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Deriving the Coordinates of the pi/6 Family

Date: 06/09/99 at 20:33:40
From:  Kelly
Subject: Deriving the coordinates of the pi/6 family on the unit 

I need to know how to derive the coordinates of the pi/6 family on the 
unit circle. I know that you use a reference triangle and use the 
definition of sin and cos to get the answer. However, I need to know 
why this works.

Date: 06/10/99 at 12:35:26
From: Doctor Rick
Subject: Re: Deriving the coordinates of the pi/6 family on the unit 

Hi, Kelly.

Do you understand why sin(pi/4) = sqrt(2)/2? I'll go over it, because 
this will lead into the angles you asked about.

Consider a 45-45-90 triangle with a hypotenuse of 1. Because 2 angles 
are equal, it is isosceles; both legs have the same length. If you 
know one leg (a), you know both legs. Let's use the Pythagorean 

               /  |
          1  /    |
           /      | a
         /        |
       /          |

   1 = sqrt(a^2 + a^2)
     = sqrt(2*a^2)
     = sqrt(2)*a

Solve for a:

   1/sqrt(2) = a

Rationalize the denominator by multiplying numerator and denominator 
by sqrt(2):

   sqrt(2)/2 = a

So these are the proportions of the sides of a 45-45-90 triangle:

                /  |
           1  /    |
            /      | sqrt(2)/2
          /        |
        /          |
     A  sqrt(2)/2   C

If A is the center of the unit circle, then B is on the unit circle at 
an angle A = 45 degrees (pi/4). The coordinates of point B are

   (sqrt(2)/2, sqrt(2)/2)

and the trig functions are

   sin(45) = BC/AB = sqrt(2)/2
   cos(45) = AC/AB = sqrt(2)/2
   tan(45) = BC/AC = 1

Now for angles of pi/6 and pi/3, we use the 30-60-90 triangle. It 
isn't isosceles like the 45-45-90 triangle, but it is half of an 
equilateral triangle:

          / | \
       1 /  |  \ 1
        /   |   \
       /   a|    \
      /     |     \
        1/2   1/2

The angles of the equilateral triangle are 60 degrees. The top angle 
is bisected, giving the 30 degree angle. And the base of the 
equilateral triangle is bisected, giving segments of length 1/2. 
That's the key. The side opposite the 30 degree angle of the 30-60-90 
triangle is half the hypotenuse.

Now we can use the Pythagorean Theorem to find our one unknown, as in 
the 45-45-90 triangle:

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

That gives us all the sides of a 30-60-90 triangle:

           / |
          /  |
       1 /   |
        /    | sqrt(3)/2
       /     |
     A   1/2  C

If A is the center of the unit circle, then B is on the unit circle at 
an angle A = 60 degrees (pi/3). The coordinates of point B are

   (1/2, sqrt(3)/2)

and the trig functions are

   sin(60) = cos(30) = BC/AB = sqrt(3)/2
   cos(60) = sin(30) = AC/AB = 1/2
   tan(60) = BC/AC = sqrt(3)
   tan(30) = AC/BC = sqrt(3)/3

I don't memorize numbers very well by themselves; I find it a lot 
easier to remember the figures I've shown and how to use Pythagoras. I 
couldn't tell you how many times I have re-derived these ratios. I 
hope this helps you.

- Doctor Rick, The Math Forum
Associated Topics:
High School Trigonometry

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