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Mathematics 20 Online
OpenStudy (anonymous):

Can someone explain how to do these two problems? http://gyazo.com/4b47d52f7388f9eee2ea59b65aa0a655

OpenStudy (loser66):

I am sorry for my answers last night.. :) now I compensate : for \(y = 4 sin(\theta/2 +\pi)-1= 4 sin ((1/2)\theta +\pi) -1\) your k = 1/2 your \(c=\pi\) hence the phaseshift is \(-\dfrac{c}{k}=\dfrac{-\pi}{1/2}= - 2\pi\) the negative sign shows the graph moves to the right.

OpenStudy (loser66):

for \(y = 4 sin(\dfrac{\theta}{4}-\dfrac{\pi}{4})-2\) your c is \(c = \dfrac{-\pi}{4}\) your k is 1/4 hence the phaseshift: \(\dfrac{\dfrac{-\pi}{4}}{\dfrac{1}{4}}=-\pi\)

OpenStudy (loser66):

again, negative sign shows the graph is shifted to the right

OpenStudy (anonymous):

What about for the original question? http://gyazo.com/4b47d52f7388f9eee2ea59b65aa0a655 @Loser66

OpenStudy (loser66):

\(y= 2sin(2x-pi/2)+3 \)

OpenStudy (anonymous):

Can you show how you got that?

OpenStudy (anonymous):

Is it the same process for #14?

OpenStudy (loser66):

we have formula to do, right? it say \(y =\pm Asin(kx +c) + verticalshift\) and PS = -c/k,

OpenStudy (loser66):

A is amplitude, nothing to do with it, just put 2 there, ok?

OpenStudy (anonymous):

Okay

OpenStudy (loser66):

k has formula , that is period = 2pi/ number in the front of x = \(\dfrac{2pi}{k}\)

OpenStudy (loser66):

hence \(period= \dfrac{2pi}{k}\\k= \dfrac{2pi}{period}\)

OpenStudy (loser66):

given period is pi, hence k = 2 , ok?

OpenStudy (loser66):

Now, so far, wy have y =2 sin(2x +c) + 3 ( 3 is vertical shift)

OpenStudy (loser66):

so , remain is c PS = -c/k. given PS = pi/4, k =2 as above, then c = -PS*k = - (pi/4)*2 =-pi/2

OpenStudy (loser66):

Combine all, y = 2 sin(2x -pi/2) +3

OpenStudy (loser66):

Take time to digest, I go for dinner.

OpenStudy (anonymous):

Is that the same process for number 14? @Loser66

OpenStudy (anonymous):

Cause that one is sin, but #14 is cos

OpenStudy (loser66):

yup, cos and sin are the same

OpenStudy (anonymous):

What if it was tan or something?

OpenStudy (loser66):

tan is another topic.

OpenStudy (loser66):

it doesn't have amplitude,

OpenStudy (anonymous):

Okay, thanks for clearing that up for me. @Loser66 I will show you my work for #14 and tell me if it's right when you get back. Enjoy your dinner.

OpenStudy (loser66):

it doesn't have phaseshift or vertical shift.

OpenStudy (anonymous):

Pi/2= 2pi/k K= 4 C= -ps*k = -(-pi)*4 C=4pi Y=1/2cos(4theta+4pi)-7

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