How do you find the power #[3(cos(pi/6)+isin(pi/6)]^3# and express the result in rectangular form?

1 Answer
May 15, 2018

#[3(cos(pi/6)+isin(pi/6)]^3=27[cos(pi/2)+isin(pi/2)]=27i#.

Explanation:

We need De' Moivre's Theorem :

#r(costheta+isintheta)^n=r^n(cos(n*theta)+isin(n*theta))#.

With #r=3, n=3 and theta=pi/6#, we have,

#[3(cos(pi/6)+isin(pi/6)]^3=3^3[cos(3*pi/6)+isin(3*pi/6)]#,

#=27[cos(pi/2)+isin(pi/2)]#,

#=27[0+i(1)]#.

Hence,

#[3(cos(pi/6)+isin(pi/6)]^3=27[cos(pi/2)+isin(pi/2)]=27i#.