Deriving The Unit Circle

The unit circle is a special circle that relates the trigonometric ratios of sine, cosine, and tangent all together. The general equation of the unit circle is as follows:

(1)
\begin{align} cos^2(\theta) + sin^2(\theta) = 1 \end{align}

The General Equation of a Circle

The unit circle equation is rather simple to derive. First, let's note that the general form of the equation of a circle is as follows:

(2)
\begin{equation} (x - h)^2 + (y - k)^2 = r^2 \end{equation}

It is not fundamentally necessary to know the general equation of a circle when we talk specifically about trigonometry at this level. Nevertheless, we will define certain elements of the general equation of a circle in order to understand the unit circle as it is an equation of a circle.

The coordinates (h, k) represent the position of the center of the circle, and r represents the radius of the circle.

Example 1

Determine the equation of a circle that has its center at coordinates (3, -2), and has a radius of 4.

In order to solve this problem and determine the equation of a circle with these specific requirements, we can simply plug in what we know into the general equation of a circle. Thus, we obtain the following:

(3)
\begin{equation} (x - 3)^2 + (y - (-2))^2 + 4^2 \end{equation}

And when simplified we obtain:

(4)
\begin{equation} (x - 3)^2 + (y + 2)^2 = 16 \end{equation}

Here is a graph of the function that we derived. Verify that the center of the circle is at (3, -2), and that the radius of the circle, r = 4:

Screen%20Shot%202013-11-23%20at%203.05.53%20AM.png

Deriving the Unit Circle

For convenience sake, we will position the center of the unit circle at the origin, which is also known as the the point (0, 0). Thus, we can begin to write the general equation of our unit circle in the following manner:

(5)
\begin{equation} (x - 0)^2 + (y - 0)^2 = r^2 \end{equation}

Which can thus be simplified to obtain the following equation:

(6)
\begin{equation} x^2 + y^2 = r^2 \end{equation}

We will now figure out the appropriate value for the radius of our unit circle. We will determine the radius of the unit circle with some specific properties of the cosine and sine functions. Remember that:

(7)
\begin{align} -1 \leq cos(\theta) \leq 1 \end{align}
(8)
\begin{align} -1 \leq sin(\theta) \leq 1 \end{align}

Thus, for any value of Θ, we will always obtain an output greater or equal to -1 and less or equal to +1 for both the functions of cosine and sine.

Hence, since we know that the possible outputs on cosΘ and sinΘ will be between -1 and 1, then we can determine that an appropriate diameter of our unit circle would be the distance between -1 and 1. Hence, we know the distance between -1 and 1 is 2 such that 1 - (-1) = 2. However, we want to find an appropriate radius for our unit circle. The fortunate thing is that we know the relationship between the diameter D of a circle and its radius r such that:

(9)
\begin{equation} D = 2r \end{equation}

Thus, we know the appropriate diameter of our unit circle will be 2. We can thus obtain an appropriate value of our radius such that:

(10)
\begin{equation} 2 = 2r \end{equation}
(11)
\begin{equation} r = 1 \end{equation}

Now we can plug in our value for r into our general equation for a unit circle to obtain:

(12)
\begin{equation} x^2 + y^2 = 1^2 \end{equation}

Or simply:

(13)
\begin{equation} x^2 + y^2 = 1 \end{equation}

We are now going to relate cosine and sine to the values of x and y in our general equation of our unit circle. Let's first geometrically solve for a relationship between the variables x and y and the trigonometric ratios cosine and sine. We're going to first construct our equation of a circle with a center at the origin and a radius of 1:

We will then begin to construct an arbitrary triangle that is inscribed within our circle. Let's begin by first creating a ray that starts at the origin and terminates where it intersects the unit circle. We will denote this ray by S. Thus, we will define Θ as the angle formed from the origin to the point (1, 0), or simply the positive x-axis and the ray, S. The diagram below illustrates what we have accomplished thus far:

Screen%20Shot%202013-11-23%20at%203.32.17%20AM.png

We will now connect ray S with some point on the x-axis to form a right triangle. Remember, a right-triangle will allow use to utilize the trigonometric ratios of sine and cosine. Hence, the line we draw from the end point of ray S will be perpendicular to some point on the x-axis. We will thus draw that line to enclose our inscribed triangle.

Screen%20Shot%202013-11-23%20at%203.32.17%20AM%281%29.png

Now we know that the trigonometric ratios of cosine and sine follow such that:

(14)
\begin{align} cos(\theta) = \frac{adjacent}{hypotenuse} \end{align}
(15)
\begin{align} sin(\theta) = \frac{opposite}{hypotenuse} \end{align}

Thus since we have a fixed location for Θ, we can assign the lengths of our triangle as being either the opposite, adjacent, or hypotenuse lengths as follows:

Screen%20Shot%202013-11-23%20at%203.32.17%20AM%282%29.png

Now it is important to recognize that for our triangle, the adjacent side of the triangle is equivalent to that of the x-coordinate associated with the point where the ray S intersects the unit circle. It is also important to notice that the hypotenuse is equal to the length of the ray S. However, S is simply the radius of the unit circle which we defined to be equal to 1. Hence we obtain that:

(16)
\begin{align} cos(\theta) = \frac{x}{1} = x \end{align}

Additionally, it is important to recognize that for our triangle, the opposite side of the triangle is equivalent to that of the y-coordinate associated with the point where the ray S intersects the unit circle. Again, the hypotenuse is equal to 1. Thus we obtain:

(17)
\begin{align} sin(\theta) = \frac{y}{1} = y \end{align}

We have now related the variables x and y with our trigonometric ratios of cosine and sine. Thus we can rewrite our general equation of the unit circle as follows:

(18)
\begin{align} (cos(\theta))^2 + (sin(\theta))^2 = 1 \end{align}

This general equation of the unit circle may also be written as:

(19)
\begin{align} cos^2(\theta) + sin^2(\theta) = 1 \end{align}

IMPORTANT: Note that the following equation is INCORRECT for representing the equation of the unit circle:

(20)
\begin{align} cos\theta^2 + sin\theta^2 = 1 = cos(\theta^2) + sin(\theta^2) \end{align}

This is because the way this equation is written squares ONLY Θ as opposed to the trigonometric ratios of cosine and sine. This is a common mistake, so beware!

Proof of the Range of Cosine and Sine

This material is supplemental knowledge for defining the unit circle. It is generally not necessary to know unless you have already been taught on graphing sinusoidal functions.

When defining the unit circle, we said that the values of cosine and sine are restricted to be between -1 and +1 inclusive, or:

(21)
\begin{align} -1 \leq cos(\theta) \leq 1 \end{align}
(22)
\begin{align} -1 \leq sin(\theta) \leq 1 \end{align}

We will now prove this. Let's let two functions relating y1 and y2 (where the 1 and 2 subscripts denotes a different function on the same cartesian plane) with the trigonometric functions of cosx and sinx. We will also show their graphs:

(23)
\begin{equation} y_{1} = cosx \end{equation}
Screen%20Shot%202013-11-23%20at%203.58.04%20AM.png
(24)
\begin{equation} y_{2} = sinx \end{equation}
Screen%20Shot%202013-11-23%20at%203.58.15%20AM.png

In both graphs we notice that the y-values are restricted between -1 and 1 inclusive. We say that the possible values of y for a function y = f(x) is known as the Range of a Function Thus, we know that for both functions:

(25)
\begin{align} -1 \leq y_{1} \leq 1, -1 \leq y_{2} \leq 1 \end{align}

But we defined that y1 = cosx, and y2 = sinx. Thus we can substitute y1 for cosx, and y2 for sinx so that we get the range to be that:

(26)
\begin{align} -1 \leq cosx \leq 1, -1 \leq sinx \leq 1 \end{align}

Which is identical to what we used to prove the equation of the unit circle earlier with the exception being the interchanging of the variable x with the variable Θ.

We sometimes denote the range in interval notation such that the range, R, of our functions y1 and y2 are as follows:

(27)
\begin{equation} R_{y_{1}}: [-1, 1] , R_{y_{2}}: [-1, 1] \end{equation}
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