# How do you evaluate tan^-1 (-sqrt3) without a calculator?

May 22, 2018

$x = \frac{2 \pi}{3} + k \pi$

#### Explanation:

$\tan x = - \sqrt{3}$. Find arc x.
Trig table and unit circle give:
$x = \frac{2 \pi}{3} + k \pi$

May 24, 2018

$\arctan \left(- \sqrt{3}\right) = {120}^{\circ} + {180}^{\circ} k \quad$integer $k$

The principal value is $- {60}^{\circ}$

#### Explanation:

Students are only expected to know two triangles, 30/60/90, and 45/45/90, and be able to figure them out in all four quadrants.

It's a goofy way to structure an entire course, but once you understand it problems become easier.

Rule of thumb, $\sqrt{3}$ means 30/60/90 and $\sqrt{2}$ means 45/45/90.

We know 30 and 60 degrees have sine and cosine $\frac{1}{2}$ and $\frac{\sqrt{3}}{2.}$

Tangent is slope, sine over cosine, so we have a sine of $- \frac{\sqrt{3}}{2}$ and a cosine of $\frac{1}{2}$ or a sine of $\frac{\sqrt{3}}{2}$ and a cosine of $- \frac{1}{2}$. So fourth or second quadrant. When the sine is bigger, that's ${60}^{\circ} ,$ so we're looking at $- {60}^{\circ}$ and ${120}^{\circ} .$

The principal value is in the fourth quadrant, $- {60}^{\circ}$. I prefer to treat $\arctan$ as the multivalued inverse,

$\arctan \left(- \sqrt{3}\right) = {120}^{\circ} + {180}^{\circ} k \quad$integer $k$