# How do you verify cos4x = 8cos(^4)x - 8cos²x + 1?

Nov 27, 2015

To verify:

$\cos 4 x = 8 {\cos}^{4} x - 8 {\cos}^{2} x + 1$

We will need the following trigonometrical identities:

1. ${\sin}^{2} x + {\cos}^{2} x = 1 \textcolor{w h i t e}{\times x} \iff {\sin}^{2} \left(x\right) = 1 - {\cos}^{2} \left(x\right)$
2. $\sin \left(x + y\right) = \sin x \cos y + \cos x \sin y$
3. $\cos \left(x + y\right) = \cos x \cos y - \sin x \sin y$

So, let's start. :-)

$\cos \left(4 x\right) = \cos \left(2 x + 2 x\right)$

$\textcolor{w h i t e}{\times \times x} = \cos \left(2 x\right) \cos \left(2 x\right) - \sin \left(2 x\right) \sin \left(2 x\right)$

$\textcolor{w h i t e}{\times \times x} = {\left(\cos \left(2 x\right)\right)}^{2} - {\left(\sin \left(2 x\right)\right)}^{2}$

$\textcolor{w h i t e}{\times \times x} = {\left(\cos \left(x + x\right)\right)}^{2} - {\left(\sin \left(x + x\right)\right)}^{2}$

$\textcolor{w h i t e}{\times \times x} = {\left(\cos x \cos x - \sin x \sin x\right)}^{2} - {\left(\sin x \cos x + \sin x \cos x\right)}^{2}$

$\textcolor{w h i t e}{\times \times x} = {\left({\cos}^{2} x - {\sin}^{2} x\right)}^{2} - {\left(2 \sin x \cos x\right)}^{2}$

$\textcolor{w h i t e}{\times \times x} = {\left({\cos}^{2} x - {\sin}^{2} x\right)}^{2} - 4 {\sin}^{2} x {\cos}^{2} x$

$\textcolor{w h i t e}{\times \times x} = {\left({\cos}^{2} x - \left(1 - {\cos}^{2} x\right)\right)}^{2} - 4 \left(1 - {\cos}^{2} x\right) {\cos}^{2} x$

$\textcolor{w h i t e}{\times \times x} = {\left(2 {\cos}^{2} x - 1\right)}^{2} - 4 {\cos}^{2} x + 4 {\cos}^{4} x$

$\textcolor{w h i t e}{\times \times x} = 4 {\cos}^{4} x - 4 {\cos}^{2} x + 1 - 4 {\cos}^{2} x + 4 {\cos}^{4} x$

$\textcolor{w h i t e}{\times \times x} = 8 {\cos}^{4} x - 8 {\cos}^{2} x + 1$

Hope that this helped!