# Thread: De Moivre, and trigonometric identities

1. ## De Moivre, and trigonometric identities

Question:
Use De Moivre's Theorem to express $\displaystyle cos6x$ and use the result to solve $\displaystyle 32x^6-48x^4+18x^2=1$.

My problems:
I was under the impression that De Moivre's theorem was something like $\displaystyle (cisx)^n=cisnx$. I don't see how it applies.

My working:
$\displaystyle cos6x = cos^23x - sin^23x$
$\displaystyle cos6x = (16cos^6x + 9cos^2 - 24cos^4x) - (16sin^6x + 9sin^2 - 24sin^4x$

What is this? lol
How do I use DeMoivre's Theorem?

2. Originally Posted by Evales
Question:
Use De Moivre's Theorem to express $\displaystyle cos6x$ and use the result to solve $\displaystyle 32x^6-48x^4+18x^2=1$.

My problems:
I was under the impression that De Moivre's theorem was something like $\displaystyle (cisx)^n=cisnx$. I don't see how it applies.

My working:
$\displaystyle cos6x = cos^23x - sin^23x$
$\displaystyle cos6x = (16cos^6x + 9cos^2 - 24cos^4x) - (16sin^6x + 9sin^2 - 24sin^4x$

What is this? lol
How do I use DeMoivre's Theorem?
$\displaystyle (\text{cis} x)^6 = \text{cis} (6x) = \cos (6x) + i \sin (6x)$ .... (1)

$\displaystyle (\text{cis} x)^6 = (\cos x + i \sin x)^6 = (\cos^6 x - 15 \cos^4 x \sin^2 x + 15 \cos^2 x \sin^4 x - \sin^6 x)$ $\displaystyle + i (6 \cos^5 x \sin x - 20 \cos^3 x \sin^3 x + 6 \cos x \sin^5 x)$ .... (2)

Equate the real components of (1) and (2) and you get an expression for $\displaystyle \cos (6x)$.

Write this expression in terms of powers of $\displaystyle \cos x$. Strangely enough, you get

$\displaystyle \cos (6 x) = 32 \cos^6 x - 48 \cos^4 x + 18 \cos^2 x - 1$.

So the equation $\displaystyle 32 x^6 - 48 x^4 + 18 x^2 - 1 = 0$ is equivalent to the equation $\displaystyle \cos (6 \theta) = 0$ where $\displaystyle x = \cos \theta \, ....$