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ganeshie8

  • one year ago

show that \(\sin(x)+\sin(x-120^{\circ})+\sin(x-240^{\circ})=0\)

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  1. ganeshie8
    • one year ago
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    my roomate and i are arguing about electric motors and im looking for some geometric explanation for this..

  2. Empty
    • one year ago
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    what's true for x=0 is always true no matter how you rotate it

  3. Empty
    • one year ago
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    \[e^{i x}(e^{i 2 \pi / 3} +e^{i 2 \pi 2/ 3} +e^{i 2 \pi 3/ 3}) =0\] the real part of that

  4. Empty
    • one year ago
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    imaginary part I mean, whatever, both are zero haha

  5. Michele_Laino
    • one year ago
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    it is a three phase system

  6. rhr12
    • one year ago
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    x + 120 = t sin(t - 120) + sin(t) + sin(t + 120) = 0 sin(t)cos(120) - sin(120)cos(t) + sin(t) + sin(t)cos(120) + sin(120)cos(t) = 0 2 * cos(120) * sin(t) + sin(t) = 0 sin(t) * (2 * cos(120) + 1) = 0 sin(t) * (2 * (-1/2) + 1) = 0 sin(t) * (-1 + 1) = 0 sin(t) * 0 = 0 So, sin(t) can equal pretty much anything t = x + 120 sin(x + 120) can be anything x + 120 can be anything x can be anything There are no wrong solutions for this one.

  7. rhr12
    • one year ago
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    That's all i got.

  8. ganeshie8
    • one year ago
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    Yes he is saying delta/star and some bs which i have no idea about... but it makes sense to think in terms of roots of unity as Empty wa suggesting.. (if im interepreting correctly..)

  9. Michele_Laino
    • one year ago
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    |dw:1438447767342:dw|

  10. Empty
    • one year ago
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    x and y components of vectors are linearly independent in 2D. So in order for vectors to be in equilibrium in 2D their components must be in equilibrium too. That's what I'm saying

  11. Empty
    • one year ago
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    complex numbers are just a cute representation of 2d vectors I am so used to just thinking with them I forget other people aren't

  12. ganeshie8
    • one year ago
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    |dw:1438448479126:dw|

  13. anonymous
    • one year ago
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    Confused

  14. ganeshie8
    • one year ago
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    welcome to the club @ruhan11

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