Coordinates on Unit Circle - Algebra-2

1. Fundamental Concepts

1.1 Definition of the Unit Circle

The unit circle is a circle in the Cartesian coordinate plane with its center at the origin $(0,0)$ and a radius $r = 1$. Its standard equation is $x^2 + y^2 = 1$. There is a core relationship between the coordinates $(x,y)$ of any point on the circle and trigonometric functions: for an angle $\theta$ rotated counterclockwise from the positive $x$-axis, the $x$-coordinate of the point is $x = \cos\theta$ and the $y$-coordinate is $y = \sin\theta$. In other words, the coordinates of the point can be expressed as $(\cos\theta, \sin\theta)$.

1.2 Basics of Special Right Triangles

The derivation of coordinates on the unit circle relies on the side length ratios of two types of special right triangles, which are the core basis for calculating coordinates of special angles:

45-45-90 Triangle: An isosceles right triangle with a side length ratio of "leg : leg : hypotenuse = ". If the hypotenuse is the radius of the unit circle (with a length of 1), each leg has a length of . This applies to the coordinates of angles such as and .
30-60-90 Triangle: A triangle with a side length ratio of "shorter leg : longer leg : hypotenuse = ". If the hypotenuse is 1, the shorter leg (opposite the angle) has a length of , and the longer leg (opposite the angle) has a length of . This is used for the coordinates of angles like , , and .

1.3 "Standard Position" of Angles

The standard position of an angle in the Cartesian coordinate plane is defined as follows: the vertex of the angle coincides with the origin, and the initial side coincides with the positive x-axis. Angles formed by counterclockwise rotation are positive angles, while those formed by clockwise rotation are negative angles. When determining the coordinates of a point on the unit circle, the standard position of the angle must be considered, and the signs of the x- and y-coordinates are determined based on the quadrant where the angle lies (e.g., in the second quadrant, and ; in the fourth quadrant, and ).

2. Key Concepts

Basic Rule: $$x^2 + y^2 = 1$$

Angle Measurement:

The unit circle has a periodicity of 360° (or radians), meaning that angles and (where k is an integer) correspond to the same coordinates on the unit circle. This property can be used to convert negative angles or angles greater than 360° into angles within the range of 0°-360° for calculation:

Conversion of negative angles: If (), it is equivalent to . For example, , and its coordinates are .
Conversion of angles greater than 360°: If (where is the remainder when is divided by 360° and lies within 0°-360°), for example, , and its coordinates are the same as those of 30°, which is .

Relationship Between Coordinates and Trigonometric Functions

The coordinates of the unit circle directly define the sine and cosine functions:

$x$-coordinate = cosine value: $x = \cos\theta$

$y$-coordinate = sine value: $y = \sin\theta$

From this, the tangent function can be further calculated as $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{y}{x}$ ($x\neq0$). Additionally, the signs of trigonometric functions in each quadrant can be determined based on the signs of the coordinates (e.g., in the third quadrant, $\sin\theta<0$ and $\cos\theta<0$, so $\tan\theta>0$).

3. Examples

3.1 Easy Difficulty (Direct Application of Special Angle Coordinates)

Question: Find the coordinates of and on the unit circle, and write their sine and cosine values.

Solution:

$60^\circ$ is in the first quadrant. According to the table of special angle coordinates, its coordinates are $(\frac{1}{2}, \frac{\sqrt{3}}{2})$. Therefore, $\cos60^\circ = \frac{1}{2}$ and $\sin60^\circ = \frac{\sqrt{3}}{2}$.

$270^\circ$ lies on the negative $y$-axis, and its coordinates are $(0, -1)$. Therefore, $\cos270^\circ = 0$ and $\sin270^\circ = -1$.

3.2 Medium Difficulty (Conversion of Negative Angles / Large Angles)

Question: Find the coordinates of and on the unit circle, and calculate the decimal values of their cosine values (rounded to two decimal places).
Solution:

Step 1: Convert the angles

$-390^\circ = -390^\circ + 2\times360^\circ = 330^\circ$ (add 2 periods to bring the angle into the range of 0°-360°);

$480^\circ = 480^\circ - 360^\circ = 120^\circ$ (subtract 1 period).

Step 2: Look up the coordinates of special angles

The coordinates of $330^\circ$ are $(\frac{\sqrt{3}}{2}, -\frac{1}{2})$, so $\cos(-390^\circ) = \cos330^\circ = \frac{\sqrt{3}}{2} \approx 0.87$;

The coordinates of $120^\circ$ are $(-\frac{1}{2}, \frac{\sqrt{3}}{2})$, so $\cos480^\circ = \cos120^\circ = -\frac{1}{2} = -0.50$.

3.3 Hard Difficulty (Finding Angles from Coordinates + Multi-Quadrant Analysis)

Question: A point $P(x, y)$ on the unit circle satisfies $y = -\frac{\sqrt{2}}{2}$ and $\tan\theta = 1$. Find the angle $\theta$ (within 0°-360°) and its corresponding coordinates.

Solution:

Step 1: From $y = \sin\theta = -\frac{\sqrt{2}}{2}$, it can be determined that $\theta$ is in the third or fourth quadrant (since the sine value is negative).

Step 2: From $\tan\theta = \frac{y}{x} = 1$, we get $x = y = -\frac{\sqrt{2}}{2}$ (if $\theta$ is in the third quadrant, both $x$ and $y$ are negative) or $x = y = \frac{\sqrt{2}}{2}$ (if $\theta$ is in the fourth quadrant, $x$ is positive and $y$ is negative, which would make $\tan\theta = -1$, a contradiction).

Step 3: Determine the angle and coordinates

In the third quadrant, since $\sin\theta = -\frac{\sqrt{2}}{2}$ and $\cos\theta = -\frac{\sqrt{2}}{2}$, the corresponding angle is $180^\circ + 45^\circ = 225^\circ$;

The coordinates are $(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2})$.

4. Problem-Solving Techniques

Visual Strategy: Use the unit circle diagram to visualize the angle and its corresponding coordinates.
Error-Proofing: Always check if the coordinates satisfy the equation $x^2 + y^2 = 1$ .
Concept Reinforcement: Practice finding coordinates for common angles such as $0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}$ .