Unizor - Trigonometry - Basic Angles
Channel: Zor Shekhtman on UNIZOR Education
Date: 11/07/2013

Certain angles with "good" characteristics are used in trigonometry quite extensively. They participate in many problems, theorems and calculations. Let's define them, position them on the unit circle and determine their coordinates.

In the first quadrant of the coordinate plane these basic angles are:
0° = 0
30° = π/6
45° = π/4
60° = π/3
90° = π/2

In other quadrants basic angles are defined by adding 90°, 180° or 270° to the above.

All of the above angles are measured, by definition of a positive direction of an angle measure, counterclockwise from the positive direction of the X-axis. The same angles measured clockwise must be considered negative. Thus an angle of, say, 315° (which is 7π/4 radians) is the same as an angle of −45° (which is −π/4 radians).

Let's determine the coordinates of points on a unit circle that form basic angles from the positive direction of the X-axis counterclockwise to vectors to these points from the beginning of coordinates.

Assume that a unit circle (that is, of a radius 1) is positioned on a coordinate plane with a center at the beginning of coordinates - point O(0,0).
Let point K(1,0) be an intersection of this unit circle with a positive direction of the X-axis, point L(0,1) be an intersection of this unit circle with a positive direction of the Y-axis, point M(−1,0) be an intersection of this unit circle with a negative direction of the X-axis and point N(0,−1) be an intersection of this unit circle with a negative direction of the Y-axis.
Let point A be a point on the unit circle such as an angle ∠KOA (the order of letters is important) measured counterclockwise is one of the basic angles mentioned above. Let coordinates of point A be (p,q).
Our task is to determine (p,q) for each basic angle.

Let's begin with the angles in the first quadrant.

(a) ∠KOA = 0° = 0 rad.
This is a trivial case, point A is (1,0).

(b) ∠KOA = 30° = π/6 rad.
Drop a perpendicular from point A to the X-axis with point B being an intersection of this perpendicular with the X-axis. Consider a right triangle ΔABO. Its acute angle ∠BOA = 30° = π/6 rad. As is known from the geometry, the cathetus AB opposite to an angle of 30° equals to half a hypotenuse. But hypotenuse equals to a radius of a unit circle, that is 1.
Therefore, an opposite cathetus AB equals to 1/2, which is an ordinate of point A.
A segment OB represents an abscissa of point A and is equal to √3/2 by Pythagorean Theorem:
OA^2 = AB^2 + OB^2
1 = (1/2)^2 + OB^2
OB^2 = 3/4
OB = √3/2
So, point A is (√3/2,1/2).

(c) ∠KOA = 45° = π/4 rad.
Drop a perpendicular from point A to the X-axis with point B being an intersection of this perpendicular with the X-axis. Consider a right triangle ΔABO. Its acute angle ∠BOA = 45° = π/4 rad. As is known from the geometry, the other acute angle in this right triangle is also equal to 45° and the catheti AB and OB are congruent. It's easy to calculate their length using Pythagorean Theorem knowing that a hypotenuse OA equals 1 as a radius of a unit circle.
Let AB=OB=x. Then
OA^2 = AB^2 + OB^2
1 = x^2 + x^2
2·x^2 = 1
x = √2/2
So, point A is (√2/2,√2/2).

(d) ∠KOA = 60° = π/3 rad.
Drop a perpendicular from point A to the X-axis with point B being an intersection of this perpendicular with the X-axis. Consider a right triangle ΔABO. Its acute angle ∠BOA = 60° = π/3 rad. As is known from the geometry, the other acute angle of this right triangle is 30°. Therefore, the cathetus OB opposite to an angle of 30° equals to half a hypotenuse. But hypotenuse equals to a radius of a unit circle, that is 1.
Therefore, an opposite cathetus OB equals to 1/2, which is an abscissa of point A.
A segment AB represents an ordinate of point A and is equal to √3/2 by Pythagorean Theorem:
OA^2 = OB^2 + AB^2
1 = (1/2)2 + AB^2
AB^2 = 3/4
AB = √3/2
So, point A is (1/2,√3/2).

(e) ∠KOA = 90° = π/2 rad.
This is a trivial case, point A is (0,1).

To analyze angles greater than 90° we will use the following theorem (watch the video for a proof).

Theorem
Consider two points, P and Q, on a circle with a center O and a diameter ST. Assume that P and Q are on opposite sides of this diameter and angles ∠POS and ∠SOQ are acute and congruent to each other (in other words, the angles are symmetrical relatively to the diameter).
Then P and Q are symmetrical relatively to this diameter.

Notice that if two points are symmetrical relatively to the Y-axis, they have the same ordinates and opposite in sign but equal in absolute value abscissas. If one has coordinates (u,v), the other has coordinates (−u,v).
Analogously, if two points are symmetrical relatively to the X-axis, they have the same abscissas and opposite in sign but equal in absolute value ordinates. If one has coordinates (u,v), the other has coordinates (u,−v).