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

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### 1 Answer

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**