Can someone explain how to do these two problems? http://gyazo.com/4b47d52f7388f9eee2ea59b65aa0a655
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.
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: \(\dfrac{\dfrac{-\pi}{4}}{\dfrac{1}{4}}=-\pi\)
again, negative sign shows the graph is shifted to the right
What about for the original question? http://gyazo.com/4b47d52f7388f9eee2ea59b65aa0a655 @Loser66
\(y= 2sin(2x-pi/2)+3 \)
Can you show how you got that?
Is it the same process for #14?
we have formula to do, right? it say \(y =\pm Asin(kx +c) + verticalshift\) and PS = -c/k,
A is amplitude, nothing to do with it, just put 2 there, ok?
Okay
k has formula , that is period = 2pi/ number in the front of x = \(\dfrac{2pi}{k}\)
hence \(period= \dfrac{2pi}{k}\\k= \dfrac{2pi}{period}\)
given period is pi, hence k = 2 , ok?
Now, so far, wy have y =2 sin(2x +c) + 3 ( 3 is vertical shift)
so , remain is c PS = -c/k. given PS = pi/4, k =2 as above, then c = -PS*k = - (pi/4)*2 =-pi/2
Combine all, y = 2 sin(2x -pi/2) +3
Take time to digest, I go for dinner.
Is that the same process for number 14? @Loser66
Cause that one is sin, but #14 is cos
yup, cos and sin are the same
What if it was tan or something?
tan is another topic.
it doesn't have amplitude,
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.
it doesn't have phaseshift or vertical shift.
Pi/2= 2pi/k K= 4 C= -ps*k = -(-pi)*4 C=4pi Y=1/2cos(4theta+4pi)-7
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