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.
The Unit Circle
The unit circle is the set of points that have a distance of 1 from the origin:
Points on the unit circle satisfy the equation
|The Unit Circle|
|The unit circle is the circle of radius 1 centered at the origin. It’s equation is:
|Show that the point is on the unit circle|
|The point can be shown to be on the unit circle by satisfying the unit circle equation.
Substituting the point into the equation gives:
And we see the equation is satisfied!
|Is the point on the unit circle?|
|Substituting the point into the equation gives:
And we see the equation is satisfied, thus, the point is indeed on the unit circle!
|The point is on the unit circle in Quadrant IV. Find its coordinate.|
|Because this point is also on the unit circle, it must solve the unit circle equation. Substituting the coordinate into the equation gives:
And this equation can be solved for :
subtracting from each side gives:
And taking the square root of each side to eliminate the square on the :
Start at the point on a unit circle moving counterclockwise a distance of units. The coordinates of this point is called the terminal point of .
If the movement is made in a clockwise direction, then is taken to be negative :
We know that the circumference of a circle is , which means that if we start at and move clockwise a distance of , we will have circumnavigated the circle, and land at the same spot. Thus, the terminal point of is .
Since the circumference of the circle is , the distance around half of the circle is . This means that the terminal point of is the point :
We can also easily see the terminal point of :
Because the terminal point occurs at the top of the circle (on the axis), the terminal point of is
There are some other values of whose terminal points are convenient to know. Consider . This is one-eighth of the way around the circle, or one-half of the way to the terminal point of along the circle. It can be seen that the terminal point should have equal and coordinates:
We can solve for the point where that is on the unit circle:
replace with (since :
And solve for :
And rationalizing the denominator:
So the terminal point of is
Another terminal point that is convenient to know is the terminal point of .
Because 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 :
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 , and note that the point on the bottom has the same coordinate as the terminal point and the opposite coordinate. These points are marked below:
Each of these distances is equal, and using the distance formula: , we can find the distance between the terminal point and the point :
We can also use the distance formula to find the distance between the terminal point and the point :
Because the two distances are equal, they create the equation
Because our points are all on the unit circle, we know that they satisfy the equation . That is, we can say that . We can then substitute this into the distance equality:
expanding the squared binomial:
Square both sides, and solving:
Thus, or . We can eliminate the negative solution, as the terminal point is in quadrant I. Thus, the coordinate of the terminal point is .
We can solve for the coordinate by substituting the found value into the unit circle equation:
And solving for :
Thus, the terminal point of is .
A similar strategy can be used to find that the terminal point of is . We can graphically show that we have found the following terminal points in quadrant I:
It can be shown that the terminal point of has the same coordinate. We will not derive this, but it can be seen graphically:
To satisfy both the unit circle equation and the fact that it is in Quadrant II, the terminal point of must have the opposite coordinate. Similar arguments can be made for the relationship between the terminal points of and and also between and . Thus, we have the following terminal points in the upper-half plane:
And of course, a similar argument can give us similar points on the lower-half of the plane:
From the above, we can see that the only difference between the terminal point of and is that the coordinate is negative for . In fact, the only difference between the terminal point of , ,, and is the signs. That is, we can state the terminal points of the other three values simply by referring to the values of . Thus, is the reference number for the values and . A reference number can be easily found by measuring the shortest distance from the terminal point, along the unit circle, to the positive axis.
|For any real number the reference number of is the shortest distance from the terminal point of along the unit circle to the axis.|
We can find the reference number for by measuring the distance along the circle between the terminal point of to the axis. Alternatively, we can see that the reference number is the difference between and :
In fact, the reference number for a number in Quadrant II can be found by .
We can also see the reference number for :
To find the reference number for in Quadrant III, we have .
Likewise, we can see that the reference number for in Quadrant IV is :
To find the reference number for first subtract off . Then find the reference angle as above. If the number is greater than then subtract as many times as is necessary to obtain a number between 0 and , and then proceed as above:
|Find the reference number for|
|Because is greater than we first will want to reduce it by :
But is still greater than , so we reduce it again:
And yet again:
|Find the reference number for .|
|Again, our is greater than , so we reduce by :
Because , then we know lies on the circle in quadrant 3. Thus, the reference number is:
|Find the terminal point for|
|We must first find the reference number for . Start by subtracting
And subtract another :
Thus, the terminal point of is that of , which is .
|Find the terminal point for|
|Again, we will first find the reference number for :
The terminal point of is . But we see that is in Quadrant IV, and so is also in Quadrant IV. We must adjust the terminal point to ensure it is in the correct quadrant. Quadrant IV has both positive coordinates, and negative coordinates, thus, the terminal point of is
|Find the terminal point of|
|which is in Quadrant II. The reference number for is . Thus, changing the coordinate of the terminal point of will give the terminal point of . That is: .|