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anonymous

  • one year ago

Can someone explain how to express -4 in trigonometric form? Really lost. Thank you!

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  1. jim_thompson5910
    • one year ago
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    Polar form: \(\Large (r,\theta)\) Trig form: \(\Large r[\cos(\theta) + i*\sin(\theta)]\) The two forms are very similar

  2. jim_thompson5910
    • one year ago
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    write -4 as -4+0i that is in the form a+bi

  3. anonymous
    • one year ago
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    Ok.

  4. jim_thompson5910
    • one year ago
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    a = -4 b = 0 plug those values into the formulas r = sqrt(a^2 + b^2) theta = arctan(b/a)

  5. anonymous
    • one year ago
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    So r = 4 and theta = 0?

  6. jim_thompson5910
    • one year ago
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    r = 4 is correct but theta = 0 is not

  7. jim_thompson5910
    • one year ago
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    |dw:1437696080574:dw|

  8. anonymous
    • one year ago
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    Isn't 0/-4 equal to 0, and the arctan(0) just 0?

  9. jim_thompson5910
    • one year ago
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    arctan(b/a) = arctan(0/(-4)) = 0 is true but we're in the wrong spot add on 180 degrees |dw:1437696139033:dw|

  10. anonymous
    • one year ago
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    So is r = 4 and theta = -4...?

  11. jim_thompson5910
    • one year ago
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    theta = 180 degrees = pi radians

  12. anonymous
    • one year ago
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    so theta = pi ?

  13. jim_thompson5910
    • one year ago
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    yeah

  14. anonymous
    • one year ago
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    Ok so let me try and see if I can get the answer.

  15. anonymous
    • one year ago
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    A. 4(cos0degrees + i sin270degrees) B. 4(cos90degrees + i sin90degrees) C. 4(cos0degrees + i sin0degrees) D. 4(cos180degrees + i sin180degrees) Would it be D?

  16. jim_thompson5910
    • one year ago
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    The trig form, in radians, is \(\Large 4[\cos(\pi) + i*\sin(\pi)]\) that's equivalent to \(\Large 4[\cos(180^{\circ}) + i*\sin(180^{\circ})]\)

  17. jim_thompson5910
    • one year ago
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    you are correct

  18. anonymous
    • one year ago
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    Thank you! I'll Medal you (:

  19. jim_thompson5910
    • one year ago
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    you're welcome

  20. anonymous
    • one year ago
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    Could you help me do the same thing for -6i?

  21. anonymous
    • one year ago
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    @jim_thompson5910

  22. jim_thompson5910
    • one year ago
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    -6i = 0 + (-6i) now a = 0 and b = -6

  23. jim_thompson5910
    • one year ago
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    use the formulas r = sqrt(a^2 + b^2) theta = arctan(b/a)

  24. jim_thompson5910
    • one year ago
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    draw a graph to make sure you get the theta in the right spot

  25. anonymous
    • one year ago
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    So would theta = pi?

  26. jim_thompson5910
    • one year ago
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    what is the value of b/a this time?

  27. anonymous
    • one year ago
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    Undefined actually

  28. jim_thompson5910
    • one year ago
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    where is tan(theta) undefined?

  29. anonymous
    • one year ago
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    Isn't it at 90 degrees and 270 degrees?

  30. jim_thompson5910
    • one year ago
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    correct

  31. jim_thompson5910
    • one year ago
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    |dw:1437697203957:dw|

  32. jim_thompson5910
    • one year ago
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    |dw:1437697245770:dw|

  33. anonymous
    • one year ago
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    So theta = 3pi/2

  34. jim_thompson5910
    • one year ago
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    yep

  35. anonymous
    • one year ago
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    Therefore making the answer: 6(cos270degrees + i sin270degrees)

  36. jim_thompson5910
    • one year ago
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    correct

  37. anonymous
    • one year ago
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    Thank you so much, could you help me with just one more? I think I understand how to do a single number, but am not sure how to do an equation?

  38. jim_thompson5910
    • one year ago
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    what's the question

  39. anonymous
    • one year ago
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    3-3i

  40. jim_thompson5910
    • one year ago
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    that's not an equation. An equation needs an = sign

  41. anonymous
    • one year ago
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    Sorry, meant expression.

  42. jim_thompson5910
    • one year ago
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    compare a+bi with 3-3i to see that a = 3 and b = -3

  43. anonymous
    • one year ago
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    Oh ok! So then r = 3SqRt2 and theta = -1 ... but I'm still confused as to how you find the theta degree value.

  44. jim_thompson5910
    • one year ago
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    theta is not -1

  45. jim_thompson5910
    • one year ago
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    b/a = -1, that is true

  46. jim_thompson5910
    • one year ago
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    arctan(b/a) = arctan(-1) = ??

  47. anonymous
    • one year ago
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    -45...

  48. anonymous
    • one year ago
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    So 315 degrees?

  49. jim_thompson5910
    • one year ago
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    yes

  50. anonymous
    • one year ago
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    So would the answer be: 3SqRt2(cos7pi/4 + i sin7pi/4) ?

  51. jim_thompson5910
    • one year ago
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    correct

  52. anonymous
    • one year ago
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    Thank you so much. I finally understand this!! (: You really helped me understand this, if I could give you more medals I definitely would.

  53. jim_thompson5910
    • one year ago
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    that would be the radian form

  54. jim_thompson5910
    • one year ago
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    you're welcome

  55. anonymous
    • one year ago
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    Yes, that is what the answers are listed as.

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