SOLUTION: Use De Moivre's Theorem to find an expression for cot(3θ)

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Question 1160753: Use De Moivre's Theorem to find an expression for cot(3θ)
Answer by KMST(5428) About Me  (Show Source):
You can put this solution on YOUR website!
De Moivre's theorem is about complex numbers.

Complex numbers have a real part and an imaginary part,
so we can consider them to be binomials, like a%2Bb or 3x%2B5y} .
%28a%2Bi%2Ab%29%5En is the power of a binomial, but i is not any variable in this case.
The i in a complex number is a specific imaginary number such that some of its powers are real numbers:
i%5E2=-1 , i%5E3=%28i%5E2%29%2Ai=-1%2Ai=-i , i%5E4=%28i%5E3%29%2Ai=%28-i%29%2Ai=-%28i%2Ai%29=-i%5E2=-%28-1%29=1 , i%5E5=i , i%5E6=-1, i%5E7=-i , i%5E8=1 , and so on.

De Moivre's theorem says that for any integer value of n and any real value of theta ,
%28cos%28theta%29%2Bi%2Asin%28theta%29%29%5En=cos%28n%2Atheta%29%2Bi%2Asin%28n%2Atheta%29 , where i is the imaginary unit that makes i%5E2=-1 .
For n=3 , that gives is

As the cube of a binomial,

However, as we know that i%5E2=-1 and i%5E3=-i , we can replace those powers of i and simplify, collecting like terms, so



The two highlighted expressions are equal to red%28%28cos%28theta%29%2Bi%2Asin%28theta%29%29%5E3%29 ,
so they must be equal to each other:
,
and because we cannot mix blue%28real%29 with green%28imaginary%29 ,
the blue%28real%29 parts (terms) must be equal to each other:
,
and the green%28imaginary%29 parts (terms) must be equal to each other:
.
cot%283theta%29%22=%22cos%283theta%29%2Fsin%283theta%29%22=%22%22=%22%22=%22%22=%22%28cot%28theta%29%29%2A%28cot%5E2%28theta%29-3%29%2F%283%2Acot%5E2%28theta%29-1%29%22=%22%28cot%5E3%28theta%29-3cot%28theta%29%29%2F%283%2Acot%5E2%28theta%29-1%29