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Geometry: How Does This Relate to the Unit Circle?

How Does This Relate to the Unit Circle?

Our trigonometric ratios were defined within the confines of a right triangle. Because the measures of the interior angles of a right triangle add up to 180º, and one of these angles has a measure equal to 90º, the other two angles of a right triangle must be acute angles. So you can only find the trigonometric ratios of acute angles. That's too limiting.

The sine, cosine, and tangent ratios can be defined for any angle, not just acute angles. But in order to do this, you must embed a right triangle into a circle. Although you could use any circle, things work out nicely if you use the “unit circle.” You might be wondering what the unit circle is and why you should use that particular circle. Well, the unit circle is a circle with radius equal to 1. Let's embed a right triangle into a unit circle and see what happens. Use Figure 20.9 as a guide.

Figure 20.9A right triangle embedded in the unit circle.

Each one of the vertices of the triangle will have some special characteristic, but the coveted properties are spread around so that one vertex isn't seen as being “better” than any other. Keep vertex A at the center of the circle. It is not allowed to move from that point. Think of vertex A as being in a trigonometric “time out.” Vertex B is constrained to lie on the circle. It can move around the circle and occupy whatever place it wants, but it must stay on the circle. The only nice property left is the triangle's right angle, so by process of elimination ∠C is the right angle. There is one last restriction. Keep the diameter that C lies on fixed (in this case, ¯XY), and let B move around the circle.

The hypotenuse of your triangle has one end point located at the center of the circle and the other end point on the circle. Because you are working with the unit circle, your hypotenuse has length 1. Recall that the sine and cosine ratios both involve the length of the hypotenuse in the denominator. You can't have a nicer denominator of a ratio than 1. That's the advantage of working on the unit circle.

Now, let B move around the circle and always embed a right triangle by dropping a perpendicular line segment from B to ¯XY. In Figure 20.10, I've shown four different locations for B and the corresponding embedded right triangles that result. Notice that the orientation of the right triangle changes, but the hypotenuse is always the line segment ¯AB.

Figure 20.10Four right triangles embedded in the unit circle, based on the location of the vertex B.

Whenever you talk about ∠A, you should always refer to ∠BAC, an interior angle of your triangle. The trigonometric ratios will always be

  • tan ∠A = a/b, sin ∠A = a/c, and cos ∠A = b/c.

These will always be positive numbers, because the length of any side is positive (by the Ruler Postulate). These ratios can be simplified a bit because you are constrained to working on the unit circle:

  • tan ∠A = a/b, sin ∠A = a, and cos ∠A = b

Now that the stage is set, I am ready to talk about the trigonometric ratios of any angle (not just acute angles)! I'll divide the circle into quarters and discuss angles that fall into each quarter one at a time. The angles that fall in the first quarter are acute angles, and I have already discussed the trigonometric ratios of acute angles.

For the second quarter, suppose you have an obtuse angle θ, as shown in Figure 20.11. Then the supplement of this obtuse angle is an acute angle. Embed a triangle like that in Figure 20.10(b) and define the trigonometric ratios of θ based on the trigonometric ratios of ∠A : sin θ = sin ∠A, cos θ = - cos A, and tan θ = - tan ∠A. Notice that the sine ratio of an obtuse angle has the same numerical value as the sine ratio of its acute supplement, whereas the cosine and tangent ratios of an obtuse angle have the same absolute value as the cosine and tangent ratios of its supplement, but they have the opposite sign.

Figure 20.11A circle with an obtuse central angle, and the corresponding embedded triangle.

Let's keep going around the circle. The third quarter involves angles whose measure is between 180º and 270º, as shown in Figure 20.12. Embed a triangle in this circle similar to the one shown in Figure 20.10(c), and again define the trigonometric ratios of these type of angles based on the trigonometric ratios of acute angles. All you need to do is change a few signs (two signs, actually): sin θ = - sin ∠A, cos θ = - cos A, and tan θ = tan ∠A. Notice that the signs of the sine and cosine ratio are negative, and the tangent ratio is the positive one.

Figure 20.12A circle with central angle between 180º and 270º, and the corresponding embedded triangle.

One more quarter and you graduate! This last quarter involves angles whose measure is between 270º and 360º, as shown in Figure 20.13. In this case, embed a right triangle like you did in Figure 20.10(d), and define the trigonometric ratios of the angle based on the trigonometric ratio of its corresponding acute interior angle. Once again, two signs will be changed: sin θ = - sin ∠A, cos θ = cos A, and tan θ = - tan ∠A. This time the cosine ratio gets to stay positive, and the sine and tangent ratios are negative.

Figure 20.13A circle with central angle between 270º and 360º, and the corresponding embedded triangle.

Eureka!

The trigonometric ratios of nonacute angles have the same magnitude as the trigonometric ratios of their corresponding acute angles. Two of the three ratios are negative, and one stays positive. As you go around the circle counter-clockwise, start with ALL ratios being positive, then the SINE ratio stays positive, next the TANGENT ratio is positive, and finally the COSINE ratio gets its turn. If you have trouble remembering which ratio is positive where, there is a famous mnemonic available to help: All Students Take Calculus—All Sine Tangent Cosine.

This is just the tip of the trigonometric iceberg. I could fill an entire book with just the material discussed in a trigonometry class! I'll get started on that book right after I finish this!

Here's your chance to shine. Remember that I am with you in spirit and have provided the answers to these questions in Answer Key.

  1. If a right triangle has an angle with tangent ratio 3/5, find the sine ratio of that angle.
  2. If a right triangle has an angle with a sine ratio of 1/2, find the tangent ratio of that angle.
  3. If a right triangle has an angle with cosine ratio 3/7, find the sine and tangent ratios of the angle.
  4. If a right triangle has an angle with sine ratio 5/9, find the tangent and cosine ratios of the angle.
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Excerpted from The Complete Idiot's Guide to Geometry © 2004 by Denise Szecsei, Ph.D.. All rights reserved including the right of reproduction in whole or in part in any form. Used by arrangement with Alpha Books, a member of Penguin Group (USA) Inc.

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