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anonymous

  • one year ago

Help! Use De Moivre's Theorem to compute the following: [3(cos(27)) + isin (27)]^-5

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  1. anonymous
    • one year ago
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    (1/(3^5))((cos(27*-5)+sin(27*-5))

  2. anonymous
    • one year ago
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    Moivre s theorem is easy ; you have to keep the formula in mind !!

  3. anonymous
    • one year ago
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    I am new to De Moivre's theorum, okay that equation you said, i continue to solve that and get the answer?

  4. anonymous
    • one year ago
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    you have just to replace 27*-5 by -135

  5. Michele_Laino
    • one year ago
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    furthermore, you have to use these identities: \[\begin{gathered} \cos \left( { - 135} \right) = \cos \left( {135} \right) \hfill \\ \sin \left( { - 135} \right) = - \sin \left( {135} \right) \hfill \\ \end{gathered} \]

  6. anonymous
    • one year ago
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    I continued to solve and got \[-(122 \sqrt(2))/243\]

  7. anonymous
    • one year ago
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    @Michele_Laino furthermore he can also use cos(135)=-sin(45) and sin(135)=cos(45)

  8. Michele_Laino
    • one year ago
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    ok!

  9. anonymous
    • one year ago
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    ;)

  10. anonymous
    • one year ago
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    Am i correct?

  11. anonymous
    • one year ago
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    absolutly noot coorrect !

  12. anonymous
    • one year ago
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    Okay I must've did something wrong :/

  13. anonymous
    • one year ago
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    you have to find a complex number (a+ib)

  14. anonymous
    • one year ago
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    Okay do i plug numbers from the equation into (a + ib)?

  15. anonymous
    • one year ago
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    try ! and show me what u get

  16. anonymous
    • one year ago
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    Okay Im guessing i is standing for the imaginary number, hmm (27 + i5) ?

  17. anonymous
    • one year ago
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    O.O how did u get thiis !

  18. anonymous
    • one year ago
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    you have to comput the value of 1/3^5cos(135) and plug it into a

  19. anonymous
    • one year ago
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    then compute the value of -1/3^5sin(135) and plug it into b

  20. anonymous
    • one year ago
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    Okay for the value should it be a decimal or a fraction?

  21. anonymous
    • one year ago
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    a fraction

  22. anonymous
    • one year ago
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    take a look ur mail box

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