## 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

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

2. anonymous

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

3. anonymous

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

4. anonymous

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

5. Michele_Laino

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

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

7. anonymous

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

8. Michele_Laino

ok!

9. anonymous

;)

10. anonymous

Am i correct?

11. anonymous

absolutly noot coorrect !

12. anonymous

Okay I must've did something wrong :/

13. anonymous

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

14. anonymous

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

15. anonymous

try ! and show me what u get

16. anonymous

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

17. anonymous

O.O how did u get thiis !

18. anonymous

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

19. anonymous

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

20. anonymous

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

21. anonymous

a fraction

22. anonymous

take a look ur mail box