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

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

Mathematics
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(1/(3^5))((cos(27*-5)+sin(27*-5))
Moivre s theorem is easy ; you have to keep the formula in mind !!
I am new to De Moivre's theorum, okay that equation you said, i continue to solve that and get the answer?

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Other answers:

you have just to replace 27*-5 by -135
furthermore, you have to use these identities: \[\begin{gathered} \cos \left( { - 135} \right) = \cos \left( {135} \right) \hfill \\ \sin \left( { - 135} \right) = - \sin \left( {135} \right) \hfill \\ \end{gathered} \]
I continued to solve and got \[-(122 \sqrt(2))/243\]
@Michele_Laino furthermore he can also use cos(135)=-sin(45) and sin(135)=cos(45)
ok!
;)
Am i correct?
absolutly noot coorrect !
Okay I must've did something wrong :/
you have to find a complex number (a+ib)
Okay do i plug numbers from the equation into (a + ib)?
try ! and show me what u get
Okay Im guessing i is standing for the imaginary number, hmm (27 + i5) ?
O.O how did u get thiis !
you have to comput the value of 1/3^5cos(135) and plug it into a
then compute the value of -1/3^5sin(135) and plug it into b
Okay for the value should it be a decimal or a fraction?
a fraction
take a look ur mail box

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