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JenniferSmart1
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graphing \[y=sin(2x\pi)\] did I graph this correctly?
 one year ago
 one year ago
JenniferSmart1 Group Title
graphing \[y=sin(2x\pi)\] did I graph this correctly?
 one year ago
 one year ago

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JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
dw:1360883013269:dw
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
I guess I can get as far as doing the phase shift but I don't know what to do with the 2 in front of the x \[y=sin(2(x\pi/2))\]dw:1360883459233:dw
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
@TuringTest
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
the phase shift is pi/2 isn't it?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
\[\sin(\pi)=0\]so the graph should start at x=0, y=0
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
what does the phase shift do then?
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
oh I see
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
no not really
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
http://www.wolframalpha.com/input/?i=y%3Dsin%282xpi%29
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
a phase shift of pi will shift the graph a halfperiod to the right, which for the sin is equivalent to turning it upsidedown
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
Oh i see. so the phase shift is pi and not pi/2
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
right, you could also see that by applying\[\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta\]
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
makes sense. Thanks!
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
welcome :D
 one year ago
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