# How can tan 4x be simplified or sec 2x?

Aug 7, 2015

Let's say we looked at $\tan 4 x$. We can use the following identities:

$\tan 4 x = \frac{\sin 4 x}{\cos 4 x}$

$\sin 2 x = 2 \sin x \cos x$
$\cos 2 x = {\cos}^{2} x - {\sin}^{2} x$

$\implies \frac{2 \sin 2 x \cos 2 x}{{\cos}^{2} 2 x - {\sin}^{2} 2 x}$

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

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

$= \textcolor{b l u e}{\frac{4 \sin x \cos x \left(1 - 2 {\sin}^{2} x\right)}{{\left(1 - 2 {\sin}^{2} x\right)}^{2} - 4 {\sin}^{2} x {\cos}^{2} x}}$

I don't know if you can get it any simpler; it's all $\sin x$ and $\cos x$ now, though.

You could also have used:
$\tan \left(2 x + 2 x\right) = \frac{\tan \left(2 x\right) + \tan \left(2 x\right)}{1 - \tan \left(2 x\right) \tan \left(2 x\right)}$

$= \frac{2 \tan \left(2 x\right)}{1 - {\tan}^{2} \left(2 x\right)}$

but that's gonna be uglier to simplify (unless you stop here).

$\sec \left(2 x\right)$ is much simpler.

$= \frac{1}{\cos \left(2 x\right)} = \frac{1}{{\cos}^{2} x - {\sin}^{2} x} = \textcolor{b l u e}{\frac{1}{1 - 2 {\sin}^{2} x}}$