The unit circle is one of the most used "laboratories" for understanding many Math concepts. The unit circle crosses Algebra (with equation of the circle), Geometry (with angles, triangles and Pythagorean Theorem) and Trigonometry (sine, cosine, tangent) in one place.

The name says it clearly: The unit circle is a circle of radius \(r=1\), which for convenience is assumed to be centered at the origin \((0, 0)\). Note that we are talking about the two-dimensional case.

## Angles and the Unit circle

The unit circle, or a circle of any radius, is a very practical way of working with angles. Let us a recall that the measure of an angle is proportional to the amount of the circumference of the circle that the angle spans.

For example, if an angle spans a quarter of the circumference, and its origin is the same as the center of the circle, then the measure of the angle is a quarter of the measure of a full angle, which is 360/4 = 90^{o} if measured in degrees, or \(2\pi/4 = \pi/2\) if measured in radians

There are other circumstances in which the origin of the angle is not the same as the center of the circle, like in the case of the graph below:

## Trigonometric Functions and the Unit Circle

Using the unit circle is super useful to work with trigonometric functions. Indeed, it turns out that if we have a point \((x,y)\) in a circle with radius \(r\), then we have that

\[\large \sin \alpha = \frac{y}{r}\] \[\large \cos \alpha = \frac{x}{r}\] \[\large \tan \alpha = \frac{y}{x}\]where \(\alpha\) is the angle shown in the figure below:

But when \(r = 1\), this is, when the radius is 1 (which is the case in the unit circle), we find that

\[\large \sin \alpha = y \] \[\large \cos \alpha = x \] \[\large \tan \alpha = \frac{y}{x}\]Therefore, the operation with trigonometric functions is much easier when the radius of a circle is 1, and then everything becomes much more visual. And we can use mnemonic rules such as "the sine of an angle is the opposite side" and "the cosine of an angle is the adjacent side".

## The Equation of the Unit Circle

For a unit circle that is centered at the origin, the equation that any point \((x, y)\) on it satisfies is:

\[\large x^2 + y^2 = 1\]Any pair \((x, y)\) that belongs to a circle of radius 1 must satisfy the above. If the point \((x, y)\) does not satisfy the above, then it does not belong to the circle.

__EXAMPLE 1__

Does the point \(\displaystyle (\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})\) belong to the unit circle?.

### ANSWER:

We need to verify that the point satisfies the equation defined above. We get:

\[\large x^2 + y^2 = \left(\frac{\sqrt 2}{2}\right)^2+ \left(\frac{\sqrt 2}{2}\right)^2 = \frac{2}{4} + \frac{2}{4} = 1 \]So in this case, the point \( \displaystyle (\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})\) does belong to the unit circle

__EXAMPLE 2__

Does the point \(\displaystyle (\frac{1}{2}, \frac{2}{3})\) belong to the unit circle?.

### ANSWER:

We need to verify whether or not the point satisfies the equation defined above. We get:

\[\large x^2 + y^2 = \left(\frac{1}{2}\right)^2+ \left(\frac{2}{3}\right)^2 = \frac{1}{4} + \frac{4}{9} = \frac{25}{36} \]So in this case, the point \( \displaystyle (\frac{1}{2}, \frac{2}{3})\) does NOT belong to the unit circle

## More About the Unit Circle

One of the questions I always get is whether or not the equation of the unit circle describes a function. The answer is NO. Indeed, the equation of the unit circle defines a relation, instead.

There are at least two ways of knowing. The favorite one for students is the "vertical line test". We have the following graph:

See in the graph above, and we can see that we have this vertical line that crosses the graph at more than one point. The conclusion is that the graph represents a relation, not a function.

Now, if you want to know what happens when the radius is not 1, and the circle is not centered at the origin, check our tutorial about the general equation of the circle, in which the general case is handled.

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