How do you evaluate tan(arccos(2/3))?

Jun 10, 2015

$\tan \left(\arccos \left(\frac{2}{3}\right)\right) = \frac{\sqrt{5}}{2}$.

Explanation:

$\alpha = \arccos \left(\frac{2}{3}\right)$.
$\alpha$ isn't a known value, but it's about 48,19°.
$\tan \left(\alpha\right) = \sin \frac{\alpha}{\cos} \alpha$
We can say something about $\cos \alpha$ and $\sin \alpha$:
$\cos \alpha = \frac{2}{3}$
$\sin \alpha = \sqrt{1 - {\left(\cos \alpha\right)}^{2}}$ (for the first fundamental relation*).
So $\sin \alpha = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3}$.

$\tan \left(\alpha\right) = \sin \frac{\alpha}{\cos} \alpha = \frac{\frac{\sqrt{5}}{3}}{\frac{2}{3}} = \frac{\sqrt{5}}{2.}$

So $\tan \left(\arccos \left(\frac{2}{3}\right)\right) = \frac{\sqrt{5}}{2}$.

*The first fundamental relation:
${\left(\cos \alpha\right)}^{2} + {\left(\sin \alpha\right)}^{2} = 1$
From which we can get $\sin \alpha$:
${\left(\sin \alpha\right)}^{2} = 1 - {\left(\cos \alpha\right)}^{2}$
$\sin \alpha = \pm \sqrt{1 - {\left(\cos \alpha\right)}^{2}}$
But in this case we consider only positive values.