# How do you evaluate tan[arccos(1/3)]?

May 3, 2016

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

#### Explanation:

arccos is the reversing of the process of cos to give the angle

$\implies \theta = \left[\arccos \left(\frac{1}{3}\right) = \arccos \left(\left(\text{adjacent")/("hypotenuse}\right)\right)\right]$

So this is giving us 2 length of sides for a right triangle. From which we can work out the tangent value.

By Pythagoras and using the notation in the diagram.

${c}^{2} = {b}^{2} + {a}^{2} \text{ " =>" } {3}^{2} = {1}^{2} + {a}^{2}$

Thus $a = \sqrt{8} = \sqrt{2 \times {2}^{2}} = 2 \sqrt{2}$

$\tan \left(\theta\right) = \left(\text{opposite")/("adjacent}\right) = \frac{a}{b} = \frac{2 \sqrt{2}}{1}$

$\tan \left(\theta\right) = 2 \sqrt{2}$

Thus: $\text{ } \tan \left[\arccos \left(\frac{1}{3}\right)\right] = 2 \sqrt{2}$