5.1 – The Unit Circle

5.1 Video 1 of 3: The Unit Circle

5.1 Video 2 of 3: Terminal Points.

5.1 Video 3 of 3: Reference Numbers.

Section Notes

The Unit Circle

The unit circle is the set of points that have a distance of 1 from the origin:

Rendered by QuickLaTeX.com

Points on the unit circle satisfy the equation $x^2+y^2=1$

The Unit Circle
The unit circle is the circle of radius 1 centered at the origin. It’s equation is: $x^2+y^2=1$

Show that the point $\left(\dfrac{2}{5},\dfrac{\sqrt{21}}{5}\right)$ is on the unit circle
The point can be shown to be on the unit circle by satisfying the unit circle equation. $x^2+y^2=1$ Substituting the point into the equation gives: $\left(\dfrac{2}{5}\right)^2+\left(\dfrac{\sqrt{21}}{5}\right)^2=1$ $\dfrac{4}{25}+\dfrac{21}{25}=1$ $\dfrac{25}{25}=1$ And we see the equation is satisfied!

Is the point $\left(\dfrac{\sqrt{3}}{4},\dfrac{\sqrt{13}}{4}\right)$ on the unit circle?
Substituting the point into the equation gives: $\left(\dfrac{\sqrt{3}}{4}\right)^2+\left(\dfrac{\sqrt{13}}{4}\right)^2=1$ $\dfrac{3}{16}+\dfrac{13}{16}=1$ $\dfrac{16}{16}=1$ And we see the equation is satisfied, thus, the point $\left(\dfrac{\sqrt{3}}{4},\dfrac{\sqrt{13}}{4}\right)$ is indeed on the unit circle!

The point $\left(\dfrac{\sqrt{5}}{6},y\right)$ is on the unit circle in Quadrant IV. Find its $y-$ coordinate.
Because this point is also on the unit circle, it must solve the unit circle equation. Substituting the $x-$ coordinate into the equation gives: $\left(\dfrac{\sqrt{5}}{6}\right)^2+y^2=1$ And this equation can be solved for $y$ : $\dfrac{5}{36}+y^2=1$ subtracting $\dfrac{5}{36}$ from each side gives: $y^2=1-\dfrac{5}{36}$ $y^2=\dfrac{31}{36}$ And taking the square root of each side to eliminate the square on the $y$ : $y=\sqrt{\dfrac{31}{36}}$ And simplifying: $y=\dfrac{\sqrt{31}}{6}$

The point $\left(\dfrac{\sqrt{5}}{6},y\right)$ is on the unit circle in Quadrant IV. Find its $y-$ coordinate.

Because this point is also on the unit circle, it must solve the unit circle equation. Substituting the $x-$ coordinate into the equation gives:
$\left(\dfrac{\sqrt{5}}{6}\right)^2+y^2=1$
And this equation can be solved for $y$ :
$\dfrac{5}{36}+y^2=1$
subtracting $\dfrac{5}{36}$ from each side gives:
$y^2=1-\dfrac{5}{36}$
$y^2=\dfrac{31}{36}$
And taking the square root of each side to eliminate the square on the $y$ :
$y=\sqrt{\dfrac{31}{36}}$
And simplifying:
$y=\dfrac{\sqrt{31}}{6}$

Terminal Points

Start at the point $(1,0)$ on a unit circle moving counterclockwise a distance of $t>0$ units. The coordinates of this point is called the terminal point $P(x,y)$ of $t$ .

Rendered by QuickLaTeX.com

If the movement is made in a clockwise direction, then $t$ is taken to be negative $(t<0)$ :

Rendered by QuickLaTeX.com

We know that the circumference of a circle is $2\pi$ , which means that if we start at $(1,0)$ and move clockwise a distance of $2\pi$ , we will have circumnavigated the circle, and land at the same spot. Thus, the terminal point of $t=2\pi$ is $(1,0)$ .

Rendered by QuickLaTeX.com

$t=\pi$

Since the circumference of the circle is $2\pi$ , the distance around half of the circle is $\pi$ . This means that the terminal point of $t=\pi$ is the point $(-1,0)$ :

Rendered by QuickLaTeX.com

We can also easily see the terminal point of $t=\dfrac{\pi}{2}$ :

Rendered by QuickLaTeX.com

Because the terminal point occurs at the top of the circle (on the $y-$ axis), the terminal point of $t=\dfrac{\pi}{2}$ is $(0,1)$

There are some other values of $t$ whose terminal points are convenient to know. Consider $t=\dfrac{\pi}{4}$ . This is one-eighth of the way around the circle, or one-half of the way to the terminal point of $t=\dfrac{\pi}{2}$ along the circle. It can be seen that the terminal point should have equal $x$ and $y$ coordinates:

Rendered by QuickLaTeX.com

We can solve for the point $(x,y)$ where $y=x$ that is on the unit circle:

$x^2+y^2=1$
replace $y$ with $x$ (since $y=x$ :

$x^2+x^2=1$

$2x^2=1$
And solve for $x$ :

$x^2=\dfrac{1}{2}$

$x=\sqrt{\dfrac{1}{2}}$
And rationalizing the denominator:

$x=\dfrac{\sqrt{2}}{2}$
So the terminal point of $t=\dfrac{\pi}{4}$ is $\left(\dfrac{\sqrt{2}}{2},\dfrac{\sqrt{2}}{2}\right)$

Another terminal point that is convenient to know is the terminal point of $t=\dfrac{\pi}{6}$ .

Rendered by QuickLaTeX.com

Because $\dfrac{\pi}{6}$ is one-twelfth of a circle, this length doubled is one-sixth of a circle, and a hexagon can be inscribed inside the circle with a point located at the terminal point of $t=\dfrac{\pi}{6}$ :

Rendered by QuickLaTeX.com

The sides of the hexagon are equal, so we will examine two sides, the right-most side and the upper right side. We will call the distance of each of these lengths $d$ , and note that the point on the bottom has the same $x-$ coordinate as the terminal point and the opposite $y-$ coordinate. These points are marked below:

Rendered by QuickLaTeX.com

Each of these distances is equal, and using the distance formula: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ , we can find the distance between the terminal point $(x,y)$ and the point $(0,1)$ :

$d=\sqrt{(x-0)^2+(y-1)^2}$

$d=\sqrt{x^2+(y-1)^2}$
We can also use the distance formula to find the distance between the terminal point $(x,y)$ and the point $(x,-y)$ :

$d=\sqrt{(x-x)^2+(y-(-y))^2}$

$d=\sqrt{0^2+(2y)^2}$

$d=2y$
Because the two distances are equal, they create the equation

$2y=\sqrt{x^2+(y-1)^2}$
Because our points are all on the unit circle, we know that they satisfy the equation $x^2+y^2=1$ . That is, we can say that $x^2=1-y^2$ . We can then substitute this into the distance equality:

$2y=\sqrt{{\color{red}x^2}+(y-1)^2}$
becomes

$2y=\sqrt{{\color{red}1-y^2}+(y-1)^2$
expanding the squared binomial:

$2y=\sqrt{1-y^2+y^2-2y+1}$
and simplify:

$2y=\sqrt{2-2y}$
Square both sides, and solving:

$4y^2=2-2y$

$4y^2+2y-2=0$

$y=\dfrac{-2\pm\sqrt{2^2-4(4)(-2)}}{2(4)}$

$y=\dfrac{-2\pm\sqrt{36}}{8}$

$y=\dfrac{-2\pm6}{8}$
Thus, $y=-1$ or $y=\dfrac{1}{2}$ . We can eliminate the negative solution, as the terminal point is in quadrant I. Thus, the $y-$ coordinate of the terminal point is $\dfrac{1}{2}$ .

We can solve for the $x-$ coordinate by substituting the found $y-$ value into the unit circle equation:

$x^2+\left(\dfrac{1}{2}\right)^2=1$
And solving for $x$ :

$x^2+\dfrac{1}{4}=1$

$x^2=1-\dfrac{1}{4}$

$x^2=\dfrac{3}{4}$

$x=\sqrt{\dfrac{3}{4}}$

$x=\dfrac{\sqrt{3}}{2}$
Thus, the terminal point of $t=\dfrac{\pi}{6}$ is $\left(\dfrac{\sqrt{3}}{2},\dfrac{1}{2}\right)$ .

A similar strategy can be used to find that the terminal point of $t=\dfrac{\pi}{3}$ is $\left(\dfrac{1}{2},\dfrac{\sqrt{3}}{2}\right)$ . We can graphically show that we have found the following terminal points in quadrant I:

Rendered by QuickLaTeX.com

Reference numbers

It can be shown that the terminal point of $t=\dfrac{2\pi}{3}$ has the same $y-$ coordinate. We will not derive this, but it can be seen graphically:

Rendered by QuickLaTeX.com

To satisfy both the unit circle equation and the fact that it is in Quadrant II, the terminal point of $t=\dfrac{2\pi}{3}$ must have the opposite $x-$ coordinate. Similar arguments can be made for the relationship between the terminal points of $t=\dfrac{3\pi}{4}$ and $t=\dfrac{\pi}{4}$ and also between $t=\dfrac{5\pi}{6}$ and $t=\dfrac{\pi}{6}$ . Thus, we have the following terminal points in the upper-half plane:

Rendered by QuickLaTeX.com

And of course, a similar argument can give us similar points on the lower-half of the plane:

Rendered by QuickLaTeX.com

From the above, we can see that the only difference between the terminal point of $t=\dfrac{\pi}{3}$ and $t=\dfrac{4\pi}{3}$ is that the $x-$ coordinate is negative for $t=\dfrac{4\pi}{3}$ . In fact, the only difference between the terminal point of $t=\dfrac{\pi}{3}$ , $t=\dfrac{2\pi}{3}$ , $t=\dfrac{4\pi}{3}$ , and $t=\dfrac{5\pi}{3}$ is the signs. That is, we can state the terminal points of the other three $t$ values simply by referring to the values of $t=\dfrac{\pi}{3}$ . Thus, $t=\dfrac{\pi}{3}$ is the reference number $\overline{t}$ for the values $t=\dfrac{2\pi}{3}, \dfrac{4\pi}{3},$ and $\dfrac{5\pi}{3}$ . A reference number can be easily found by measuring the shortest distance from the terminal point, along the unit circle, to the positive $x-$ axis.

Reference Number
For any real number $t$ the reference number $\overline{t}$ of $t$ is the shortest distance from the terminal point of $t$ along the unit circle to the $x-$ axis.

We can find the reference number for $t=\dfrac{5\pi}{6}$ by measuring the distance along the circle between the terminal point of $t=\dfrac{5\pi}{6}$ to the $x-$ axis. Alternatively, we can see that the reference number is the difference between $t=\pi$ and $t=\dfrac{5\pi}{6}$ :

Rendered by QuickLaTeX.com

In fact, the reference number for a number $t<2\pi$ in Quadrant II can be found by $\overline{t}=\pi-t$ .

We can also see the reference number for $t=\dfrac{5\pi}{4}$ :

Rendered by QuickLaTeX.com

To find the reference number for $t<2\pi$ in Quadrant III, we have $\overline{t}=t-\pi$ .

Likewise, we can see that the reference number for $t<2\pi$ in Quadrant IV is $t=2\pi-t$ :

Rendered by QuickLaTeX.com

To find the reference number for $2\pi<t<4\pi$ first subtract off $2\pi$ . Then find the reference angle as above. If the number $t$ is greater than $4\pi$ then subtract $2\pi$ as many times as is necessary to obtain a number between 0 and $2\pi$ , and then proceed as above:

Rendered by QuickLaTeX.com

Find the reference number for $t=\dfrac{27\pi}{4}$
Because $\dfrac{27\pi}{4}$ is greater than $2\pi$ we first will want to reduce it by $2\pi$ : $\dfrac{27\pi}{4}-2\pi$ $\dfrac{27\pi}{4}-\dfrac{8\pi}{4}$ $\dfrac{19\pi}{4}$ But $\dfrac{19\pi}{4}$ is still greater than $2\pi$ , so we reduce it again: $\dfrac{19\pi}{4}-2\pi$ $\dfrac{19\pi}{4}-\dfrac{8\pi}{4}$ $\dfrac{11\pi}{4}$ And yet again: $\dfrac{11\pi}{4}-\dfrac{8\pi}{4}$ $\dfrac{3\pi}{4}$ $t=\dfrac{3\pi}{4}$ is in the second quadrant, so the reference number is $\pi-\dfrac{3\pi}{4}$ $\dfrac{4\pi}{4}-\dfrac{3\pi}{4}$ $=\dfrac{\pi}{4}$

Find the reference number for $t=\dfrac{27\pi}{4}$

Because $\dfrac{27\pi}{4}$ is greater than $2\pi$ we first will want to reduce it by $2\pi$ :
$\dfrac{27\pi}{4}-2\pi$
$\dfrac{27\pi}{4}-\dfrac{8\pi}{4}$
$\dfrac{19\pi}{4}$
But $\dfrac{19\pi}{4}$ is still greater than $2\pi$ , so we reduce it again:
$\dfrac{19\pi}{4}-2\pi$
$\dfrac{19\pi}{4}-\dfrac{8\pi}{4}$
$\dfrac{11\pi}{4}$
And yet again:
$\dfrac{11\pi}{4}-\dfrac{8\pi}{4}$
$\dfrac{3\pi}{4}$ $t=\dfrac{3\pi}{4}$ is in the second quadrant, so the reference number is

$\pi-\dfrac{3\pi}{4}$

$\dfrac{4\pi}{4}-\dfrac{3\pi}{4}$

$=\dfrac{\pi}{4}$

Find the reference number for $t=\dfrac{18\pi}{5}$ .
Again, our $t$ is greater than $2\pi$ , so we reduce by $2\pi$ : $\dfrac{18\pi}{5}-2\pi$ $\dfrac{18\pi}{5}-\dfrac{10\pi}{5}$ $\dfrac{8\pi}{5}$ Because $\dfrac{5\pi}{5}<\dfrac{8\pi}{5}<\dfrac{10\pi}{5}$ , then we know $\dfrac{8\pi}{5}$ lies on the circle in quadrant 3. Thus, the reference number is: $\dfrac{8\pi}{5}-\pi$ $\dfrac{8\pi}{5}-\dfrac{5\pi}{5}$ $=\dfrac{3\pi}{5}$

Find the terminal point for $t=\dfrac{19\pi}{3}$
We must first find the reference number for $t=\dfrac{19\pi}{3}$ . Start by subtracting $2\pi$ $\dfrac{19\pi}{3}-2\pi$ $\dfrac{19\pi}{3}-\dfrac{6\pi}{3}$ $\dfrac{13\pi}{3}$ And subtract another $2\pi$ : $\dfrac{13\pi}{3}-\dfrac{6\pi}{3}$ $\dfrac{7\pi}{3}$ And another: $\dfrac{7\pi}{3}-\dfrac{6\pi}{3}$ $\dfrac{\pi}{3}$ Thus, the terminal point of $t=\dfrac{19\pi}{3}$ is that of $\dfrac{\pi}{3}$ , which is $\left(\dfrac{1}{2},\dfrac{\sqrt{3}}{2}\right)$ .

Find the terminal point for $t=\dfrac{19\pi}{3}$

We must first find the reference number for $t=\dfrac{19\pi}{3}$ . Start by subtracting $2\pi$
$\dfrac{19\pi}{3}-2\pi$
$\dfrac{19\pi}{3}-\dfrac{6\pi}{3}$
$\dfrac{13\pi}{3}$
And subtract another $2\pi$ :
$\dfrac{13\pi}{3}-\dfrac{6\pi}{3}$
$\dfrac{7\pi}{3}$
And another:
$\dfrac{7\pi}{3}-\dfrac{6\pi}{3}$
$\dfrac{\pi}{3}$
Thus, the terminal point of $t=\dfrac{19\pi}{3}$ is that of $\dfrac{\pi}{3}$ , which is $\left(\dfrac{1}{2},\dfrac{\sqrt{3}}{2}\right)$ .

Find the terminal point for $t=\dfrac{23\pi}{4}$
Again, we will first find the reference number for $t=\dfrac{23\pi}{4}$ : $\dfrac{23\pi}{4}-2\pi$ $\dfrac{23\pi}{4}-\dfrac{8\pi}{4}$ $\dfrac{15\pi}{4}$ $\dfrac{15\pi}{4}-\dfrac{8\pi}{4}$ $\dfrac{7\pi}{4}$ $\dfrac{7\pi}{4}$ is found in Quadrant IV, so its reference number is found by: $2\pi-\dfrac{7\pi}{4}$ $\dfrac{8\pi}{4}-\dfrac{7\pi}{4}$ $\dfrac{\pi}{4}$ The terminal point of $t=\dfrac{\pi}{4}$ is $\left(\dfrac{\sqrt{2}}{2},\dfrac{\sqrt{2}}{2}\right)$ . But we see that $t=\dfrac{7\pi}{4}$ is in Quadrant IV, and so $t=\dfrac{23\pi}{4}$ is also in Quadrant IV. We must adjust the terminal point to ensure it is in the correct quadrant. Quadrant IV has both positive $x-$ coordinates, and negative $y-$ coordinates, thus, the terminal point of $t=\dfrac{23\pi}{4}$ is $\left(\dfrac{\sqrt{2}}{2},-\dfrac{\sqrt{2}}{2}\right)$

Find the terminal point for $t=\dfrac{23\pi}{4}$

Again, we will first find the reference number for $t=\dfrac{23\pi}{4}$ :
$\dfrac{23\pi}{4}-2\pi$
$\dfrac{23\pi}{4}-\dfrac{8\pi}{4}$
$\dfrac{15\pi}{4}$

$\dfrac{15\pi}{4}-\dfrac{8\pi}{4}$
$\dfrac{7\pi}{4}$ $\dfrac{7\pi}{4}$ is found in Quadrant IV, so its reference number is found by:

$2\pi-\dfrac{7\pi}{4}$

$\dfrac{8\pi}{4}-\dfrac{7\pi}{4}$

$\dfrac{\pi}{4}$
The terminal point of $t=\dfrac{\pi}{4}$ is $\left(\dfrac{\sqrt{2}}{2},\dfrac{\sqrt{2}}{2}\right)$ . But we see that $t=\dfrac{7\pi}{4}$ is in Quadrant IV, and so $t=\dfrac{23\pi}{4}$ is also in Quadrant IV. We must adjust the terminal point to ensure it is in the correct quadrant. Quadrant IV has both positive $x-$ coordinates, and negative $y-$ coordinates, thus, the terminal point of $t=\dfrac{23\pi}{4}$ is $\left(\dfrac{\sqrt{2}}{2},-\dfrac{\sqrt{2}}{2}\right)$

Find the terminal point of $t=\dfrac{26\pi}{3}$
$\dfrac{26\pi}{3}-\dfrac{24\pi}{3}=\dfrac{2\pi}{3}$ which is in Quadrant II. The reference number for $t=\dfrac{26\pi}{3}$ is $\dfrac{\pi}{3}$ . Thus, changing the $x-$ coordinate of the terminal point of $\dfrac{\pi}{3}$ will give the terminal point of $\dfrac{26\pi}{3}$ . That is: $\left(-\dfrac{1}{2},\dfrac{\sqrt{3}}{2}\right)$ .

PreCalculus I

5.1 – The Unit Circle

5.1 Video 1 of 3: The Unit Circle

5.1 Video 2 of 3: Terminal Points.

5.1 Video 3 of 3: Reference Numbers.

Section Notes

The Unit Circle

Terminal Points

Reference numbers

Modal title