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bleck19

  • one year ago

Write the complex number in the form a + bi sqrt6(cos 315° + i sin 315°) PLEASE HELP

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  1. bleck19
    • one year ago
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    options are:

  2. bleck19
    • one year ago
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    @jim_thompson5910 could you help with this last one :/ ? Im so sorry

  3. jim_thompson5910
    • one year ago
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    what is the cosine of 315 degrees

  4. bleck19
    • one year ago
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    sqrt2/2

  5. bleck19
    • one year ago
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    @jim_thompson5910

  6. jim_thompson5910
    • one year ago
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    what about the sine of 315 degrees

  7. bleck19
    • one year ago
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    -1/sqrt2 @jim_thompson5910

  8. jim_thompson5910
    • one year ago
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    or -sqrt(2)/2

  9. bleck19
    • one year ago
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    sorry about that youre right ./.... typo @jim_thompson5910

  10. jim_thompson5910
    • one year ago
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    so this means that we have this so far \[\Large \sqrt{6}\left(\cos(315) + i\sin(315)\right)\] \[\Large \sqrt{6}\left(\frac{\sqrt{2}}{2} - i*\frac{\sqrt{2}}{2}\right)\]

  11. bleck19
    • one year ago
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    so like (2-2i) sqrt3 @jim_thompson5910

  12. jim_thompson5910
    • one year ago
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    how did you get that

  13. bleck19
    • one year ago
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    I have no clue ... I messed up on the way Im thinking it was either A or C but Im leaning towards A

  14. bleck19
    • one year ago
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    @jim_thompson5910

  15. jim_thompson5910
    • one year ago
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    one sec

  16. bleck19
    • one year ago
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    okay

  17. jim_thompson5910
    • one year ago
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    \[\Large \sqrt{6}\left(\cos(315) + i\sin(315)\right)\] \[\Large \sqrt{6}\left(\frac{\sqrt{2}}{2} - i*\frac{\sqrt{2}}{2}\right)\] \[\Large \sqrt{6}*\left(\frac{\sqrt{2}}{2}\right) - i*\sqrt{6}*\left(\frac{\sqrt{2}}{2}\right)\] \[\Large \frac{\sqrt{6}*\sqrt{2}}{2} - i*\frac{\sqrt{6}*\sqrt{2}}{2}\] \[\Large \frac{\sqrt{6*2}}{2} - i*\frac{\sqrt{6*2}}{2}\] \[\Large \frac{\sqrt{12}}{2} - i*\frac{\sqrt{12}}{2}\] \[\Large \frac{\sqrt{4*3}}{2} - i*\frac{\sqrt{4*3}}{2}\] \[\Large \frac{\sqrt{4}*\sqrt{3}}{2} - i*\frac{\sqrt{4}*\sqrt{3}}{2}\] \[\Large \frac{2*\sqrt{3}}{2} - i*\frac{2*\sqrt{3}}{2}\] \[\Large \sqrt{3} - i*\sqrt{3}\] \[\Large \sqrt{3} - \sqrt{3}*i\]

  18. bleck19
    • one year ago
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    Thank you !!!

  19. jim_thompson5910
    • one year ago
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    np

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