# Thread: Proof of trig identity

1. ## Proof of trig identity

Establish the following identity: $\cos (\theta)= 4 \cos^3(\theta/3) - 3\cos(\theta/3)$ Our hint is to use DeMoivre's formula, $\left(\cos x + i \sin x\right)^k = \cos\left(kx\right) + i \sin\left(kx\right) \,$ I have no idea where to begin and any help would be appreciated!

2. Originally Posted by Chief65
Establish the following identity: $\cos (\theta)= 4 \cos^3(\theta/3) - 3\cos(\theta/3)$ Our hint is to use DeMoivre's formula, $\left(\cos x + i \sin x\right)^k = \cos\left(kx\right) + i \sin\left(kx\right) \,$ I have no idea where to begin and any help would be appreciated!
$(\cos{x} + i\sin{x})^k = \cos{(kx)} + i\sin{(kx)}$

$(\cos{x} + i\sin{x})^3 = \cos{(3x)} + i\sin{(3x)}$

$\cos^3{x} + 3i\cos^2{x}\sin{x} + 3i^2\cos{x}\sin^2{x} + i^3\sin^3{x} = \cos{(3x)} + i\sin{(3x)}$

$\cos^3{x} + 3i\cos^2{x}\sin{x} - 3\cos{x}\sin^2{x} - i\sin^3{x} = \cos{(3x)} + i\sin{(3x)}$

$\cos^3{x} + 3i(1 - \sin^2{x})\sin{x} - 3\cos{x}(1 - \cos^2{x}) - i\sin^3{x} = \cos{(3x)} + i(3\sin{x} - 4\sin^3{x})$

$\cos^3{x} + 3i\sin{x} - 3i\sin^3{x} - 3\cos{x} + 3\cos^3{x} - i\sin^3{x} = \cos{(3x)} + 3i\sin{x} - 4i\sin^3{x}$

$4\cos^3{x} - 3\cos{x} = \cos{(3x)}$.

Now let $x = \frac{\theta}{3}$.

3. A shorter method:

Using De Moivre's Theorem
$(\cos x + i \sin x)^3 = (\cos 3x + i \sin 3x)$

Using binomial expansion:
$(\cos x + i \sin x)^3 = \cos^3x + 3i \cos^2x \sin x - 3\cos x \sin^2x - i\sin^3x$

Equating the real parts:

$\cos 3x = \cos^3x - 3\cos x \sin^2x$
$\cos 3x = \cos^3x - 3\cos x (1 - \cos^2x)$
$\cos 3x = 4\cos^3x - 3\cos x$

Substituting $x = \frac{\theta}{3}$
$\cos \theta = 4\cos^3 \frac{\theta}{3} - 3\cos \frac{\theta}{3}$ as required

4. Originally Posted by Gusbob
A shorter method:

Using De Moivre's Theorem
$(\cos x + i \sin x)^3 = (\cos 3x + i \sin 3x)$

Using binomial expansion:
$(\cos x + i \sin x)^3 = \cos^3x + 3i \cos^2x \sin x - 3\cos x \sin^2x - i\sin^3x$

Equating the real parts:

$\cos 3x = \cos^3x - 3\cos x \sin^2x$
$\cos 3x = \cos^3x - 3\cos x (1 - \cos^2x)$
$\cos 3x = 4\cos^3x - 3\cos x$

Substituting $x = \frac{\theta}{3}$
$\cos \theta = 4\cos^4 \frac{\theta}{3} - 3\cos \frac{\theta}{3}$ as required
That's probably the preferred method, as I'm sure the next part of the question is to prove the triple/third-angle identity for sine, which I already used in my proof...