# How do you find the value of tan [(1/2) cos^-1 (2/3)]?

Oct 4, 2015

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

#### Explanation:

From the pythagorean identity we have that

${\sin}^{2} x + {\cos}^{2} x = 1$

Dividing both sides by ${\cos}^{2} x$ we have

${\tan}^{2} x + 1 = \frac{1}{\cos} ^ 2 x$

Which means that, if we isolate the tangent we have

${\tan}^{2} x = \frac{1}{\cos} ^ 2 x - 1$

So for $x = \arccos \frac{\frac{2}{3}}{2}$ we have

${\tan}^{2} \left(\arccos \frac{\frac{2}{3}}{2}\right) = \frac{1}{\cos} ^ 2 \left(\arccos \frac{\frac{2}{3}}{2}\right) - 1$

${\cos}^{2} \left(\arccos \frac{\frac{2}{3}}{2}\right)$ can be calculated using the half angle formula, so we know that

${\cos}^{2} \left(\arccos \frac{\frac{2}{3}}{2}\right) = \frac{1 + \cos \left(\arccos \left(\frac{2}{3}\right)\right)}{2}$

Since $\cos \left(\arccos \left(x\right)\right) = x$ we can say that

${\cos}^{2} \left(\arccos \frac{\frac{2}{3}}{2}\right) = \frac{1 + \frac{2}{3}}{2} = \frac{\frac{5}{3}}{2} = \frac{5}{3} \cdot \frac{1}{2} = \frac{5}{6}$

Putting that back on the formula

${\tan}^{2} \left(\arccos \frac{\frac{2}{3}}{2}\right) = \frac{1}{\frac{5}{6}} - 1 = \frac{6}{5} - 1 = \frac{6 - 5}{5} = \frac{1}{5}$

Taking the root

$\tan \left(\arccos \frac{\frac{2}{3}}{2}\right) = \pm \sqrt{\frac{1}{5}}$

Knowing that during the range of the arccosine, the tangent is only negative if the cosine's also negative we have that

$\tan \left(\arccos \frac{\frac{2}{3}}{2}\right) = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}$