# How do you find the exact value of inverse trig functions?

May 7, 2018

Students are only expected to memorize the trig functions of the 30/60/90 triangle and the 45/45/90 triangle, so really only have to remember how to evaluate "exactly":

$\arccos \left(0\right) , \arccos \left(\pm \frac{1}{2}\right) , \arccos \left(\pm \frac{\sqrt{2}}{2}\right) , \arccos \left(\pm \frac{\sqrt{3}}{2}\right) , \arccos \left(1\right)$

Same list for $\arcsin$

$\arctan \left(0\right) , \arctan \left(\pm 1\right) , \arctan \left(\pm \sqrt{3}\right) , \arctan \left(\pm \frac{1}{\sqrt{3}}\right)$

#### Explanation:

Except for a handful of arguments, the inverse trig functions won't have exact values.

The dirty little secret of trig as taught is that the students are really expected to deal with only two triangles "exactly." Those are of course 30/60/90 and 45/45/90. Learn the trig functions of the multiples of ${30}^{\circ}$ and ${45}^{\circ}$; those are pretty much the only one's a student will be asked to invert "exactly."

You already know them, e.g. $\sin {30}^{\circ} = \cos {60}^{\circ} = \frac{1}{2} ,$ $\cos {30}^{\circ} = \sin {60}^{\circ} = \setminus \frac{\sqrt{3}}{2}$ and $\sin {45}^{\circ} = \cos {45}^{\circ} = \frac{\sqrt{2}}{2.}$ The tangents are $\tan {30}^{\circ} = \frac{1}{\sqrt{3}} ,$ $\tan {45}^{\circ} = 1 ,$ and $\tan {60}^{\circ} = \sqrt{3} .$ There are also the multiples of ${90}^{\circ}$ (easy) and the other quadrants, which involve some sign twiddling. It's really not that much to remember.

So a student will be expected to do "exactly":

$\arctan \left(1\right) , \arctan \left(\sqrt{3}\right) , \arctan \left(\frac{1}{\sqrt{3}}\right) , \arctan \left(0\right)$

$\arcsin \left(\frac{1}{2}\right) , \arcsin \left(\frac{\sqrt{2}}{2}\right) , \arcsin \left(\frac{\sqrt{3}}{2}\right) , \arcsin \left(0\right) , \arcsin \left(1\right)$

$\arccos$ of the same set.

These can appear with a negative sign as well..