# How do you simplify f(theta)=-tan4theta+sin2theta+cos4theta to trigonometric functions of a unit theta?

Jun 10, 2018

$- \frac{4 \sin \theta \cos \theta \left(1 - 2 {\sin}^{2} \theta\right)}{1 - 8 {\sin}^{2} \theta {\cos}^{2} \theta}$
$q \quad q \quad - 2 \sin \theta \cos \theta + 1 - 8 {\sin}^{2} \theta {\cos}^{2} \theta$

#### Explanation:

We start with the double angle identities :

$\sin \left(2 \theta\right) = 2 \sin \theta \cos \theta$

and

$\cos \left(2 \theta\right) = {\cos}^{2} \theta - {\sin}^{2} \theta = 2 {\cos}^{2} \theta - 1 = 1 - 2 {\sin}^{2} \theta$

Now

$\cos \left(4 \theta\right) = \cos \left(2 \times 2 \theta\right)$
$q \quad q \quad q \quad = 1 - 2 {\sin}^{2} \left(2 \theta\right)$
$q \quad q \quad q \quad = 1 - 2 {\left(2 \sin \theta \cos \theta\right)}^{2}$
$q \quad q \quad q \quad = 1 - 8 {\sin}^{2} \theta {\cos}^{2} \theta$

and

$\sin \left(4 \theta\right) = \sin \left(2 \times 2 \theta\right)$
$q \quad q \quad q \quad = 2 \sin \left(2 \theta\right) \cos \left(2 \theta\right)$
$q \quad q \quad q \quad = 2 \left(2 \sin \theta \cos \theta\right) \left(1 - 2 {\sin}^{2} \theta\right)$
$q \quad q \quad q \quad = 4 \sin \theta \cos \theta \left(1 - 2 {\sin}^{2} \theta\right)$

$\tan \left(4 \theta\right) = \sin \frac{4 \theta}{\cos} \left(4 \theta\right)$
$q \quad q \quad q \quad = \frac{4 \sin \theta \cos \theta \left(1 - 2 {\sin}^{2} \theta\right)}{1 - 8 {\sin}^{2} \theta {\cos}^{2} \theta}$
$f \left(\theta\right) = - \tan \left(4 \theta\right) + \sin \left(2 \theta\right) + \cos \left(4 \theta\right)$
$q \quad \setminus \quad = - \frac{4 \sin \theta \cos \theta \left(1 - 2 {\sin}^{2} \theta\right)}{1 - 8 {\sin}^{2} \theta {\cos}^{2} \theta}$
$q \quad q \quad - 2 \sin \theta \cos \theta + 1 - 8 {\sin}^{2} \theta {\cos}^{2} \theta$