# What is Arctan (-1/3 * sqrt3)?

May 5, 2018

$\arctan \left(- \frac{1}{3} \cdot \sqrt{3}\right) = - {30}^{\circ}$

#### Explanation:

We need to remember what the tan is. See the following two right triangles:

$T a n \left(B\right) = \frac{A C}{A B} = - \frac{\sqrt{3}}{3}$
and $T a n \left(F\right) = \frac{D E}{D F} = - \frac{1}{\sqrt{3}}$
These two angles are the same since $- \frac{\sqrt{3}}{3}$ =$- \frac{1}{\sqrt{3}}$

Pytagoras says that the hypotenuse squares is the sum of the square of the sides, i.e.
$E {F}^{2} = D {E}^{2} + F {D}^{2} = 1 + 3 = 4$
Therefore $E F = \sqrt{4} = 2$

That means $E F = 2 D E$.

Hopefully you remember that in an equilateral triangle all sides are equal, and that each angle is ${60}^{\circ}$ like this:

Here $G J = \frac{1}{2} G K$, which is exactly the situation we have here.

This means that $\angle D E F = {60}^{\circ}$
$\angle D F E = {30}^{\circ}$

Except that we have been working with positive values, so
$\angle D F E = - {30}^{\circ}$

May 21, 2018

$\arctan \left(- \frac{\sqrt{3}}{3}\right) = - {30}^{\circ} + {180}^{\circ} k \quad$ integer $k$

#### Explanation:

There are really very few non-trivial trig function values that student is expected to have at their fingertips. Besides the multiples of ${90}^{\circ}$ they're all from either 45/45/90 or 30/60/90 triangles.

We hopefully can already recognize

$\cos {30}^{\circ} = \sin {60}^{\circ} = \frac{\sqrt{3}}{2}$

$\cos {60}^{\circ} = \sin {30}^{\circ} = \frac{1}{2}$

$\cos {45}^{\circ} = \sin {45}^{\circ} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$

Now we learn the tangents.

$\tan {30}^{\circ} = \frac{\sin {30}^{\circ}}{\cos {30}^{\circ}} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

That's pretty close to the one we need for this problem; let's enumerate the others.

$\tan {60}^{\circ} = \frac{\sin {60}^{\circ}}{\cos {60}^{\circ}} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}$

Complementary angles have reciprocal tangents.

$\tan {45}^{\circ} = \frac{\sin {45}^{\circ}}{\cos {45}^{\circ}} = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = 1$

OK, we learned to recognize

$\tan {30}^{\circ} = \frac{\sqrt{3}}{3}$

So,

$\tan \left(- {30}^{\circ}\right) = - \frac{\sqrt{3}}{3}$

The general solution to $\tan x = \tan x$ is $x = a + {180}^{\circ} k \quad$ integer $k$.

$\tan x = \tan \left(- {30}^{\circ}\right)$

$x = - {30}^{\circ} + {180}^{\circ} k \quad$ integer $k$

That's $- {30}^{\circ}$ in the fourth quadrant and ${150}^{\circ}$ in the second.