TRIGONOMETRY

During the course of this tutorial, you will, from time to time, hear this sound (sound). This is the signal to advance to the next numbered statement in the book which accompanies this tape. I will, on occasion, ask you to stop the tape to work out a brief problem. On restarting the tape, the correct answer will be provided.

Panel 1 The purpose of this tutorial is to review with you the elementary properties of the trigonometric functions. Facility with this subject is essential to success in all branches of science, and you are strongly urged to review and practice the concepts presented here until they are mastered. Let us consider the right-angle triangle shown in Panel 1. The angle at c is a right angle and the angle at a we will call Ø. The lengths of the sides of the triangle we will denote as p, q and r. Now, from your elementary geometry, you know several things about this triangle. For example, you know the Pythagorian relation, q² = p² + r². That is, the square of the length of the side opposite the right angle, which we call the hypotenuse, is equal to the sum of the squares of the lengths of the other two sides.

We know other things. For example, we know that if the lengths of the three sides of any triangle p, q and r are specified, then the whole triangle is determined, angles included. If you think about this for a moment, you will see it is correct. If I give you three sticks of fixed length and told you to lay them down in a triangle, there's only one triangle which you could make. What we would like to have is a way of relating the angles in the triangle, say Ø, to the lengths of the sides.

Now, it turns out that there's no simple analytic way to do this. Even though the triangle is specified by the lengths of the three sides, I cannot give you a simple formula which will allow you to calculate the angle Ø. We must specify it in some new way. To do this, we define three ratios of the sides of the triangle.

One ratio we call the sine of theta, written sin(Ø), and it is defined as the ratio of the side opposite Ø to the hypotenuse, that is r/q.

The cosine of Ø, written cos(Ø), is the side adjacent to Ø over the hypotenuse, that is, p/q.

This is really enough, but because it simplifies our mathematics later on, we define the tangent of Ø, written tan(Ø), as the ratio of the opposite to the adjacent sides, that is r/p. This is not an independent definition since you can readily see that the tangent of Ø is equal to the sine of Ø divided by the cosine of Ø. Stop the tape to verify for yourself that this is correct.

In order to make these functions useful in calculations, we need numerical values of them for the different values of Ø . Such values are given in trigonometric tables such as those you see on the left-hand page. The tables shown here are for the sine function. Let's for a moment look at these in order to see how to use them. Down the left-hand column of each of the two tables is the value of Ø, the angle in degrees, and across the top is the fraction of the angle in minutes, from 0 minutes to 54 minutes, in steps of 6 minutes. You will remember that there are 60 minutes in a degree, so the next 6 minutes step takes you to the next degree. The angles 0º to 44º are in the first table and 45º to 90º in the second.

Let's suppose I want the sine of 24º. On the first page, I find in the column 24º 0 minutes the value .4067. Therefore, the sine of 24º is .4067. That is, in a triangle like panel 1 where Ø = 24º, the ratio of the sides r to q is .4067. If you look again in the table, you will see that the sine of 24º 36 minutes is .4163. Stop the tape and check this.

Obviously, looking up a value in the tables is very simple. Look up the sine of 42º 24 minutes. The sine of 42º 24 minutes is .6743. Did you get this result? If not, stop the tape and check it again.

Panel 8 The use of cosine and tangent tables is in exactly the same way. It is now possible for us to solve the simple problem concerning triangles. For example, in Panel 8, the length of the hypotenuse is 3 cm and the angle Ø is 24º. What is the length of the opposite side r? The sine of 24º as we saw is .4067 and it is also by definition r/3. So, sine of 24º = .4067 = r/3, therefore, r = 3 x .4067 = 1.22 cm.

Panel 10 Conversely, suppose you knew that the opposite side was 2 cm long and the hypotenuse was 3 cm long, as in panel 10, what is the angle Ø? Stop the tape while you determine the sine of Ø in this case.
The sine of Ø is 2/3, which equals .6667. Now stop the tape and determine what angle has .6667 as its sine.
You will have found that an angle Ø = 41º48' is very close to the angle you want.

Panel 12 One use of these functions which is very important is in the calculation of components. In panel 12 is shown a line oa in an xy reference frame. We would like to find the y component of this line. That is, the projection ob of the line on the y axis. Obviously, ob = ca and ca/oa = sine(Ø), so ca = oa sin(Ø). Similarly, the x-component of oa is oc. And oc/oa = cos(Ø) so oc = oa cos(Ø).

Panel 15 There are many relations among the trigonometric functions which are important, but one in particular you will find used quite often. Panel 1 has been repeated as Panel 15 for you. And let us look at the sum cos² + sine². From the figure, this is (p/q)² + (r/q)², which is [(p² + r²) / (q²)]. The Pythagorean theorem tells us that p² + r² = q² so we have [(p² + r²) / q²] = (q²/q²) = 1. Therefore, we have in Expression 18
cos² + sin² = 1.

Our discussion so far has been limited to angles between 0 and 90º. Indeed, that is all the tables cover. How could we deal with, say, the sine of 140º? Panel 19 will help us here.

Panel 19 In this xy reference frame, the angle Ø is clearly between 90º and 180 º, and clearly, the angle , which is 180 - Ø (I've marked with a double arc) can be dealt with. In this case, we say that the magnitude of sine, cosine, and tangent of Ø are those of the supplement and we only have to examine whether or not they are positive or negative.
We always assume that the hypotenuse q is positive, but r and p have the sign appropriate to their direction with respect to the origin. Clearly, in Panel 19, r is positive and p is negative. Therefore, we can write

Notice that only the sin is positive.

For example, what is the sine, cosine and tangent of 140º? The supplement is 180º - 140º = 40º. Stop the tape and look up the sine of 40º.

If you have your own trig tables with you, look up also the cosine and the tangent of 40º.

Now we know that sin(140º) = 0.6428, cos(140º) = -0.7660, and tan(140º) = -0.8391.

Panel 8	The use of cosine and tangent tables is in exactly the same way. It is now possible for us to solve the simple problem concerning triangles. For example, in Panel 8, the length of the hypotenuse is 3 cm and the angle Ø is 24º. What is the length of the opposite side r? The sine of 24º as we saw is .4067 and it is also by definition r/3. So, sine of 24º = .4067 = r/3, therefore, r = 3 x .4067 = 1.22 cm.
Panel 10	Conversely, suppose you knew that the opposite side was 2 cm long and the hypotenuse was 3 cm long, as in panel 10, what is the angle Ø? Stop the tape while you determine the sine of Ø in this case. The sine of Ø is 2/3, which equals .6667. Now stop the tape and determine what angle has .6667 as its sine. You will have found that an angle Ø = 41º48' is very close to the angle you want.
Panel 12	One use of these functions which is very important is in the calculation of components. In panel 12 is shown a line oa in an xy reference frame. We would like to find the y component of this line. That is, the projection ob of the line on the y axis. Obviously, ob = ca and ca/oa = sine(Ø), so ca = oa sin(Ø). Similarly, the x-component of oa is oc. And oc/oa = cos(Ø) so oc = oa cos(Ø).
Panel 15	There are many relations among the trigonometric functions which are important, but one in particular you will find used quite often. Panel 1 has been repeated as Panel 15 for you. And let us look at the sum cos² + sine². From the figure, this is (p/q)² + (r/q)², which is [(p² + r²) / (q²)]. The Pythagorean theorem tells us that p² + r² = q² so we have [(p² + r²) / q²] = (q²/q²) = 1. Therefore, we have in Expression 18 cos² + sin² = 1.

TRIGONOMETRY

Continue to: Trigonometry, Part 2