+2592 Find the equation of a sine function that has a period of pi, amplitude 5, a vertical shift of zero and passes through (pi/6, 5/2).
Thank you CPhill, I already know you are gonna answer this.
+118687 Find the equation of a sine function that has a period of pi, amplitude 5, a vertical shift of zero and passes through (pi/6, 5/2).
\(y=asin(n(x-k))\)
a = amplitude = 5
period = 2pi/n, so n=2
k is the horizontal phase shift which we have not been given so
y=5sin[2(x-k)]
and it passes through (pi/6, 5/2)
sub to find k
\(\frac{5}{2} =5sin[2(\frac{pi}{6}-k)]\\~\\ \frac{1}{2} =sin[2(\frac{pi}{6}-k)]\\~\\ \frac{\pi}{6}=2(\frac{pi}{6}-k)\\~\\ \frac{\pi}{6}=\frac{\pi}{3}-2k\\~\\ k=\frac{\pi}{12}\\~\\ k\approx\;\;0.26\qquad (\mbox{ I added this for the benefit of the diagram)}\\~\\ so\\~\\ y=5sin[2(x-\frac{\pi}{12})] \)
The phase shift pi/12 in the positive direction.
I forgot to include the graph
https://www.desmos.com/calculator/ksryxoioez
(sorry my graph was slightly wrong - I have replaced it.)
+118687 Find the equation of a sine function that has a period of pi, amplitude 5, a vertical shift of zero and passes through (pi/6, 5/2).
\(y=asin(n(x-k))\)
a = amplitude = 5
period = 2pi/n, so n=2
k is the horizontal phase shift which we have not been given so
y=5sin[2(x-k)]
and it passes through (pi/6, 5/2)
sub to find k
\(\frac{5}{2} =5sin[2(\frac{pi}{6}-k)]\\~\\ \frac{1}{2} =sin[2(\frac{pi}{6}-k)]\\~\\ \frac{\pi}{6}=2(\frac{pi}{6}-k)\\~\\ \frac{\pi}{6}=\frac{\pi}{3}-2k\\~\\ k=\frac{\pi}{12}\\~\\ k\approx\;\;0.26\qquad (\mbox{ I added this for the benefit of the diagram)}\\~\\ so\\~\\ y=5sin[2(x-\frac{\pi}{12})] \)
The phase shift pi/12 in the positive direction.
I forgot to include the graph
https://www.desmos.com/calculator/ksryxoioez
(sorry my graph was slightly wrong - I have replaced it.)
+2592 Hey Melody, long time no see.
Thank you for this explanation.
I shall study it for the next 1/2 hour because i still do not understand how i am to do this my self.( without a calculator)
