# What does cos(arctan((3pi)/2))-2sin(arcsec(pi/4))  equal?

Jan 21, 2016

$\frac{2}{\sqrt{9 {\pi}^{2} + 4}} - 2 \cdot \frac{\sqrt{{\pi}^{2} - 16}}{\pi}$

#### Explanation:

If we make $\arctan \left(\frac{3 \pi}{2}\right) = \theta$ then
$\tan \theta = \frac{3 \pi}{2}$ => $\sin \frac{\theta}{\cos} \theta = \frac{3 \pi}{2}$ => $\sqrt{1 - {\cos}^{2} \theta} = \frac{3 \pi}{2} \cdot \cos \theta$ =>$1 - {\cos}^{2} \theta = \frac{9 {\pi}^{2}}{4}$ => ${\cos}^{2} \theta \cdot \frac{9 {\pi}^{2} + 4}{4} = 1$ => $\cos \theta = \frac{2}{\sqrt{9 {\pi}^{2} + 4}}$ => $\theta = \arccos \left(\frac{2}{\sqrt{9 {\pi}^{2} + 4}}\right)$

If we make $a r c \sec \left(\frac{\pi}{4}\right) = \phi$ then
$\sec \phi = \frac{\pi}{4}$ => $\cos \phi = \frac{4}{\pi}$
$\sin \phi = \sqrt{1 - {\cos}^{2} \phi} = \sqrt{1 - \frac{16}{\pi} ^ 2}$ => $\sin \phi = \frac{\sqrt{{\pi}^{2} - 16}}{\pi}$ => $\phi = \arcsin \left(\frac{\sqrt{{\pi}^{2} - 16}}{\pi}\right)$

Using the results in the original expression
$\cos \theta - 2 \sin \phi =$
$\cos \left(\arccos \left(\frac{2}{\sqrt{9 {\pi}^{2} + 4}}\right)\right) - 2 \cdot \sin \left(\arcsin \left(\frac{\sqrt{{\pi}^{2} - 16}}{\pi}\right)\right) =$
$= \frac{2}{\sqrt{9 {\pi}^{2} + 4}} - 2 \cdot \frac{\sqrt{{\pi}^{2} - 16}}{\pi}$