How do you use DeMoivre's Theorem to simplify $(2(cos(pi/8)+isin(pi/8)))^6$ ?

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Answer 1 · 2016-08-10T03:52:33

$=-32sqrt2(1-i)$ .

De Moivre's Theorem states # :

{r(costheta+isintheta)}^n=r^n(cos(ntheta)+isin(ntheta))#.

Using it, with, $n=6, r=2, and theta=pi/8$ , we have,

${2(cos(pi/8)+isin(pi/8)}^6=2^6*{cos(6pi/8)+isin(6pi/8)}$

$=64(cos(3pi/4)+isin(3pi/4))$

$=64{cos(pi-pi/4)+isin(pi-pi/4)}$

$=64{-cos(pi/4)+isin(pi/4)}$

$=64(-1/(sqrt2)+i/sqrt2)$

$=-32sqrt2(1-i)$ .

How do you use DeMoivre's Theorem to simplify (2(cos(pi/8)+isin(pi/8)))^6(2(cos(pi/8)+isin(pi/8)))^6?