How do you verify (2cos^2x-1)^2/(cos^4x-sin^4x)=1-2sin^2x?

Apr 30, 2018

Cosine double angle identity:
$\cos \left(2 x\right) = {\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right)$

Next, substitute rearranged ${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$

Substitute ${\sin}^{2} \left(x\right) = 1 - {\cos}^{2} \left(x\right)$:
$2 {\cos}^{2} \left(x\right) - 1$
Sub ${\cos}^{2} \left(x\right)$:
$1 - 2 {\sin}^{2} \left(x\right)$

$\frac{{\left(\cos \left(2 x\right)\right)}^{2}}{{\cos}^{4} \left(x\right) - {\sin}^{4} \left(x\right)} = \cos \left(2 x\right)$

The denominator ${\sin}^{4} \left(x\right) - {\cos}^{4} \left(x\right)$ is a difference of squares:
$\left({\sin}^{2} \left(x\right) - {\cos}^{2} \left(x\right)\right) \left({\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right)\right)$
$= \left(\cos \left(2 x\right)\right) \cdot 1$

Therefore:

${\cos}^{2} \frac{2 x}{\cos} \left(2 x\right) = \cos \left(2 x\right)$

Factors cancel, leaving:
$\cos \left(2 x\right) = \cos \left(2 x\right)$