anonymous
  • anonymous
graphing \[y=sin(2x-\pi)\] did I graph this correctly?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1360883013269:dw|
anonymous
  • anonymous
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|
anonymous
  • anonymous
@TuringTest

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anonymous
  • anonymous
the phase shift is pi/2 isn't it?
TuringTest
  • TuringTest
\[\sin(-\pi)=0\]so the graph should start at x=0, y=0
anonymous
  • anonymous
what does the phase shift do then?
anonymous
  • anonymous
oh I see
anonymous
  • anonymous
no not really
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=y%3Dsin%282x-pi%29
TuringTest
  • TuringTest
a phase shift of -pi will shift the graph a half-period to the right, which for the sin is equivalent to turning it upside-down
anonymous
  • anonymous
Oh i see. so the phase shift is pi and not pi/2
TuringTest
  • TuringTest
right, you could also see that by applying\[\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta\]
anonymous
  • anonymous
makes sense. Thanks!
TuringTest
  • TuringTest
welcome :D

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