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JenniferSmart1

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

  • one year ago
  • one year ago

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  1. JenniferSmart1
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    |dw:1360883013269:dw|

    • one year ago
  2. JenniferSmart1
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    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
  3. JenniferSmart1
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    @TuringTest

    • one year ago
  4. JenniferSmart1
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    the phase shift is pi/2 isn't it?

    • one year ago
  5. TuringTest
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    \[\sin(-\pi)=0\]so the graph should start at x=0, y=0

    • one year ago
  6. JenniferSmart1
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    what does the phase shift do then?

    • one year ago
  7. JenniferSmart1
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    oh I see

    • one year ago
  8. JenniferSmart1
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    no not really

    • one year ago
  9. JenniferSmart1
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    http://www.wolframalpha.com/input/?i=y%3Dsin%282x-pi%29

    • one year ago
  10. TuringTest
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    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

    • one year ago
  11. JenniferSmart1
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    Oh i see. so the phase shift is pi and not pi/2

    • one year ago
  12. TuringTest
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    right, you could also see that by applying\[\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta\]

    • one year ago
  13. JenniferSmart1
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    makes sense. Thanks!

    • one year ago
  14. TuringTest
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    welcome :D

    • one year ago
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