# Question #37bc4

Apr 6, 2017

${\left(2 {\cos}^{2} \theta - 1\right)}^{2} / \left({\cos}^{4} \theta - {\sin}^{4} \theta\right) = 1 - 2 {\sin}^{2} \theta$
${\cos}^{2} \frac{2 \theta}{\left({\cos}^{2} \theta + {\sin}^{2} \theta\right) \left({\cos}^{2} \theta - {\sin}^{2} \theta\right)} = 1 - 2 {\sin}^{2} \theta$
${\cos}^{2} \frac{2 \theta}{\cos} \left(2 \theta\right) = 1 - 2 {\sin}^{2} \theta$
$\cos \left(2 \theta\right) = 1 - 2 {\sin}^{2} \theta$
$1 - 2 {\sin}^{2} \theta = 1 - 2 {\sin}^{2} \theta$
Q.E.D.

#### Explanation:

The original statement, with $\theta$ added for sensibility.
${\left(2 {\cos}^{2} \theta - 1\right)}^{2} / \left({\cos}^{4} \theta - {\sin}^{4} \theta\right) = 1 - 2 {\sin}^{2} \theta$

Let's work the left hand side.

Using one form of the cosine double angle formula, $\cos \left(2 \theta\right) = 2 {\cos}^{2} \theta - 1$, the left side becomes
${\left(\cos \left(2 \theta\right)\right)}^{2} / \left({\cos}^{4} \theta - {\sin}^{4} \theta\right)$, or equivalently ${\cos}^{2} \frac{2 \theta}{{\cos}^{4} \theta - {\sin}^{4} \theta}$

Then, notice that the denominator is a difference of squares, it can be factored as follows: ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$. In this case $a = {\cos}^{2} \theta$ and $b = {\sin}^{2} \theta$. So the left side is now
${\cos}^{2} \frac{2 \theta}{\left({\cos}^{2} \theta + {\sin}^{2} \theta\right) \left({\cos}^{2} \theta - {\sin}^{2} \theta\right)}$

Now we can use the Pythagorean Identity, ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$ to get
${\cos}^{2} \frac{2 \theta}{\left(1\right) \left({\cos}^{2} \theta - {\sin}^{2} \theta\right)}$, or equivalently ${\cos}^{2} \frac{2 \theta}{{\cos}^{2} \theta - {\sin}^{2} \theta}$

And if we can use another form of the cosine double angle formula, $\cos \left(2 \theta\right) = {\cos}^{2} \theta - {\sin}^{2} \theta$, we can see that
${\cos}^{2} \frac{2 \theta}{\cos} \left(2 \theta\right)$

Simplifying this gives a friendly
$\cos \left(2 \theta\right)$

And using the third and final form of the cosine double angle formula, $\cos \left(2 \theta\right) = 1 - 2 {\sin}^{2} \theta$, the left hand side finally becomes
$1 - 2 {\sin}^{2} \theta$

And since $1 - 2 {\sin}^{2} \theta = 1 - 2 {\sin}^{2} \theta$, the left side is equal to the right side and the identity is proven. Q.E.D.