Special Angles

Date: 01/18/2002 at 19:52:37
From: Patricia Henry
Subject: Trigonometry

I am taking a 1st year math course at university. I have been through 
the text, my notes, and math software (The Princeton Review Math 
Library) but I cannot figure this out. In the text, a sample problem 
includes:

   tan(theta) = -sqrt(3)

   theta = 2pi/3

I cannot figure out how the solution for theta was found.  Can you 
tell me?  I need to know in order to answer the questions at the end 
of the chapter.

Thanks,
Patricia

Date: 01/18/2002 at 22:54:32
From: Doctor Peterson
Subject: Re: Trigonometry

Hi, Patricia.

This is one of the "special angles" whose trig functions should be 
familiar 
to you:

   30-60-90 and 45-45-90 Triangles
   http://mathforum.org/dr.math/problems/kristina3.15.99.html   

   Remembering Trig Functions
   http://mathforum.org/dr.math/problems/kim.09.27.01.html   

If you don't recall those offhand, but suspect (since they asked) that 
the answer must be simple, you can work backward and construct the 
triangle. First, ignore the signs and just make a right triangle where 
the ratio of the legs is sqrt(3):

          +
         /|
        / |
       /  |sqrt(3)
      /   |
     /A   |
    +-----+
       1

How can we see if this is a familiar triangle? Use Pythagoras to find 
the hypotenuse:

    1^2 + sqrt(3)^2 = 1 + 3 = 4

so the hypotenuse is 2. Hmmm ... that's twice the bottom leg, so if we 
double the right triangle, we get ...

          +
         /|\
        / | \
      2/  |  \2
      /   |   \
     /A   |    \
    +-----+-----+
       1     1

... an equilateral triangle! So angle A must be 60 degrees.

Now you have to handle the sign. To do this, picture a unit circle. 
The tangent will be negative in the second and fourth quadrants, 
so we have

           +
           |\
           | \
    sqrt(3)|  \
           |   \
           |  60\ theta
           +-----o-----------
             -1

So theta is 180-60 = 
120 degrees; or, in radians, pi - pi/3 = 2pi/3.


- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/