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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
circle

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
circle

Hi, Kelly.

Do you understand why sin(pi/4) = sqrt(2)/2? I'll go over it, because

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
Theorem:

/|
/  |
1  /    |
/      | a
/        |
/          |
/____________|
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:

B
/|
/  |
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:

B
/|
/ |
/  |
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
http://mathforum.org/dr.math/
```
Associated Topics:
High School Trigonometry

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