Calculate the product p = cos20*cos40*cos80 using complex numbers

We'll consider the complex number put into the polar
form:

z = cos 20 + i*sin 20
(1)

We'll write the conjugate of
z:

z^-1 = cos 20 - i*sin
20

1/z = cos 20 - i*sin 20
(2)

We'll add (1) + (2):

z +
1/z = cos 20 + i*sin 20 + cos 20 - i*sin 20

We'll combine
and eliminate like terms:

z + 1/z = 2*cos
20

We'll multiply z by z:

(z^2
+ 1)/z = 2*cos 20

We'll use symmetric
property:

2*cos 20 = (z^2 +
1)/z

We'll divide by
2:

cos 20 = (z^2 + 1)/2z
(3)

cos 40 = cos 2*(20) = 2 (cos 20)^2 -
1

We'll substitute cos 20 by
(3):

cos 40 = 2 (z^2 + 1)^2/4z^2 -
1

cos 40 = (z^2 + 1)^2/2z^2 -
1

We'll raise to square the binomial z^2 +
1:

cos 40 = (z^4 + 2z^2 + 1)/2z^2 -
1

We'll multiply -1 by
2z^2:

cos 40 = (z^4 + 2z^2 + 1 -
2z^2)/2z^2

We'll eliminate like
terms:

cos 40 = (z^4 + 1)/2z^2
(4)

cos 80 = cos 2*(40) = 2 (cos 40)^2 - 1
(5)

We'll substitute (4) in
(5):

cos 80 = 2 (z^4 + 1)^2/4z^4 -
1

cos 80 = (z^4 + 1)^2/2z^4 -
1

We'll raise to square the binomial z^4 +
1:

cos 80 = (z^8 + 2z^4 + 1)/2z^4 -
1

We'll multiply -1 by
2z^4:

cos 80 = (z^8 + 2z^4 + 1 -
2z^4)/2z^4

We'll eliminate like
terms:

cos 80 = (z^8+
1)/2z^4

Now, we'll calculate the
product:

P = [(z^2 + 1)/2z]*[(z^4 + 1)/2z^2]*[(z^8+
1)/2z^4]

P =  (z^2 + 1)*(z^4 + 1)*(z^8+
1)/8z^7

If z^2 + z + 1 = 0 and z^3 =
1

P = (-z)(-z^2)(-z)/8z

P =
-z/8z

P =
-1/8