.Multiples of 18 Degrees
Trigonometry texts always include material early in the course on finding the
exact values of trig functions of the angles 0°, 30°, 45°, 60°, and 90°.
It is also true that by a similar argument, exact values of trig functions of
the angles 18°, 36°, 54°, and 72° may also be found. These angles are
equivalent to unit circle arc lengths that are multiples of
.
Trig Values of Multiples of
:
| θ | sin θ | cos θ | tan θ | cot θ | sec θ | csc θ |
|
 |
 |
 |
 |
 |
 |
 |
 |
 |
 |
 |
 |
 |
 |
|
|
|
|
|
|
 |
|
|
|
|
|
|
Proof: Let A be the point
on the unit circle,
and choose points B, C, D, and E on the unit circle
counterclockwise from A so that arc lengths AB,
BC, CD, DE, and EA are each
. Let
be the coordinates of
point B, and let
be the coordinates of point C. Then, by symmetry, the coordinates
of point D are 
, and
the coordinates of point E are
. Since all of
these points are on the unit circle, the equations
and
are both true.
Since the line segments AB and BC are equal, we apply the
distance formula to obtain the equation
, which simplifies to
. Similarly, the
equality of the line segments AC and BD will produce an equation
which simplifies to 
.
Equating these results gives
. The equality of
the line segments AB and CD will produce an equation which
simplifies to 
.
Using to substitute
for the variable w, then using
to substitute for the
variable v, then removing fractions, we obtain the polynomial equation
. Since 1 is a
solution of this equation, the polynomial factors into a linear and a quadratic
expression. Since point B is in the first quadrant, the value of
x must be positive, and the quadratic formula produces
. The values of
the variables y, v, and w then can be produced by
substitution.
The values of the sine and cosine functions are then related to the variables through symmetry arguments as follows:
The values of the other four trigonometric functions can be found with the ratio and reciprocal identities.♦
Alternate Proof: Suppose triangle ABC is isosceles, with congruent sides AB and AC each of length 1, and the measure of angle BAC is 36°. Then angles ABC and ACB are also congruent, and since the sum of the angles of a triangle is 180°, the measure of each of these two angles is 72°.
Now bisect angle ABC, and let point D be the intersection of side AC with the bisector of angle ABC. Now angle ABD is 36°, so triangle ABD is isosceles, with congruent sides AD and BD. Also, since angle DBC is 36° and angle BCD is 72°, we find that angle BDC must be 72°, therefore triangle BDC is isosceles with congruent sides BD and BC. Note that triangles ABC and BCD are similar triangles, by AAA Similarity.
Let x be the length of side BC (which is also the length of BD and of
AD). Then by similar triangles, we obtain the proportion
, which is equivalent to
.
Solving this equation by cross-multiplying, using the quadratic formula and
discarding the negative solution, we obtain
.
Now bisect angle BAC, and let point E be the intersection of side BC with the bisector of angle BAC. Since triangle ABC is isosceles, triangles ABE and ACE are congruent, and by corresponding parts CE is half the measure of BC, and angle AEC is a right angle.
Triangle ACE is now observed to be a right triangle whose acute angles
measure 18° and 72°, and with sides of lengths AC = 1 and
. By the definitions of sine and cosine, we have
.
All other results can be obtained through the use of
sum, difference, and
double
angle formulas, and the Pythagorean identity.♦