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)#.