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

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Answer 1 · 2018-05-15T09:28:40

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

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#.