# 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 \(x^2 + y^2 = r^2\) where *r* is the radius of the circle. For a unit circle, \(x^2 + y^2 = 1\). Therefore, we have the most famous identity of trigonometry: \( \sin^{2}\theta + \cos^{2}\theta = 1 \).

For our example, \(\sin\theta = 0.8 = \frac{4}{5}\) and \(\cos\theta = 0.6 = \frac{3}{5}\) for *θ* = 53.13°. (This is a 3-4-5 right triangle.) And our trigonometric identity holds up: \( \left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2 = \frac{9+16}{25} = 1 \).

### 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 \(\tan\theta = \frac{\sin\theta}{\cos\theta}\). The segment EP represents the tangent value of *θ*. There are many similar right triangles in Figure 1. I have shown some of the equal angles that can be proved by noting that the angles are complementary angles of a right triangle.

To prove EP equals tangent of *θ*, consider similar triangles PXE and PYO. Therefore, we have the following relationship: \(\frac{\text{PY}}{\text{PO}} = \frac{\text{PX}}{\text{PE}}\).

The segment PO = 1 because it is the radius of the unit circle. Therefore, \(\frac{\text{PY}}{1} = \frac{\text{PX}}{\text{PE}}\). Solving for PE, we have the following relationship: \(\text{PE} = \frac{\text{PX}}{\text{PY}}\). However, we already know that PX is the sine value and PY is the cosine value. Therefore, \(\text{PE} = \frac{\sin\theta}{\cos\theta}\).

For our example, \(\tan\theta = \frac{0.8}{0.6} = \frac{4}{3} \approx 1.333\).

The cotangent value is equal to \(\frac{\cos\theta}{\sin\theta} \). Using similar triangles PXO and PYF, we have \(\frac{\text{PY}}{\text{PF}} = \frac{\text{PX}}{\text{PO}}\). Solving for PF, we get \(\text{PF} = \frac{\text{PY}}{\text{PX}} = \frac{\cos\theta}{\sin\theta} = \cot\theta\).

For our example, \(\cot\theta = \frac{0.6}{0.8} = \frac{3}{4} = 0.75\).

As you may have guessed, tangent and cotangent values are inverses.

### The Secant Value

The secant is defined as the inverse of cosine or \( \frac{1}{\cos\theta} \). The segment OE is the secant value of angle *θ*.

Triangles PXO and EOF are similar right triangles. This gives us the relationship:

(i) \(\frac{\text{PO}}{\text{PX}} = \frac{\text{EF}}{\text{EO}}\)

(ii) \(\frac{1}{\sin\theta} = \frac{\tan\theta + \cot\theta}{\text{EO}}\)

(iii) \( \text{EO} = \sin\theta\left(\frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta}\right) \)

(iv) \( \text{EO} = \frac{\sin^2\theta}{\cos\theta} + \cos\theta \)

(v) \( \text{EO} = \frac{\sin^2\theta + \cos^2\theta}{\cos\theta} = \sec\theta \)

For the example above, \(\sec\theta = \frac{1}{0.6} = \frac{5}{3} \approx 1.667\).

### The Cosecant Value

The cosecant is defined as the inverse of sine or \( \frac{1}{\sin\theta} \). The segment OF is the cosecant value of angle *θ*.

Triangles PYO and FOE are similar right triangles. This gives us the relationship:

(i) \(\frac{\text{PO}}{\text{PY}} = \frac{\text{EF}}{\text{OF}}\)

(ii) \(\frac{1}{\cos\theta} = \frac{\tan\theta + \cot\theta}{\text{OF}}\)

(iii) \( \text{FO} = \cos\theta\left(\frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta}\right) \)

(iv) \( \text{FO} = \sin\theta + \frac{\cos^2\theta}{\sin\theta} \)

(v) \( \text{FO} = \frac{\sin^2\theta + \cos^2\theta}{\sin\theta} = \csc\theta \)

For the example above, \(\csc\theta = \frac{1}{0.8} = \frac{5}{4} = 1.25\).

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 *x*² 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 \(\sqrt{xy}\). If *x* = 1, then the altitude equals \(\sqrt{y}\) as shown in Figure 3. That means the segment *y* is the square of the altitude.

### Can We Really?

It may seem from the above construction that we can construct the \(x^2\) value or the \(\sqrt{x}\) value of any given segment of length *x*. Actually, it is impossible!

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 \(\cos\theta\). For the example, we have been using *θ* ≈ 53.13° but *θ* can be any angle. Because the circle has a radius of 1, OB = 1. Therefore, we will make OX the altitude and OB will be the segment of the hypotenuse that is equal to 1. We just need to draw a perpendicular to BX, making X the vertex of the right triangle. The perpendicular at X will cross the *y*-axis at point A, making AB the hypotenuse.

Therefore, in Figure 4, OA is the value \(\cos^{2}\theta\).

For the example we have been using, \(\cos^{2}(53.13^{\circ}) = 0.36\), which is indeed \(0.6^2\).

### 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 \(\sin\theta\). For our example, \(\sin(53.13^{\circ}) = 0.8\). We will translate this length onto the *y*-axis first. Therefore, PX = AO = sin*θ*.

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, \(0.8^2 = 0.64\), which is confirmed by the Geogebra calculation.

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 \(\sin^{2}x + \cos^{2}x = 1\), then once you find one, you automatically find the other. For example, in Figure 5, OB equals sine squared. That means the distance from B to the intersection of the circle with the *x*-axis is cosine squared because of the squares identity and the radius of the circle is 1: \(\text{BC}^2 = 1 - \text{OB}^2\) or \(\sin^{2}x = 1 - \cos^{2}x\).

### 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, \(\tan(53.13^{\circ}) = \frac{4}{3} \approx 1.333\). The tangent squared value is \(\frac{16}{9} \approx 1.778\). This confirms with the Geogebra calculation.

Curiously, the segment AB is equal to \(\tan\theta\cdot\sec\theta\). For our example this is \(\frac{4}{3}\cdot\frac{5}{3} = \frac{20}{9} \approx 2.222\).

The proof is a simple calculation. Because APB is a right triangle, we have the relationship: \(\text{AB}^2 = \text{AP}^2 + \text{PB}^2\) or \(\text{AB}^2 = \tan^{2}\theta + \tan^{4}\theta = \) \( \tan^{2}\theta(1+\tan^{2}\theta) = \) \( \tan^{2}\theta\cdot\sec^{2}\theta \). Thus, \(\text{AB} = \tan\theta\cdot\sec\theta\).

### 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, \(\sec(53.13^{\circ}) = \frac{5}{3}\). Therefore, \(\sec^{2}(53.13^{\circ}) = \frac{25}{9} \approx 2.778\). This checks with the Geogebra calculation.

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 \(\sec^{2}\theta = 1 + \tan^{2}\theta\). The number “1” that shows up in the identities is clearly the radius of the circle. Figure 8 below shows the other secant squared segment and the tangent squared segment. Secant squared is simply the segment OD because PD is the tangent squared segment.

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 \(\tan^{4}(49.939^{\circ}) = 2\). Curiously, the angle at which the tangent to the fourth power equals 2 is approximately 49.939° (approximately 0.87161 radians - that’ about \(\frac{\pi}{3.604}\)) and the segment PB is the diameter of the circle.

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 \(\text{PB} = \tan^{4}\theta\).

Remember, point B will not be on the circle if the angle is not approximately 49.939° (or to be precise \(\arctan(\sqrt[4]{2})\). That was a special case. Otherwise we will have a scenario like below where point B is inside the circle for 0 < *θ* < 49.939°:

If \(\frac{\pi}{2}\) > *θ* > 49.939°, point B will lie outside the circle.

## Conclusion

These activities were a great way to visulize the basic trigonometric values and to extend them with the squares of the values.