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anonymous
 3 years ago
graphing \[y=sin(2x\pi)\] did I graph this correctly?
anonymous
 3 years ago
graphing \[y=sin(2x\pi)\] did I graph this correctly?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1360883013269:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the phase shift is pi/2 isn't it?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what does the phase shift do then?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh i see. so the phase shift is pi and not pi/2

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