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

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

`Tan(B) =(AC)/(AB) = -sqrt(3)/3`
and `Tan(F) = (DE)/(DF) = -1/sqrt(3)`
These two angles are the same since `-sqrt(3)/3` =`-1/sqrt(3)`

Pytagoras says that the hypotenuse squares is the sum of the square of the sides, i.e.
`EF^2= DE^2 + FD^2 = 1+3=4`
Therefore `EF=sqrt(4)=2`

That means `EF = 2DE`.

Hopefully you remember that in an equilateral triangle all sides are equal, and that each angle is `60^@` like this:

Here `GJ = 1/2 GK`, which is exactly the situation we have here.

This means that `/_DEF=60^@`
`/_DFE=30^@`

Except that we have been working with positive values, so
`/_DFE=-30^@`

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 = sqrt{3}/2`

`cos 60^circ = sin 30^circ = 1/2`

`cos 45^circ = sin 45^circ = 1/sqrt{2} = sqrt{2}/2`

Now we learn the tangents.

`tan 30^circ = {sin 30^circ}/{cos 30^circ} = {1/2} / {sqrt{3}/2} = 1/sqrt{3} = sqrt{3}/3`

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

`tan 60^circ = {sin 60^circ}/{cos 60^circ} = {sqrt{3}/2}/{1/2}=sqrt{3}`

Complementary angles have reciprocal tangents.

`tan 45^circ = {sin 45^circ}/{cos 45^circ}={1/sqrt{2}}/{1/sqrt{2}}=1`

OK, we learned to recognize

`tan 30^circ = sqrt{3}/3`

So,

`tan (-30^circ) = -sqrt{3}/3`

The general solution to `tan x=tan x` is `x=a + 180^circ k quad` integer `k`.

`tan x= tan (-30^circ)`

`x= -30^circ + 180^circ k quad` integer `k`

That's `-30^circ` in the fourth quadrant and `150^circ` in the second.