{C}{B}\)

Answer 19

so to use say, the sine function, which is really the one drawn there, it'd end up more or less like \(\bf 4sin\left[\left(\cfrac{4\pi}{3} = \cfrac{2\pi}{B}\right)x+\left(\cfrac{\pi}{3} = \cfrac{C}{B}\right)\right]\)