Visualizing the Trigonometric Measurements with a Circle and a Tangent
All of the trigonometric measurements can be visualized with a circle and a tangent line to the circle. We will use the following diagram of a unit circle (i.e. the radius is assumed to be 1) and a tangent at point P to locate all the trigonometric values represented by the lengths of the various segments.
Sine and Cosine Values
The basic trigonometric values are the sine and cosine. For a unit circle, the x coordinate represents the cosine value of the angle θ and the y coordinate represents the sine value of the angle θ. In Figure 1, cosine is equal to the segment OX and sine is equal to OY for the angle created by OP and the x-axis.
The figure shows the angle meausure of approximately 53.13°, which corresponds to the angle created by the point (0.6, 0.8). We will use this coordinate as an example for the rest of the trigonometric values.
For any circle, any point (x, y) on the circle has the relationship
For our example,
Tangent and Cotangent Values
The tangent value is probably the third most well-known trigonometric value. For the tangent value, we need to draw a tangent line at P. That’s probably why it is called the tangent function.
The tangent is defined as
To prove EP equals tangent of θ, consider similar triangles PXE and PYO. Therefore, we have the following relationship:
The segment PO = 1 because it is the radius of the unit circle. Therefore,
For our example,
The cotangent value is equal to
For our example,
As you may have guessed, tangent and cotangent values are inverses.
The Secant Value
The secant is defined as the inverse of cosine or
Triangles PXO and EOF are similar right triangles. This gives us the relationship:
(i)
(ii)
(iii)
(iv)
(v)
For the example above,
The Cosecant Value
The cosecant is defined as the inverse of sine or
Triangles PYO and FOE are similar right triangles. This gives us the relationship:
(i)
(ii)
(iii)
(iv)
(v)
For the example above,
We have covered all 6 basic trigonometric values in one diagram. Next, we will construct the lengths of the squares of the trigonometric values.
The Squares of the Trigonometric Values
Prerequisite
To construct the squares of the trigonometric values, we first need to be able to construct the x² value given the x value. We will need the help of the following right triangle.
Figure 2 shows a right triangle with an altitude drawn to the hypotenuse. The altitude divides the hypotenuse into segments of x and y and creates 3 similar triangles. By similar triangles, we can determine that the altitude is equal to
Can We Really?
It may seem from the above construction that we can construct the
But did we not just do that above? The answer is no. We actually constructed the square root of a product of x and y. And x just happens to equal 1 because we let it equal 1. If we are given a segment of length x, then squaring or square-rooting x is impossible.
Given 2 segments of 2 lengths, then we can find the square root of their product but we cannot square or square-root either of them individually. It should be clear now that we will always need one segment to equal 1 to construct the square of the trigonometric values. With that said, we continue.
To construct the measurements of cosine squared and sine squared, we just need to make the sine and cosine values the altitude of a right triangle and one part of the hypotenuse equal to 1. Try this yourself before continuing.
Cosine Squared Value
For the cosine squared value, we have to identify a length of 1 and an altitude that is the cosine value. This is actually quite easy. Consider Figure 4 below.
In Figure 4, OX is of course
Therefore, in Figure 4, OA is the value
For the example we have been using,
Sine Squared Value
There are two ways (maybe there are more than two ways) to construct the sine squared segment. We will look at one way shown below.
The segment PX is equal to
Next, we draw segment AB which intersects the circle at B so that OB = 1, the radius of the circle. We now have a segment of the hypotenuse equal to 1 and the sine value as the altitude.
Next, draw a perpendicular to AB at point A to complete the right triangle. Segment OB is equal to the sine squared value. For the example we have been using,
The alternate way, if we keep the sine segment in place, is to construct a length of 1 from X toward the negative x-axis. That will keep the sine segment as the altitude in place and the hypotenuse segment equal to 1. The rest is for the reader to figure out.
Wait a Minute!
We found sine squared and cosine squared separately. However, when you realize that
The Tangent Squared
If constructing the values of sine and cosine squared seemed difficult, you may think the construction of tangent squared should be even harder. As it turns out, it is easier than one thinks.
You may be able to figure it out from Figure 6 below yourself.
We have already established that PA is equal to the tangent value. Conveniently, because it is tangent to the circle, a line drawn from P to the center of the circle forms a right angle. Therefore, APO is a right angle.
Therefore, we have AP equal to the tangent value as the altitude and PO equal to 1 as one segment of the hypotenuse. We simply draw a perpendicular to the x-axis at A and the intersection of the perpendicular line and the ray OP is our point of interest.
Therefore, PB is the tangent squared segment. For our example,
Curiously, the segment AB is equal to
The proof is a simple calculation. Because APB is a right triangle, we have the relationship:
The Secant Squared
Lastly, we will construct the secant squared value. The cosecant squared value will be left for the reader.
First, draw a tangent to the circle at point P which crosses the x-axis at point A. The segment OA represents the secant value of θ as was established earlier.
Next, draw segment AC, where C is a point on the circle at (0, –1). We have OA, the altitude of our right triangle, as the secant value and OC = 1.
Lastly, draw a perpendicular to AC at A. This line crosses the y-axis at point B. The segment OB represents the secant squared value we seek.
For our example,
In the Figure 6 caption, I posed a question: Do you see the secant squared segment? Figure 6 shows a really easy way to identify the secant squared segment becaues of the identity
The two segments that represent secant squared are OJ and OD. We can draw a circle through J and D because both points are the radii of the circle with center at O.
The segments EJ and PD are also equal to tangent squared, of course.
Bonus: Tangent to the Fourth Power
With the right triangle and altitude method, we can find some interesting values. One such value is the tangent to the fourth power. Figure 8 below shows segment PB, which is equal to
For angles other than 49.939°, point B will not be on the circle but the segment will cross the center of the circle.
Here are the construction steps for tangent to the fourth power.
First, construction PD which is equal to tangent squared. The steps were shown earlier.
Next, find point E that lies on the tangent at P such that PE = PD by drawing a circle with radius of PD and center at P.
Next, find point F such that PF = 1 by drawing a circle with radius PO and center at P.
Next, connect F and E with a segment.
Lastly, draw a perpendicular to FE at E. The intersection point B is the point we seek.
The triangle FEB is a right triangle with FP = 1 and PE equal to tangenent squared. That gives us
Remember, point B will not be on the circle if the angle is not approximately 49.939° (or to be precise
If
Conclusion
These activities were a great way to visulize the basic trigonometric values and to extend them with the squares of the values.