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graphing \[y=sin(2x-\pi)\] did I graph this correctly?

Mathematics
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|dw:1360883013269:dw|
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|

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Other answers:

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

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