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

2 Answers
May 5, 2018

#arctan (-1/3*sqrt(3))=-30^@#

Explanation:

We need to remember what the tan is. See the following two right triangles:
enter image source here
#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:
enter image source here
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^@#

May 21, 2018

# arctan(-sqrt{3}/3) = -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 = 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.