Answer to Question #117712 in Algebra for Julia Oduro

Question

12. Use De Moivre's theorem to simplify the following
a. ( cos(-30)+isin(-30))-⁴
b. (Cos20+isin20)-³
c.(cos(-30)+isin(-30))-⁴

Answer 1

1.    

a.  $(cos(-30\degree) + i.sin(-30\degree))^{-4}$   

De Moivre’s theorem states that

$(cosx+i.sinx)^n=cos (nx)+i.sin(nx)$

Substitute for $x=-30\degree$  and $n=-4$

$(cos(-30\degree) + i.sin(-30\degree))^{-4}=cos(-4*(-30\degree))+i.sin(-4*(-30\degree))$

$=cos120\degree+i.sin120\degree$

$=-1/2 +(i\sqrt{\smash[b]{3}})/2$

Answer: $=(-1 +i\sqrt{\smash[b]{3}})/2$

b.    $(cos 20\degree+i.sin20\degree)^{-3}$

De Moivre’s theorem states that

$(cosx+i.sinx)^n=cos (nx)+i.sin(nx)$

Substitute for  $x=20\degree$ and $n=-3$

$(cos 20\degree+i.sin20\degree)^{-3}=cos(-3*20\degree)+i.sin(-3*20\degree)$

$=cos(-60\degree)+i.sin(-60\degree)$

$=1/2 -(i\sqrt{\smash[b]{3}})/2$

Answer: $=(1 -i.\sqrt{3})/2$

c.  $(cos(-30\degree) + i.sin(-30\degree))^{-4}$

De Moivre’s theorem states that

$(cosx+i.sinx)^n=cos (nx)+i.sin(nx)$

Substitute for $x=-30\degree$ and $n=-4$

$(cos(-30\degree) + i.sin(-30\degree))^{-4}=cos(-4*(-30\degree))+i.sin(-4*(-30\degree))$

$=cos120\degree+i.sin120\degree$

$=-1/2 +(i\sqrt{\smash[b]{3}})/2$

Answer: $=(-1 +i\sqrt{\smash[b]{3}})/2$