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metalslayer

  • one year ago

I know how to graph a cosine equation in degrees, but how do I graph a sine equation in degrees?

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  1. metalslayer
    • one year ago
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    I know how to graph a cosine equation. Basically the amplitude (minus/plus the vertical shift) is the upper bound, and I know how to get the lower bound. For sine equations I can find the upper lower bounds and the sinuoidal axis. BUT, HOW DO I FIND WERE TO EXACTLY START THE EQUATION ACCORDING TO THE PERIOD?

  2. welshfella
    • one year ago
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    upper and lower bounds of sin x are -1 and 1 and sin 0 = 0 so the curve goes through the origin

  3. metalslayer
    • one year ago
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    I know that, i'm just confused on the graphing of it. For example: Y=20sin1/2(x-120 degree) -10

  4. metalslayer
    • one year ago
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    If it was a cosine equation, I could just graph the point on (120 deg, 10) and that would be the upper bound point.

  5. metalslayer
    • one year ago
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    But for sine, were exactly should I start at the x axis? Basically I now the Y axis and how high/low it would be, but were do I start is my question? Thanks.

  6. anonymous
    • one year ago
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    y = sin x passes through the origin. Use the phase shift value to shift the function horizontally. If that's what you're asking

  7. metalslayer
    • one year ago
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    So basically were on x would the point be located at? Since the phase shift is 120, should I do 120-90 or 120+90

  8. metalslayer
    • one year ago
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    I can attach the book answer, the graph approximately starts at 180 degrees horizontally. Even thought the phase shift is 120

  9. anonymous
    • one year ago
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    Not sure where the 90 is coming from. Shift it 120° to the right This is y = sin x|dw:1442167291189:dw|

  10. anonymous
    • one year ago
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    so this would be sin (x - 120°) |dw:1442167349251:dw|

  11. metalslayer
    • one year ago
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    yes I figured that too, but please look the the image, I will attatch it now

  12. anonymous
    • one year ago
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    ok

  13. metalslayer
    • one year ago
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  14. metalslayer
    • one year ago
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    Problem: 3. Again the equation is: 20sin1/2(x-120deg) -10

  15. anonymous
    • one year ago
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    ok. I see. The midline of the sine function passes through the origin. that's what's at 120°. so your first point should be (120°, -10)

  16. metalslayer
    • one year ago
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    The point that the graph passes through is 180 not 120. Or is it 120?

  17. metalslayer
    • one year ago
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    nevermind, I see it

  18. metalslayer
    • one year ago
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    Oh that makes sense now

  19. amistre64
    • one year ago
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    i recall this as \[A\sin(\psi(t-\phi))+B\]

  20. anonymous
    • one year ago
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    yes, so the point that is at the origin in y = sin x is translated to \((\phi,B)\)

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