Question

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

Answer 1

(1/(3^5))((cos(27*-5)+sin(27*-5))

Answer 2

Moivre s theorem is easy ; you have to keep the formula in mind !!

Answer 3

I am new to De Moivre's theorum, okay that equation you said, i continue to solve that and get the answer?

Answer 4

you have just to replace 27*-5 by -135

Answer 5

furthermore, you have to use these identities:
\[\begin{gathered}
\cos \left( { - 135} \right) = \cos \left( {135} \right) \hfill \\
\sin \left( { - 135} \right) = - \sin \left( {135} \right) \hfill \\
\end{gathered} \]

Answer 6

I continued to solve and got
\[-(122 \sqrt(2))/243\]

Answer 7

@Michele_Laino furthermore he can also use cos(135)=-sin(45) and sin(135)=cos(45)

Answer 8

ok!

Answer 9

;)

Answer 10

Am i correct?

Answer 11

absolutly noot coorrect !

Answer 12

Okay I must've did something wrong :/

Answer 13

you have to find a complex number (a+ib)

Answer 14

Okay do i plug numbers from the equation into (a + ib)?

Answer 15

try ! and show me what u get

Answer 16

Okay Im guessing i is standing for the imaginary number, hmm (27 + i5) ?

Answer 17

O.O how did u get thiis !

Answer 18

you have to comput the value of 1/3^5cos(135) and plug it into a

Answer 19

then compute the value of -1/3^5sin(135) and plug it into b

Answer 20

Okay for the value should it be a decimal or a fraction?

Answer 21

a fraction

Answer 22

take a look ur mail box