# How do you verify the following identity?

## $\frac{\cos \left(3 x\right)}{\cos \left(x\right)} = 1 - 4 {\sin}^{2} \left(x\right)$

Jul 4, 2016

Use a few trig identities and a lot of simplifying. See below.

#### Explanation:

When dealing with things like $\cos 3 x$, it helps to simplify it to trigonometric functions of a unit $x$; i.e. something like $\cos x$ or ${\cos}^{3} x$. We can use the sum rule for cosine to accomplish this:
$\cos \left(\alpha + \beta\right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$

So, since $\cos 3 x = \cos \left(2 x + x\right)$, we have:
$\cos \left(2 x + x\right) = \cos 2 x \cos x - \sin 2 x \sin x$
$= \left({\cos}^{2} x - {\sin}^{2} x\right) \left(\cos x\right) - \left(2 \sin x \cos x\right) \left(\sin x\right)$

Now we can replace $\cos 3 x$ with the above expression:
$\frac{\cos 3 x}{\cos} x = 1 - 4 {\sin}^{2} x$
$\frac{\left({\cos}^{2} x - {\sin}^{2} x\right) \left(\cos x\right) - \left(2 \sin x \cos x\right) \left(\sin x\right)}{\cos} x = 1 - 4 {\sin}^{2} x$

We can split this larger fraction up into two smaller fractions:
$\frac{\left({\cos}^{2} x - {\sin}^{2} x\right) \left(\cos x\right)}{\cos} x - \frac{\left(2 \sin x \cos x\right) \left(\sin x\right)}{\cos} x = 1 - 4 {\sin}^{2} x$

Note how the cosines cancel:
$\frac{\left({\cos}^{2} x - {\sin}^{2} x\right) \cancel{\cos x}}{\cancel{\cos x}} - \frac{\left(2 \sin x \cancel{\cos x}\right) \left(\sin x\right)}{\cancel{\cos}} x = 1 - 4 {\sin}^{2} x$

$\to {\cos}^{2} x - {\sin}^{2} x - 2 {\sin}^{2} x = 1 - 4 {\sin}^{2} x$

Now add a ${\sin}^{2} x - {\sin}^{2} x$ into the left side of the equation (which is the same thing as adding $0$). The reasoning behind this will become clear in a minute:
${\cos}^{2} x - {\sin}^{2} x - 2 {\sin}^{2} x + \left({\sin}^{2} x - {\sin}^{2} x\right) = 1 - 4 {\sin}^{2} x$

Rearrange terms:
${\cos}^{2} x + {\sin}^{2} x - \left({\sin}^{2} x + {\sin}^{2} x + 2 {\sin}^{2} x\right) = 1 - 4 {\sin}^{2} x$

Use the Pythagorean Identity ${\sin}^{2} x + {\cos}^{2} x = 1$ and combine the ${\sin}^{2} x$s in the parentheses:
$1 - \left(4 {\sin}^{2} x\right) = 1 - 4 {\sin}^{2} x$

You can see that our little trick of adding ${\sin}^{2} x - {\sin}^{2} x$ has allowed us to use the Pythagorean Identity and collect the ${\sin}^{2} x$ terms.

And voila:
$1 - 4 {\sin}^{2} x = 1 - 4 {\sin}^{2} x$

Q.E.D.