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The Mathematics That Every Secondary School Math Teacher Needs to Know

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622 Chapter 12 Trigonometry 6 (C) Solve the launch question. 7* Using the damped oscillation function presented earlier after Example 12.17, y = eÀ2t(À5 + 10 sin 2πt), ﬁnd the times that the spring passes its equilibrium point. 8 When a projectile is ﬁred from the ground at an angle of θ degrees with the horizontal, the horizontal distance it travels is given by HðyÞ ¼ v02 sin 2y : 32 (a)* If v0 is the initial velocity and v0 = 1100 ft/s what values of θ to the nearest degree will make H(θ) = 1125 feet? (b) What value of θ will make H(θ) a maximum? 9 When a beam of light is shined into the water at an angle of 20 to the horizontal, it can be shown using Snell’s Law for refraction of light, that the beam bends towards the perpendicular to the water’s surface at an angle of θ degrees where θ is a solution to the equation: sin y ¼ 0:752 487 319 305 7: sin 70 Find θ to the nearest degree. 10* Suppose that a rocket could be launched from the ground with a constant velocity of 400 feet per minute and that an observer is standing 100 feet from the rocket on level ground with his eye on the bottom of the rocket. He keeps his eye on the bottom as the rocket rises straight up. What angle will his eye make with the horizontal after 2 minutes if the man’s eye is 6 feet off the ground? 11* Given a rectangular sheet of paper 6 inches by 9 inches. Suppose that someone grabs one corner of the 6 inch side and carries it to the other side and presses down forming a fold as shown in Figure 12.81. θ L 9 in 6 in Figure 12.81 Find L in terms of θ. 12 Show by using right triangles, that if a is acute, sinÀ1a þ cosÀ1a ¼ p : Is this still true if a is not 2 acute? How do you know?

12.8 Trigonometric Identities 623 12.8 Trigonometric Identities LAUNCH In the following three equations, graph both the right side of the equation and the left side of the equation on the same set of axes. So, for example, if the equation is 1 + tan2θ = sec2θ, then graph the two equations: y = 1 + tan2θ and y = sec2θ. In each of the following cases, determine if the graphs are the same or different. If they are the same, state what that means. (a) sec x þ csc x ¼ sin x þ cos x: tan x þ cot x (b) 1 À cos x ¼ cos x sin x þ cos 1 x (c) cos(3x) = 4 cos3 x À 3 cos x Now that you have graphed the equations in the launch question, you are probably trying to recall what it might mean when both sides of the original equation have the same graph. You might remember other relationships where this has happened, such as: sin2 y þ cos2 y ¼ 1 cosðyÞ ¼ cosðÀyÞ sinðÀyÞ ¼ ÀsinðyÞ: We begin this section by asking you how you would refer to the following relationships. Do you now recall how they are referred to? Are they formulas or equations? Actually, they fall into a category of relationships known as identities. Identities may be thought of as different ways to represent the same expressions. We are somewhat familiar with this idea from arithmetic. For example, 1 ¼ ,100 50% and 2 200 0.5 all represent the same quantity. However, in different situations, one representation makes more sense or is more useful than another. The same is true for these trigonometric identities. More formally, a trigonometric identity is an equation involving trigonometric functions of an angle, say θ, which is valid for all values of θ for which the functions on both sides of the equation are deﬁned. Here is a list of identities that we have discussed so far either in the text or in the Student Learning Opportunities. They constitute the basic identities given whenever one studies trigonometry. sin 2 y þ cos 2 y ¼ 1 cos ðyÞ ¼ cos ðÀyÞ sinðÀyÞ ¼ ÀsinðyÞ sinðp À yÞ ¼ sin y cosðp À yÞ ¼ ÀcosðyÞ sin2p À cosðyÞ ¼ y tan y ¼ sin y when cos y ¼6 0 cos y cosðy1 Æ y2Þ ¼ cos y1 cos y2 Ç sin y1 sin y2

624 Chapter 12 Trigonometry sinðy1 Æ y2Þ ¼ sin y1 cos y2 Æ cos y1 sin y2 ð12:34Þ sin2y ¼ 2 sin y cos y ð12:35Þ cos2y ¼ 1 À 2 sin2 y ¼ 2 cos2 y À 1: ð12:36Þ Based on these identities, it is possible to generate other useful identities, such as the following. 1 þ tan 2 y ¼ sec2 y 1 þ cot2 y ¼ csc2 y coty ¼ cos y when sin y 6¼ 0 sin y sin2 y ¼ 1 À cos 2y ð12:37Þ 2 cos2 y ¼ 1 þ cos 2y ð12:38Þ 2 sin 2x ¼ 2 sin x cos x: ð12:39Þ We ask you to prove these identities in the Student Learning Opportunities. Part of the secondary school curriculum deals with trigonometric identities, but unfortunately, most secondary school students fail to appreciate their value. This might be due to the fact that they rarely get to see how extremely useful they are in all kinds of applications. For example, many real-life problems necessitate the use of calculus, which involves the evaluation of certain integrals that without the use of trigonometric identities, would be too difﬁcult to do. Identities are especially useful in Fourier series, which forms the foundation for so much of our current tech- nology. We saw them used repeatedly in the chapter on transformations to get some interesting and practical results. We also used them earlier in the chapter in solving certain kinds of trigonometric equations used to model real-life situations, and we will use them again later in this chapter. Before calculators were so readily available, students, mathematicians, and scientists used trig- onometric tables to ﬁnd the values they needed. These tables were ﬁrst created by the Hellenistic mathematician Hipparchus, and were later reﬁned by Ptolemy, the mathematician we spoke about in Chapter 5. Ptolemy was aware of certain trigonometric values that could be established geometrically, and used these, together with trigonometric identities, to build a table of trigono- metric functions for all angles. Speciﬁcally, he used the half angle formulas of equations (12.37) and (12.38) to cut angles into smaller and smaller pieces to ﬁnd their sines and cosines, and then combined these results and used the trigonometric identities for sin(A ± B) and cos(A ± B) to ﬁnd the trigonometric functions of many other angles formed from the ones he knew. In that way, he built the trigonometric tables that were used by astronomers of his day. It is worth noting that Ptolemy had to do his work in terms of chords in a circle, since the terminology we use today for our identities was not yet available. To get a better idea of how he did this, check back to Chapter 5 to see how chords of a circle can be expressed in terms of sines and cosines. Of course, his famous theorem given in Section 4 of Chapter 5, Theorem 5.19, was the basis of much of his work. So now let us get to our study of trigonometric identities. There is an enormous number of trig- onometric identities that one can create—in fact, an inﬁnite number. Some of the more obscure

12.8 Trigonometric Identities 625 identities have strong uses in applications and can be deduced from the identities we gave earlier. We will now only give a few identities and then ask you to do others in the Student Learning Opportunities. We also point out some of the common mistakes that students make when proving identities. Generally speaking, when you are asked to prove a trigonometric identity, you are given an equation of the form A = B where A and B usually are trigonometric expressions, and you have to show that the left side equals the right side (for all values of θ for which both sides make sense), usually by manipulating only one side of the equation. Example 12.24 Prove that sec x þ csc x ¼ sin x þ cos x: tan x þ cot x Solution: Since the right side of this equation is expressed only in terms of sin x and cos x, it makes sense to express the left hand side (LHS) in a similar way. Using the facts that sec x, csc x, and cot x are the reciprocals of cos x, sin x, and tan x, respectively and using the fact that tan x ¼ ,sin x we have cos x LHS ¼ secx þ cscx tan x þ cotx ¼ 1 þ 1 x : cos x þ sin x sin x cos cos x sin x Combining the fractions in the numerator and denominator of the overall fraction we get sin x þ cos x sin x cos x x : sin 2x þ cos 2 sin x cos x Inverting the bottom fraction and multiplying by the top fraction we get sin x þ cos x sin x cos x ¼ sin x þ cos x Á sin x cos x ¼ sin x þ cos x sin 2x þ cos 2x sin x cos x sin 2x þ cos 2x sin 2x þ cos 2x sin x cos x and since sin2 x + cos2 x = 1, this last fraction simpliﬁes to sin x þ cos x which is the right hand side of the identity we were trying to prove. So, we are done. Example 12.25 Prove that 1 2 tan x ¼ sin 2x: þ tan2x

626 Chapter 12 Trigonometry Solution: The denominator of the fraction stands out. We know that, according to one of the pre- vious identities, 1 + tan2 x = sec2 x so that should be our starting point. Our left hand side (LHS) is LHS ¼ 2 tan x 1 þ tan 2x ¼ 2 tan x : sec2x Now since tan x ¼ sin x and sec x ¼ 1 x, this last equation becomes cos x cos sin x 2 x cos LHS ¼ 1 cos 2x which, when we invert and multiply and divide common factors, yields LHS ¼ 2 sin x cos x: But we know that 2 sin x cos x = sin 2x by equation (12.39). So we have shown that the left hand side and the right hand side of the identity we were trying to prove are the same, and we are done. Note that, in both of these examples, we expressed all expressions in terms of sines and cosines. This is an effective strategy for students to use. They need to also realize that it is best to only change one side of the identity at a time. A common mistake that students make is illustrated in the following “solution” to show an identity is valid. Example 12.26 Show that 1 À cos x ¼ sin x : sin x þ cos 1 x Student’s Mistaken Solution: Cross multiply (or equivalently multiply both sides of the equa- tion by sin x(1 + cos x) to get 1 À cos2 x ¼ sin2 x which we know is true since sin2 x + cos2 x = 1. Done! So what is wrong? The student is using the fact that both sides of an equation can be multiplied by the same quantity. But we don’t know the result is an equality yet! That is the point of the exercise! So, when you multiply both sides of the “equation” by sin x(1 + cos x), you are assuming they are already equal. That is, you are assuming what you are trying to prove in the proof and that is illegal! A correct way to solve the problem follows: Solution: Work only on the left side and to get it to look like the right side, we multiply the numerator and denominator by 1 + cos x LHS ¼ 1 À cos x sin x ¼ 1 À cos x Á 1 þ cos x sin x 1 þ cos x ¼ 1 À cos 2x xÞ sin xð1 þ cos ¼ sin sin 2x xÞ xð1 þ cos ¼ 1 sin x x þ cos ¼ RHS: We are done!

12.8 Trigonometric Identities 627 Identities can often be used to explain mysterious behavior. For example, suppose that we graph y = 2 sin x and y = 4 cos x on the same set of axes as shown in Figure 12.82. y4 2 0 –5 –2.5 0 2.5 5 x –2 –4 Figure 12.82 If we were to add these two functions to get y = 2 sin x + 4 cos x, what would you guess the resulting curve would look like? Might you guess at something like the graph shown in Figure 12.83? y 1.75 1.5 1.25 1 0.75 0.5 0.25 –5 –2.5 0 2.5 5 x Figure 12.83 Or would you expect to get something like the following graph shown in Figure 12.84? y5 2.5 0 –5 –2.5 0 2.5 5 x –2.5 –5 Figure 12.84 Most people would guess that it probably looks more like the ﬁrst graph than the second. The second graph is just too regular and looks like a sine curve. Surprisingly, y = 2 sin x + 4 cos x looks

628 Chapter 12 Trigonometry like the graph in Figure 12.84. Other examples like this seem to indicate that, when we graph equa- tions like y = A sin x + B cos x, we get a sine-like curve. But why? The following identity, stated as a theorem, explains it. This identity is very useful to electrical engineers. Theorem 12.27 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ where θ satisﬁes cos y ¼ pﬃﬃﬃﬃﬃAﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ and sin A sin x þ B cos x ¼ A2 þ B2 sinðx þ yÞ A2 þ B2 y ¼ pﬃﬃﬃﬃﬃBﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ : A2 þ B2 Proof. We use a creative approach to this problem that is surprising. Rewrite A sin x + B cos x as pﬃAﬃﬃﬃ2ﬃﬃﬃþﬃﬃﬃﬃBﬃﬃﬃ2ﬃﬃpﬃﬃﬃﬃﬃAﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ sin x þ pﬃﬃﬃﬃﬃBﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ cos ð12:40Þ A2 þ B2 A2 þ B2 x: Now if we can ﬁnd θ such that cos y ¼ pﬃﬃﬃﬃﬃAﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ and sin y ¼ pﬃﬃﬃﬃﬃBﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ then the expression A2 þ B2 A2 þ B2 (12.40) can be written as pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ A2 þ B2ðcos y sin x þ sin y cos xÞ which by our identity (12.34) (taking θ1 = θ and θ2 = x) is the same as pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ A2 þ B2ðsinðx þ yÞÞ: pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ Thus, the graph of y ¼ A sin x þ B cos x ¼ A2 þ B2sinðx þ yÞ should look like a sine curve! & Let us illustrate this with an example. Example 12.28 Write Àsin x + cos x as a sine function. pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃ Solution: Here A = À1 and B = 1. Hence, A2 þ B2 ¼ 2 and we must ﬁnd a θ such that cos y ¼ pﬃﬃﬃﬃﬃAﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ pÀ1ﬃﬃﬃ and sin y ¼ A2 þ B2 2 pﬃﬃﬃﬃﬃBﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ p1ﬃﬃﬃ : A2 þ B2 2 The fact that sin θ is positive and cos θ is negative tells us that our θ should be in the second quad- rant, and that our reference angle is 45. Thus, θ = 135 will work. Our function Àsin x þ cos x ¼ pﬃﬃﬃ 2 sinðx þ 135Þ. One can check on the graphing calculator that the graphs of y = Àsin x + cos x pﬃﬃﬃ and y ¼ 2sinðx þ 135Þ are one and the same and look like the graph shown in Figure 12.85. y 1 0.5 0 –5 –2.5 0 2.5 5 x –0.5 –1 Figure 12.85

12.8 Trigonometric Identities 629 Student Learning Opportunities 1* (C) One of your students proves the equation is an identity by doing the following work: cot x + tan x = sec x csc x cos x þ sin x ¼ 1 x : 1 x sin x cos x cos sin cos2x þ sin2x ¼ 1 sin x cos x cos x sin x 1 ¼ cos 1 x : cos x sin x x sin He is asserting that it is okay when proving an identity to work on both sides of the equal sign, ma- nipulate each side, and then show that the manipulated results are the same on each side. Is the student right? 2* Using your calculator, ﬁrst determine if each of the following appears to be an identity. If it does, prove it. (a) tan(90Àu) = cot u (b) sin2a + cos2a + tan2a = sec2a (c) sin 3x = 3 sin x (d) sec4xÀtan4x = sec2 x + tan2 x [Hint: Factor the left side.] (e) cos(x + 30) + sin(xÀ60) = 0 (f) sin 2x + cos 2x = 1. (g) (tan x + cot x)2 = sec2 x + csc2 x (h) sin sin x ¼ 1 tan x if cos x 6¼ 0 x þ cos x þ tan x (i) 1 sin x ¼ 1 tan x þ cos x þ tan x 3 Using the formulas for sin(θ1±θ2) show that: (a) sin y1 cos y2 ¼ 1 ½sinðy1 þ y2Þ þ sinðy1 À y2Þ 2 (b) cos y1 sin y2 ¼ 1 ½sinðy1 þ y2Þ À sinðy1 À y2Þ 2 (c) cos y1 cos y2 ¼ 1 ½cosðy1 þ y2Þ þ cosðy1 À y2Þ 2 (d) sin y1 sin y2 ¼ 1 ½cosðy1 À y2Þ À cosðy1 þ y2Þ 2 These are called the product to sum formulas. 4 Replacing θ1 in each of the identities from the previous question by x þ y and θ2 by x À y, show 2 2 that: (a) sin x þ sin y ¼ 2 sinx þ y cos x À y 2 2 cosx y sin x y (b) sin x À sin y ¼ 2 þ À 2 2 2 cosx y cos x y (c) cos x þ cos y ¼ þ À 2 2 sinx y sin x y (d) cos x À cos y ¼ À2 þ À 2 2 These are known as sum to product formulas.

630 Chapter 12 Trigonometry 5 Using the formulas in Student Learning Opportunity 4(b) and 4(c) show that sin 3x À sin x ¼ tan x: cos 3x þ cos x 6* Using the appropriate formula from Question 3, write the product sin2xcos3x as a sum. 7 Using the appropriate formula from Question 4, write the sum sin5x + sin3x as a product. 8 Using the identities listed in this section, show that (a) tan 2x ¼ 1 À cos 2x 1 þ cos 2x (b) cos(3x) = 4 cos3 x À 3 cos x [Hint: 3x = 2x + x] (c) sin 4x ¼ 4 cos x cos 2x sin x (d) tanðx Æ yÞ ¼ tan x Æ tan y and hence tan2 x ¼ 2 tan x 1 Ç tan x tan y 1 À tan2x (e) sin 10x ¼ cos 5x [Hint: The results of Student Learning Opportunity 4 might help.] sin 9x þ sin x cos 4x 9 Express each of the following as a function of sin x. 1 pﬃﬃﬃ 2 3 (a)* sin x þ 2 cos x pﬃﬃﬃ (b) À 3sin x þ cos x (c) 3 sin x + 4 cos x 10 Corroborate graphically the identity that we proved in Example 12.26, 1 À cos x ¼ 1 sin x x. sin x þ cos 12.9 Solutions of Cubic Equations Using Trigonometry LAUNCH Solve the cubic equation x3 = 15x + 4 by using algebraic methods or by trial and error (Hint: check integer values of x between 0 and 5). (a) What solutions did you ﬁnd? (b) How many roots does a cubic equation have? Were you able to ﬁnd them all? If not, why not? Now that you have tried solving what appeared to be a relatively simple cubic equation and you were only able to ﬁnd one real root, you are probably wondering how on earth you might ﬁnd the other roots. If you are thinking about using the graphing calculator to see where the graph crosses the x-axis, that certainly would work. But, is there a way to solve this problem without using tech- nology? Would you ever imagine that trigonometry could play a role in its solution? Most people wouldn’t. But, you will be surprised to ﬁnd out as you read this section that indeed, trigonometry can play a major role in solving such cubic equations as this one. Read on to ﬁnd out how it is done.

12.9 Solutions of Cubic Equations Using Trigonometry 631 While you are reading this, you will be exposed to a most fascinating aspect of the history of solving cubic equations. In Chapter 3, we discussed how one ﬁnds a solution to the general cubic equation.. From the material in that chapter, it follows that the formula for a solution of the cubic equation x3 À px À q ¼ 0 where p and q are nonnegative ð12:41Þ is ð12:42Þ x ¼ vuut3 qﬃﬃﬃﬃÆﬃﬃﬃﬃﬃqﬃﬃﬃﬃ2ﬃqﬃﬃﬃﬃﬃ2ﬃﬃﬃﬃﬃﬃÀﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ42ﬃﬃpﬃﬃ7ﬃﬃ3ﬃﬃþuuvt3 qﬃﬃﬃﬃÇﬃﬃﬃﬃﬃqﬃﬃﬃﬃﬃ2ﬃqﬃﬃﬃﬃﬃ2ﬃﬃﬃﬃﬃﬃÀﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ4ﬃ2ﬃﬃpﬃﬃ7ﬃﬃ3ﬃﬃ: This formula led to some strange kinds of answers to cubic equations, which involved square roots of negative numbers for cubic equations whose roots are known to be real. This kind of situation was what motivated the study of imaginary numbers, as we have pointed out in the chapter on imaginary numbers, Chapter 9, and somewhat in Chapter 3. It is interesting to note that, after these formulas were discovered, the French mathematician Francois Viète (1540–1603) (who was a lawyer by trade and did mathematics on the side), discov- ered a formula that would give a solution of the cubic equation (12.41) that involved trigonometric functions. We discuss that now, since the solution process used a trigonometric identity and hence used material from this chapter. The identity he used was cos(3θ) = 4 cos 3θ À3 cos θ, which can be derived as follows: cosð3yÞ ¼ cosð2y þ yÞ ¼ cosð2yÞcos y À sin2y sin y ¼ ðcos 2 y À sin 2 yÞcos y À 2 sin y cos y sin y ðUsing ð12:35Þ and ð12:36ÞÞ ¼ cos 3 y À sin 2 y cos y À 2sin 2 y cos y ¼ cos 3 y À 3 cos ysin 2 y ¼ cos 3 y À 3ðcos yÞð1 À cos 2 yÞ ¼ cos 3 y À 3 cos y þ 3cos 3 y ¼ 4cos 3 y À 3 cos y: Viète observed the following: From the identity cos(3θ) = 4cos3θ À 3cos θ we can solve for cos3θ to get cos3 y ¼ 3 cos y þ 1 cos ð3yÞ: ð12:43Þ 44 Now he wrote equation (12.41) as x3 ¼ px þ q ð12:44Þ and since he assumed that p and q were nonnegative, he observed that we couldrwﬃrﬃﬃite p = 3a2 for p 3q some a and q = a2b for some b. (This is genius at work. To see this, just take a ¼ 3 and b ¼ p :) Then equation (12.44) becomes x3 ¼ 3a2x þ a2b: ð12:45Þ Then out of the clear blue he said, “Let us suppose there is a solution of the form x = 2a cos θ to (12.45).” (His goal was to show that his assumption is true by actually ﬁnding the θ that works. Let’s

632 Chapter 12 Trigonometry bear with him.) From x = 2a cos θ we get that cos y ¼ x : Substituting this into the identity equa- 2a tion (12.43) we get x3 ¼ 3 x þ 1 cosð3yÞ; 8a3 4 2a 4 and multiplying both sides of this equation by 8a3 we get x3 ¼ 3a2x þ 2a3 cosð3yÞ; which is our equation (12.45), provided a2b = 2a3cos(3θ). Thus, x = 2acos θ will be a solution pro- vided a2b = 2a3 cos(3θ) or equivalently if b = 2acos(3θ). Solving this for cos(3θ) we get that cos 3y ¼ b : 2 a qﬃﬃ ð12:46Þ Using the facts that a ¼ p and p ¼ 3q and putting everything in equation (12.46) in terms of p’s 3 b and q’s we have, after doing some algebraic simpliﬁcations, that pﬃﬃﬃ! 3 qpp3ﬃpﬃ cosð3yÞ ¼ 2 : ð12:47Þ In summary, we have found a solution to equation (12.44), namely rﬃﬃﬃ p x ¼ 2a cos y ¼ 2 3 cos y; ð12:48Þ where θ satisﬁes (12.47). This is a remarkable way of solving cubic equations by using trigonometry. Let us return to the launch problem. Example 12.29 Solve the cubic equation x3 = 15x + 4 using Viète’s formula (12.47). Solution: Here p = 15 and q = 4. Substituting this into (12.47) we get that pﬃﬃﬃ ! cos 3y ¼ 12pﬃﬃ3ﬃﬃﬃﬃ ¼ 0:17889: 30 15 Hence 3θ = cosÀ1 0.17889 = 1.3909 radians, so θ = 0.46363 radians. From our experience with solving trigonometric equations, we get our other solutions by solving 3θ = 1.3909 ± 2πk, k = 0, 1, 2,. . . or just θ = 0.46363 ± 2πk/3. Substituting k = 0, 1, and 2, we get θ = 0.46363, θ = 0.46363 ± 2π/3 = 2p.5ﬃ5ﬃﬃ8, and θ = 4.6524. Evalupatﬃﬃiﬃng equation (12.48) at each of pthﬃﬃﬃese three values, we get x = 2 5cos 0.46363 = 4.0, x = 2 5cos2.558 = À3.7319, and x = 2 5cos 4.6524 = À0.26812. Not only did we get one solution, we got all 3 solutions of our equation without even having to deal with the complicated expression in equation (12.42)! (Recall, a cubic polynomial can only have 3 solutions.) This really is impressive. Of course, you might have wondered what would happen in Viete’s solution if the right hand side of equation (12.47) was more than 1? Well, we won’t get into it, but that just means that we have some imaginary roots. Surprisingly, equation (12.47) does have solutions in that case, but they are imaginary and a discussion of that is left to a course in Complex Analysis, since it is beyond the scope of this book. There is, however, another way to get the real roots of a cubic like equation (12.41) when cos 3θ > 1. It uses the notion of hyperbolic cosine instead of cosine. The development parallels what we did in this section. Some of the details are given in the Student Learning Opportunities.

12.10 Vectors 633 Student Learning Opportunities 1 Using the Viète formula, ﬁnd one or more real roots of the following cubic equations. (a) x3Àx = 0 (b) y3À3y = 2 (c) z3À5z = 17 2 The deﬁnition of hyperbolic sine (sinh) and hyperbolic cosine (cosh) follow: sinh x ¼ f ðxÞ ¼ ex À eÀx 2 coshx ¼ gðxÞ ¼ ex þ eÀx 2 (a)* Show that cosh2 xÀsinh2 x = 1. (b) Show that cosh2x = cosh2 x + sinh2 x. (c) Show that sinh2x = 2sinh xcosh x. (d) Show that cosh(2x) = 1 + 2 (sinh(x))2. (e) Show that cosh(x + y) = cosh(x) cosh(y) + sinh(x) sinh(y). (f) Show that cosh(3θ) = 4 cosh3 θ + 3 cosh θ and hence that cosh 3 y ¼ 3 cosh y À 1 cosh ð3yÞ. 4 4 (g) Mimicking what we did in this section, solve the cubic equation given in equation (12.44) terms of coshθ when the right side of equation (12.46) is more than 1. 3 (C) As you have read in this chapter, Francois Viète was quite a genius. Use the Internet to ﬁnd out more about him. What other areas of mathematics was he involved in? What were his other professions? What were some of the notable things he did? 12.10 Vectors LAUNCH Many people who want to experience the excitement of being a pilot use the Internet to access ﬂight simulation programs. They undoubtedly encounter situations where the wind affects the motion of the plane. Let us ask a few questions that you can probably answer. 1 If the plane is traveling due north and the wind is blowing in an easterly direction, how will it affect the direction in which the plane is traveling? How will it affect the speed at which the plane is traveling? 2 If the plane is traveling due north and the wind is blowing in a northerly direction, how will it affect the direction in which the plane is traveling? How will it affect the speed at which the plane is traveling?

634 Chapter 12 Trigonometry Even if you have never accessed a ﬂight simulation program, you probably have an intuitive idea of what happens to an airplane when it is affected by wind. In actuality, you have an idea of how vectors work together. You will learn more details about this in the upcoming section. 12.10.1 Basic Vector Algebra In Section 11.2 you have seen how vectors are used in trigonometry and applied in physics. Although we deﬁned vectors as anything that has magnitude and direction, we have not yet studied them from a purely geometric or algebraic standpoint. In this section we will do just this by expanding on Section 11.2 and reviewing the topic of vectors that most of you have probably learned in linear algebra or calculus. We will then turn to some interesting applications that are most likely new to you. As you will recall, vectors can be represented by arrows whether they are forces or other things that have magnitude and direction. The length of the arrow, just as in forces, represents the mag- nitude of the vector and the angle the arrow makes with the horizontal, the direction. The arrowhead is the tip of the vector, while the other end is the tail. Thus, if a plane is ﬂying at 600 miles per hour northeast, we could denote this velocity as a vector which we show in Figure 12.86. 600 Figure 12.86 As in Section 12.2, vectors will be represented by bold letters, while their lengths will be repre- sented without bold. Two vectors are considered equal if they have the same magnitude and direc- tion. In two dimensions this means that arrows representing them are both parallel and congruent. So, each of the vectors, V1, V2, and V3 shown in Figure 12.87 are the same because they are parallel and have the same length. y v1 v2 v3 x Figure 12.87 When a vector is moved parallel to itself so that its tail is at the origin, we say the vector is in standard form. When a vector is in standard form, its tip will lie at some point (a, b) and a and b are called the components of the vector. If the tail of a vector is at P(x1, y1) and the tip is at Q(x2, y2)

12.10 Vectors 635 then the vector when put in standard form has components (x2 À x1, y2 À y1) as shown in Fig- ure 12.88. y Q (x2, y2) y2– y1 x2 – x1 P (x1 , y1 ) x Figure 12.88 This vector is abbreviated as Q À P or P~Q. The components of a vector are analogous to the components of the forces that we described in Section 12.2, but now we are speaking of vectors that need not be forces. So, a vector representing the velocity of a car can be broken into components, just like was done for a force. The components are called velocity in the x direction and velocity in the y direction. Let us now see how to add vectors. It can be done in two ways. Suppose for example we have two vectors, V1 and V2 shown in Figure 12.89. V1 V2 Figure 12.89 One way to add these vectors and get V1 + V2 is to move the tail of V1 to the tip of V2 and draw the arrow from the tail of V2 to the tip of the moved V1. See Figure 12.90. V1 V1 +V2 movedV1 V2 Figure 12.90 Another way to add two vectors V1 and V2 is to move V1 so that its tail coincides with the tail of V2 and then complete the parallelogram having V1 and V2 as sides. The diagonal of the parallelogram will be the V1 + V2 as the diagram in Figure 12.91 shows.

636 Chapter 12 Trigonometry V1 movedV1 V1 +V2 V2 Figure 12.91 You may be wondering if addition of vectors is commutative. Well, in a similar manner to what we did earlier, we can construct V2 + V1 by moving the tail of V2 to the tip of V1 and drawing an arrow from the tail of V1 to the tip of the moved V2. We will get the same diagonal as we did earlier for the sum. So V1 + V2 = V2 + V1. We now explore the subtraction of vectors. If V1 and V2 are vectors and we draw the vector from the tip of V1 to the tip of V2, what vector is that? (See Figure 12.92.) x V1 V2 Figure 12.92 Well, if we call that vector x, then V1 + x = V2 and it seems logical to call x = V2ÀV1. Indeed, that is what we call it, and we now have our deﬁnition of subtraction of vectors:V2ÀV1 is the arrow drawn from the tip of V1 to the tip of V2. Given a vector V, the vector ÀV is the arrow with the same length but opposite direction as shown in Figure 12.93. V –V Figure 12.93

12.10 Vectors 637 Now we examine the concept of multiplication of vectors. If V is a vector and λ is a scalar, then λV is the vector |λ| times as long as V and in the same direction if λ > 0 and in the opposite direction if λ < 0. Thus, the vector 3V is three times as long as V as is the vector À3V. Only 3V is in the same direction as V while À3V is in the opposite direction. See Figure 12.94 where 3V is shown. 3V V Figure 12.94 Let us practice a bit with “arrow algebra.” Although we have said that the result of adding two vectors can be interpreted as the diagonal of a parallelogram, in practice, this is rarely used. Phys- icists simply represent vectors by arrows and then use the geometric deﬁnition of adding arrows. Namely, to add V1 and V2 we move V1 so that its tail is at the tip of V2 and then draw the arrow from the tail of V2 to the tip of the moved V1. The next example clariﬁes this. Example 12.30 Using the vectors u, v, and w in Figure 12.95, w V u Figure 12.95 draw each of the following vectors (a) u + v + w, (b) u À v, (c) v À u. Solution: To add u + v + w, ﬁrst add u + v as shown in Figure 12.96,

638 Chapter 12 Trigonometry w u +v+w v u+v u Figure 12.96 and then add w. We notice all we need to do to get our ﬁnal result is to move v to the tip of u and then move w to the tip of the moved v. Once we draw an arrow from the tail of u to the tip of the moved w, we are done. (b and c) u À v is the vector drawn from the tip of v to the tip of u, and v À u is the vector going from the tip of u to the tip of v. See Figure 12.97. v– u u v Figure 12.97 Example 12.31 Suppose that we have any triangle ABC and turn the sides into vectors as shown in Figure 12.98. C A B Figure 12.98 Show that the sum of the vectors is the zero vector, (0,0).

12.10 Vectors 639 Solution: To simplify this problem we need to situate one of the vertices at the origin. We can do this by moving the triangle parallel to itself so that A is at the origin. Since we are moving the entire ﬁgure parallel to itself, none of the vectors are changed. Using the result of the previous problem part (a), the sum of the vectors, is where the ﬁnal tip lands when the vectors are put “tail to tip” as they were. This place is (0, 0) as Figure 12.99 shows. C AB Figure 12.99 So, the sum of the vectors is (0, 0), the zero vector. We will take this geometric interpretation of vectors to an interesting conclusion in subsection 3. There are many useful applications of addition of vectors. For example, if we have many forces acting on an object (each represented by a vector) and we want to know if we could replace the many forces by a single force acting on the object, the answer is “Yes,” and this single force is the sum of the vectors representing the individual forces. As alluded to in the launch question, vectors are also used by pilots in steering their planes when confronted by wind forces so that they maintain a certain speed and direction, allowing them to get to their destinations on time. We will see an application of this in the next subsection. 12.10.2 Components of Vectors We just explained how to add and subtract two vectors and multiply a vector by a scalar. For addi- tion, we used the notationV1 + V2, the same as we used for adding forces. But, in Section 12.2, when we added forces, we added components of the vectors. Thus, we seem to have used the same nota- tion for two different things which is confusing, unless of course, they are really the same. Does the “tail to tip” geometric approach that we used to add vectors have anything to do with adding the components? The answer is “Yes.” In fact, the two approaches are equivalent, which we will now show. In Figure 12.100 we let V1 = (a, b) and V2 = (c, d). To add V1 and V2 we move V1, so that its tail is at R and draw horizontal line segment RT and also draw QT perpendicular to RT. Triangle OPD is congruent to triangle RQT and we see by the diagram, the point, Q, where the tip of V1 + V2 lies, when it is put in standard form, is (a + c, b + d). Thus, V1 + V2 = (a + c, b + d). So, to add two vectors, we just add their components, exactly as we did with forces!

640 Chapter 12 Trigonometry y Q(a +c, b +d ) P (a,b) b V1 a V2 T R (c,d ) O aD x Figure 12.100 Using components it can be shown that if V2 = (c, d) and V1 = (a, b), then V2ÀV1 = (cÀa, dÀb) = V2 + (ÀV1). For practice, you can show that if V = (a, b), then ÀV = (Àa, Àb). One also notices a similarity between vectors and imaginary numbers. In Chapter 9 we found that imaginary numbers could be represented by arrows and that their sum is found by adding the components. Thus, imaginary numbers can also be considered vectors. How do we ﬁnd the components of a vector? We do it exactly as in Section 12.2. The magnitude of a vpeﬃcﬃﬃtﬃﬃoﬃﬃﬃrﬃﬃﬃVﬃﬃﬃﬃ1 = (a, b) is the length of the vector, which we can see from Figure 12.101 is just V1 ¼ a2 þ b2. y (a, b) V1 b ax Figure 12.101 (Remember, we are using the convention that boldface represents the vector, while unbolded rep- resents the length of the vector.) Using trigonometry we see that a ¼ cos y and that b ¼ sin y, so V1 V1 a ¼ V1 cos y and b ¼ V1 sin y

12.10 Vectors 641 just as with forces. Furthermore, tan y ¼ b : a We can again make the connection with imaginary numbers, where, as shown in Chapter 9 we had the same results. There, a was called the real part of the imaginary number and b the complex part. Here is a problem similar to one we did with forces earlier, only this time it involves velocities. At ﬁrst glance you will probably think this problem looks quite different. Example 12.32 An airplane starts out traveling at 300 miles per hour north, but a wind traveling at a direction N 40E is blowing at 60 miles per hour takes it in another direction. (a) What will be the resulting speed and direction of the airplane? (b) With what speed and in what direction must the plane ﬂy so that, taking the wind into account, it ﬂies at 300 miles per hour north? Solution: (a) Figure 12.102 shows us our picture. plane 300 mph 40° wind 60 mph Figure 12.102 We are really looking for the vector which results from adding these two vectors. The velocity of the plane is the vector V1 = (0, 300). To ﬁnd the components of the vector, V2, the velocity of the wind, we need to use ðV2Þx ¼ V2 cos 50 ¼ 60 cos 50 ¼ 38:567 miles per hour ðV2Þy ¼ V2 sin 50 ¼ 60 sin 50 ¼ 45:963 miles per hour Note that the 50 is the angle the wind vector makes with the horizontal. So V2 = (38.567, 45.963). Adding V1 and V2 we get V1 þ V2 ¼ ð38:567; 345:963Þ: pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ The magnitude of this vector is V1 þ V2 ¼ 38:5672 þ 345:9632 ¼ 348:11 miles per hour which is the actual speed of the plane with the help of the wind. The direction of the plane is found from tan y ¼ b ¼ 345:963 ¼ 8:9704: a 38:567 Thus, y ¼ tanÀ1ð8:9704Þ ¼ 83:64 is the direction the plane is ﬂying relative to the horizontal.

642 Chapter 12 Trigonometry (b) Suppose the plane ﬂies with velocity vector V = (a, b). Here is our picture where we want the diagonal of the parallelogram (the resulting velocity) to be (0, 300). (See Figure 12.103.) 300 mph V 40° wind 60 mph Figure 12.103 From our picture, V þ velocity of the wind ¼ ð0; 300Þ: Thus, ða; bÞ þ ð38:567; 45:963Þ ¼ ð0; 300Þ and so, ða; bÞ ¼ ðÀ38: 567; 254: 037Þ: To ﬁnd θ, the direction of V, we solve tan y ¼ b ¼ 254: 037 ¼ À6:5869: a À38: 567 We get θ = 98.634. Sqo, ﬃtﬃﬃhﬃﬃﬃeﬃﬃﬃﬃpﬃﬃﬃlﬃaﬃﬃﬃnﬃﬃﬃeﬃﬃﬃmﬃﬃﬃﬃﬃuﬃﬃﬃsﬃﬃtﬃﬃﬃﬂﬃﬃﬃyﬃﬃﬃﬃaﬃﬃtﬃﬃﬃaﬃﬃﬃnﬃ angle of 98.634 with the horizontal, or N 8.634 W and with a speed of ðÀ38: 567Þ2 þ 254: 0372 % 257 miles per hour. You may have noticed that vectors behave much like matrices with respect to addition and sub- traction. In fact 1 × 2 and 1 × 3 matrices are used to represent vectors in two and three dimensions. Since we have talked about the geometry of vectors, it is natural to ask geometric questions about vectors. For example, how can we tell if two vectors V1 = (a, b) and V2 = (c, d) are parallel? One simple way is to realize that if they are parallel, then when brought into standard position, they will both emanate from the origin and will make the same angle θ with the positive x-axis. Thus, tan θ is the same for both vectors. That is, tan y ¼ b ¼ d : ð12:49Þ a c If we substitute tan θ with λ, then (12.49) says that b ¼ d ¼ l a c which can be rewritten as c ¼ d ¼ l: a b From this, it follows that c = aλ and d = bλ. Thus, (c, d) = (aλ, bλ) = λ(a, b). Remembering that V1 = (a, b)

12.10 Vectors 643 and V2 = (c, d), this last statement says that V2 ¼ lV1: Thus, we have shown that if V2 is parallel to V1 then V2 = λV1 for some constant λ. For practice, we are leaving the converse for you to prove. Thus, we have Theorem 12.33 Two nonzero vectors V1 and V2 are parallel, if and only if there is some constant λ such that V2 ¼ lV1: Another way of saying this is that two vectors are parallel if and only if one is a scalar multiple of the other. For example, the vector, V1 = (2, 2) is parallel to V2 = (8, 8) since (8, 8) is 4V1. But V1 is not parallel to (3, 4) since no multiple of V1 will give (3, 4). Notice that every vector is considered parallel to itself. (Just take λ = 1.) Now that we know how to tell if two vectors are parallel (they have the same slope or one is a multiple of the other), it is natural to ask how we know if two vectors are perpendicular. This too is not too difﬁcult to answer, since in two dimensions if two vectors, V1 = (a, b) and V2 = (c, d) are perpendicular, then according to what we learned in secondary school, the product of their slopes is À1. Thus, the slope of V1 Á slope of V2 ¼ À1 or, put another way b Á d ¼ À1: a c Multiplying both sides of this equation by ac we get bd ¼ Àac or ac þ bd ¼ 0: We have shown that if two vectors (a, b) and (c, d) are perpendicular, then ac + bd = 0. Just reversing the steps we can show that, if ac + bd = 0 then the product of the slopes of the vectors is À1 and thus, they are perpendicular. So what we have shown is that the following theorem holds. Theorem 12.34 Two nonzero vectors u = (a, b) and v = (c, d) are perpendicular if and only if ac + bd = 0. The quantity ac + bd given in the theorem is known as the dot product of the vectors u and v and is abbreviated as uÁv (read u dot v). Example 12.35 Are the following pairs of vectors perpendicular? (a) u = (À3, 4) and v = (4, 3) (b) u = (1, 2) and v = (À5, 3)

644 Chapter 12 Trigonometry Solution: (a) Here uÁv = (À3)(4) + (4)(3) = 0, so u and v are perpendicular. (b) Here uÁv = (1)(À5) + (2)(3) ¼6 0. So these vectors are not perpendicular. Most pre-service teachers have been exposed to dot products either in their linear algebra courses, or in their calculus courses, so we will just state some rules for dot products that we will use and that you can verify. uÁv¼vÁu ð12:50Þ u Á ðv þ rÞ ¼ u Á v þ u Á r ð12:51Þ u Á u ¼ u2 ð12:52Þ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ Note that the third result is immediate once you see that if u = (a.b) then u ¼ a2 þ b2 and uÁu = a2 + b2 so certainly uÁu = u2. 12.10.3 Using Vectors to Prove Geometric Theorems We have shown how vectors can be used in physics and navigation. We now turn to a very inter- esting consequence of the geometric interpretation of vectors and their role in proof. You have probably seen a proof of the fact that the diagonals of a parallelogram bisect each other in a second- ary school geometry course. We prove it here using vector methods. Notice its elegance. Example 12.36 Prove that the diagonals of a parallelogram bisect each other. Solution: Before starting any geometric proof, we must represent the geometric ﬁgure using vectors. In this case we start with parallelogram ABCD and consider each side as a vector as shown in Figure 12.104. Note that it does not matter how you choose the direction of your vectors. We draw one diagonal and indicate its midpoint by E. The goal is to show that E is also the midpoint of DB. We will do that by showing that DE ¼ 1 DB. 2 DC E bb Aa B Figure 12.104 Now, diagonal AC = a + b, and so AE = AE ¼ 1 ða þ bÞ . Furthermore, DE ¼ AE À AD ¼ 2 1 ða þ bÞ À b ¼ 1 ða À bÞ: But the diagonal DB = a À b. So we have shown that DE ¼ 1 DB and we 2 2 2 are done! We hope you appreciated this interesting use of vectors. Let’s try another geometric proof using vectors, to reinforce the approach. Example 12.37 Using vectors, show that an angle inscribed in a semicircle is a right angle.

12.10 Vectors 645 Solution: In Figure 12.105 you see angle ACB inscribed in a semicircle. Our goal is to show that angle C = 90. C AO B Figure 12.105 We draw CO and then represent the sides using vectors. (As we said, you may choose any direc- tions for your vectors just as long as the relationships you write are consistent with your drawing.) Figure 12.106 shows our new diagram using vectors. C u rv Aw Ow B Figure 12.106 We need to show that u Á v = 0, from which it will follow that AC is perpendicular to CB and angle C is 90 degrees. Observe, that according to our diagram, AO = w and OB = w since the arrows have the same length (both are radii) and point in the same direction. From our diagram we have u = w + r and r + v = w. So v = w À r. Therefore, uÁv ¼ ðw þ rÞ Á ðw À rÞ ¼ w Á w À w Á r þ r Á w À r Á r ðBy repeated use of equation ð12:51Þ:Þ ¼ w Á w À r Á r ðBy equation ð12:50Þ:Þ ¼ w2 À r2 ¼ 0 ðSince w2 is the square of the radius and so is r2: Þ We have shown that u Á v = 0, hence AC is perpendicular to BC and angle C = 90. We hope that you enjoy doing more of these geometric proofs in some of the Student Learning Opportunities that follow. Student Learning Opportunities 1* If an airplane starts out ﬂying southeast at 300 miles per hour, but a tailwind of 50 miles per hour south acts on it, at what speed will the plane ﬂy and in what direction? 2* A boat needs to travel at 20 knots per hour east, but the current is pulling it north at 3 knots per hour. At what speed and direction must the boat travel to accomplish this?

646 Chapter 12 Trigonometry 3* Find, if possible, scalars, λ, μ such that λ(2, 5) + μ(À1, 3) = (À8, À9). 4* Find all values of r such that the magnitude of r(3, 5) is 1. 5 Determine if the following pairs of vectors, (given in terms of their components) are parallel, perpendicular, or neither? Explain. (a)* (À2, 4) and (4, À8) (b) (4, 6) and (6, 9) (c) (À2, 5) and (5, 2) (d) (3, 4) and (5, 7) (e) (a, b) and (9a, 9b) where (a, b) 6¼ (0, 0) (f) (c, d) and (Àd, c) where (c, d) 6¼ (0, 0) 6 (C) One of your students, Georgia, knows how to use slopes to ﬁgure out if a triangle is a right triangle and wants to know whether it can be done using vectors, and if it can, how do the two methods compare? How do you respond? [Hint: It is a good idea to begin by asking Georgia to examine a speciﬁc case, say a triangle MAT, where M, A, and T, are the points M = (4, 10), A = (8, 2), and T = (2, 4).] 7 Using the parallelogram shown in Figure 12.107, express each of the following in terms of x and/or y. BC x A yD Figure 12.107 (a)* BC (b)* CD (c) AC (d) BD (e) AM, where M is the intersection of the diagonals of the parallelogram. 8 Using the diagram shown in Figure 12.108, B y x D E 5y A C Figure 12.108 express each of the given vectors in terms of x and y, given that E is the midpoint of AD.

12.10 Vectors 647 (a)* AD (b) AE (c) BE (d) EC (e) CE 9 (C) Your students ask if there is a way to use vectors to prove the Pythagorean Theorem. How do you do it? [Hint: Draw your vectors so c = a + b then dot c with itself.] 10 (C) You told your students that vectors can be used to simplify certain proofs. They want you to prove your point by using vectors to prove the well known theorem in geometry that says that the line joining the midpoints of two sides of a triangle is parallel to the third side and half its length. Show how you would prove this theorem by using Figure 12.109. B x y E D y x A C Figure 12.109 11* A well-known theorem from geometry says that if the diagonals of a quadrilateral ABCD bisect each other (that is, have the same midpoint), then the quadrilateral is a parallelogram. Prove this using vectors. You will need the following: If P and Q are points in the plane, then the vector joining the points P and Q is denoted by Q À P. (This is really an abbreviation for the vector OQ-OP where O is the origin.) From secondary school the midpoint of the line segment joining P and Q is obtained from averaging the x coordinates and averaging the y coordinates and this can be abbreviated by 1 ðP þ QÞ. [Hint: Begin with quadrilateral ABCD 2 shown in Figure 12.110. BC AD Figure 12.110

648 Chapter 12 Trigonometry Saying the diagonals have the same midpoint is the same as saying 1 ðA þ CÞ ¼ 1 ðB þ DÞ. Show 2 2 that this implies that both B À A = C À D and that C À B = D À A.] 12 In this problem we guide you through the proof that the altitudes of a triangle are concurrent, that is, meet at a point. We use the conventions stated in the previous problem, namely that Q À P is the vector joining P to Q. Begin with triangle ABC which has been put on the coordi- nate plane. Draw altitudes AD and CE. They of course meet in some point F, as Figure 12.111 shows. B ED C F A Figure 12.111 (a) Show (F À A) Á (B À C) = 0. (b) Show (F À C) Á (A À B) = 0. (c) Show that BF extended to AC is an altitude by showing that the vector BF is perpendicular to the vector AC. [Hint: Expand the expressions in (a) and (b), add the results and then factor, realizing that F, for example, is really an abbreviation for the vector OF where O is the origin.] 13 Using vectors, show that if we have a parallelogram with legs of lengths a and b, and diagonals with lengths d1 and d2, then 2a2 þ 2b2 ¼ d12 þ d22. 14 Explain why each of the following holds. (a)* u Á v = v Á u (b) (u + v) + w = u + (v + w) (c) u Á (v + r) = u Á v + u Á r (d) (u + v)Á(u + v) = u Á u + 2u Á v + v Á v (e) 5(u + v) = 5u + 5v (f) Àv = À1(v) 15 Show that if (u + v) Á (u À v) = 0, then u and v have the same magnitude. 16 In this problem we guide you through the derivation of the formula cos y ¼ uÁv ðuÞðvÞ which gives the cosine of the smaller angle between two vectors.

12.10 Vectors 649 Suppose that V1 = (a, b) and V2 = (c, d) are two vectors, and we are interested in the smaller angle, θ, between them. Figure 12.112 gives us our picture. Figure 12.112 Draw the vector from the tip of u to the tip of v forming a triangle as shown in Figure 12.113. Figure 12.113 Begin by using the Law of Cosines in the triangle: ðu À vÞ2 ¼ u2 þ v2 À 2 uv cos y and recall that u2 = u Á u, v2 = v Á v and (u À v)2 = (u À v) Á (u À v). Express everything in the Law of Cosines in terms of dot products, expand and simplify and then solve for cos θ. 17 Using the previous exercise, ﬁnd the angle between each of the following vectors. (a)* u = (À2, 3) and v = (4, 5) (b)* u = (À4, 3) and v = (3, 2) (c) u = (À4, 3) and v = (3, 4) 18* (C) Your students have accepted all they have learned about the algebra of vectors. But, they still want to know if it is the case that if u Á v = u Á r then v = r (where u, v, and r are non zero vectors). What is your reply? Explain.

650 Chapter 12 Trigonometry 12.11 Lissajous Curves LAUNCH The three curves in Figures 12.114–12.116 are called Lissajous curves. Examine them carefully and describe what their similarities and differences are. Figure 12.114 Figure 12.115 Figure 12.116

12.11 Lissajous Curves 651 In looking at these curves, you probably have noticed that they contain curves that look similar to the graphs of trigonometric functions, perhaps on their sides. In this section you will learn more about them and how they can be used in practical settings. We mentioned earlier that sound is believed to be wavelike. Where did this idea come from? The French physicist Jules Lissajous (1822–1880) is given credit for this discovery. He would strike tuning forks, hear sound and wanted to be able to “see” the sound. After numerous experi- ments, he happened upon the “right” way to “see” sound. He would place the tuning fork in front of a series of carefully placed mirrors. He would then strike it, hear the sound, and shine a beam of light on it. The image went from one mirror to the next, to a screen on the wall. Suddenly, he saw images of waves dancing on the screen as the tuning fork vibrated. He actually saw the sound waves! He then started playing with images formed by striking two tuning forks positioned at right angles to one another and discovered some beautiful patterns now known as Lissajous curves (also known as Bowditch curves). For these experiments he received a Nobel prize. The Lissajous curves can be described by using two trigonometric functions, one for x and the other for y. Since the x and y axes are perpendicular, we can simulate what he did in the laboratory. Following are typical equations for x and y, where A, B, a, b, d and e are constants. (There are several different variations on this.) x ¼ A sinðat þ dÞ y ¼ B sinðbt þ eÞ Lissajous curves have applications in physics, astronomy, and other sciences. Students can graph these parametric equations on their graphing calculators by setting the machine to parametric mode. The pictures one gets are very sensitive to the ratio of a/b and can be very pretty when done with a good grapher program. Following are two such pictures which are not so elaborate. Our ﬁrst picture shown in Figure 12.84 is generated by the parametric equations x = sin(2t) and y = sin (3t). It looks a lot like the shield of the Atomic Energy Commission (Figure 12.117). y1 0.5 0 –1 –0.5 0 0.5 1 x –0.5 –1 Figure 12.117

652 Chapter 12 Trigonometry The second, shÀown Áin Figure 12.118 is the logo of television channel ABC, and is generated by x ¼ sinðtÞ; y ¼ sin3 t À p . 2 y1 0.5 0 –1 –0.5 0 0.5 1 x –0.5 –1 Figure 12.118 A check of websites dealing with Lissajous curves will show you some much nicer and more interesting pictures. Student Learning Opportunities 1 Draw each of the following Lissajous curves by using your graphing calculator or ﬁnding a site on the Internet that does graphing. (a) x = 2 sin(3t + π), y = À2 sin (4t À π) (b) x = 4 sin(πt + π/2), y = 2 sin (2t) (c) x = 5 sin(3t), y = 4 sin (5t) pﬃﬃﬃ p (d) x ¼ sinð5 t À 6Þ; y ¼ 2sin t À 2 2 Make up your own values and using the Internet or your graphing calculator, ﬁnd some inter- esting Lissajous curves. Print out at least two of the ones you like best and bring them to class. Describe how you made them.

CHAPTER 13 DATA ANALYSIS AND PROBABILITY 13.1 Introduction Students begin studying probability and statistics early in their schooling, ﬁrst very informally, and later with increasing abstraction. This area of mathematics can be most interesting to learn and to teach, as the applications pervade our daily life. For example, life insurance companies decide on their rates based on the percentage of people that they believe will die in the coming year, the same way auto insurance companies base their rates on the percent of their subscribers who they feel will be in accidents. How can they tell what percent of their subscribers will have acci- dents? The answer is, they can’t. But what they can do is estimate what should happen, based on the percent of drivers who have had accidents in the past, because what insurance companies have found is that there is, for the most part, a certain regularity in the percent of their subscribers involved in accidents from year to year. This is based on many years of data from millions of drivers. Thus, their data come from a very large population. Sometimes the percents do vary. For example, when the seat belt laws were implemented, the percent of deaths due to car accidents decreased, and that, in turn, affected their analysis. The study of probability is, the study of the likelihood of events occurring and this is why prob- ability is often deﬁned as the study of chance. When we say the chance of rain is 45% today, we are saying that, in the past, when conditions were similar to what they are now, rain resulted about 45% of the time. Of course, that guarantees nothing about what will happen today, but it does give us data upon which we can make decisions. For example, should we carry an umbrella today? For some people, a 45% chance of rain is not enough to carry an umbrella. For others, it is. In this chapter we will discuss some basic concepts and point out some interesting and thought-provoking ideas and issues of both probability and statistics. We will begin by giving a quick review of the basics of probability. 13.2 Basic Ideas of Probability LAUNCH Maxine loves to play the lottery. She claims she has a better chance of winning if she continues to bet the same set of ﬁve lottery numbers every time. But Molly says that it is better if you randomly

654 Chapter 13 Data Analysis and Probability select any set of ﬁve numbers for the same period of time. Sophie says it really doesn’t matter how you pick your numbers. Who is right? Why? People often have very strong opinions about issues concerning probability. It is most interesting that many of these opinions are based on fundamental misconceptions about probabilistic events. We hope that in doing the launch question you noticed one of these misconceptions. This next section will address many of these issues. 13.2.1 Different Approaches to Probability Every one has heard a statement like, “If we toss a fair coin, the probability that ‘heads’ shows up is 1/2.” What does this mean? Does it mean that we can be assured that if we toss a coin ten times that 50% of the time heads will show up? The answer is, “Of course not.” Any child knows that when we toss a coin anything can show up and in any order. For example, if we toss a coin 5 times, we may very well get 5 heads in a row. Or, we may gets head–tail–head–tail–head. Saying that the probability of getting a head is 1/2 means that, if we toss the coin over and over and over a large number of times and measure the ratio of times that a head comes up to the total number of tosses, we would expect that ratio to be close to 1/2. So, if we toss the coin say, a million times, we would expect about 500 thousand of the ﬂips to turn up heads. Will this in fact happen? Isn’t it possible that we can get a million ﬂips all of which turn up heads? Not likely, but possible! But when lots and lots of people ﬂip a coin a million times, and we total the number of ﬂips and the number of heads that come up, we are saying we believe heads will come up about 50% of the time. So how do we check this assertion? Do we just toss coins day in and day out and check our assertion that way? We could, but even if we did ﬁnd our ratio of heads to ﬂips is 21, there is no guarantee that, when others do the same, their results will be the same, or when we put all our results together that we will be close to 50%. So, in answer to our ques- tion, “How do we check this assertion?” the answer is, we don’t. When we say that the probability of getting a head is 1/2, we are using a model which expresses our belief that many tosses will result in a ratio of heads to total number of tosses that is roughly 1/2. This model is known as the frequency approach. Now, take note that not all models are necessarily good. But this model has been around for several hundred years, and is based on what we have observed in the past, and what we believe will always happen in a large number of ﬂips. What is different about this model is that we can never really check our beliefs, for even though in a million ﬂips the ratio of heads to ﬂips might be close to 21, unlikely as it is, by the time we get to a billion ﬂips, things might change. It is not like Hooke’s law (see Chapter 9) for a spring, where we can actually test spring after spring and see that Hooke’s law is true. Probability is a different kind of model. In general, if we wish to determine the probability of an event occurring during an experiment using the frequency approach, we perform the experiment many times. Each time we perform the experiment, we have performed what is called a trial. Each time the event we are interested in occurs, we say that we have a success. If n is the number of trials performed and S is the number of successes, the probability, p, of the event is deﬁned to be p ¼ lim S : ð13:1Þ n!1 n

13.2 Basic Ideas of Probability 655 Because of our inability to check statements like “If we toss a coin a large number of times, the percent of heads is always close to 50%,” some people prefer the classical approach, which we will discuss shortly. Using the frequency approach, ﬁnding the probability of an event requires that we take a limit as n ! 1. Obviously, we can’t perform a huge number of trials, so the practical person needs to be able to estimate the probability of an event happening. This is done by performing an experiment a ﬁxed, but large number of times, calculating the ratio of successes to the total number of trials, and using that ratio as an estimate of the true probability of the event occurring. This ratio is called the experimental probability or empirical probability of an event. The person then uses this number in his or her calculations. As it turns out, there are many situations where using experimental prob- ability is the only way that a probability can be calculated. For example, insurance companies use experimental probability all the time to price their life insurance policies. If they want to determine the probability that, say a man 40 years old who smokes and drinks will die of a heart attack by the age of 60, they use the data gathered from extensive records of people who died by the age of 60 and fell into this category, to price the life insurance policy for a 40-year old man who smokes and drinks and asks to be insured. Also, only experimental probability can be used to ﬁnd the probability of an irregular object landing in a certain position. As a simple case, suppose one had a cup and for some reason wanted to know the probability that it would land on its side when tossed and allowed to fall. He or she could toss the cup up and allow it to fall, say 1000 times. If 914 times the cup landed on its side, that person would say that the probability of this happening is approximately .914 1000 From an instructional perspective, asking students to compute experimental probabilities gives them a feel for the problem and an inkling into what is a reasonable result for the theoretical prob- ability. For example, a teacher might ask a student to toss a pair of dice say 50 times and count the number of times the faces on the dice add up to 7 before ever computing the theoretical probability of this event. If later they do compute the theoretical probability, they can determine whether or not the result they arrived at was reasonable. One problem with experimental probability is that, if different people perform the experiment 1000 times, the results can vary widely. So, each will have his or her own probability to use in com- putations. The theoretical probability discussed soon, by contrast, is a ﬁxed number and does not vary from person to person. Now, suppose we have a dart board, part of which is colored yellow and the rest of which is colored blue, and we asked “What is the probability that a dart thrown at random which hits the board, lands on yellow?” One might say, the probability is 1/2, the reason being that there are only two possibilities, one of which is landing on yellow. This may seem reasonable. But suppose that we have the dart board shown in Figure 13.1. yellow blue Figure 13.1

656 Chapter 13 Data Analysis and Probability Now would you say that the chances or probability of a dart hitting the yellow region is 1/2? Look at how much greater the yellow area is than the blue area. This leads to the notion of likelihood. Landing on yellow in the dartboard seems more likely than landing on blue, provided the dart hits the dart board. Thus, just because we have two outcomes does not mean that the probability of each outcome is 1/2. As another example, suppose that one tosses a pair of dice. The sum of the numbers that come up when the dice fall can be any number from 2 to 12 for a total of 11 possibilities. If one were to ask, “What is the probability of a sum of 7 coming up on the dice?”, one might answer “1/11, the same as the probability of a sum of 2 coming up.” But the probability of getting a 2 is not the same as the probability of getting a 7, since there is only one way to get a 2, namely, each die falls with 1 dot up which we denote by the ordered pair (1, 1), and there are several ways to get a sum of 7 (e.g., (1, 6), (2, 5), (3, 4), (6, 1), and so on.) Thus, the event of getting a sum of 7 is more likely to occur than the event of getting a 2. This brings us to what is called the classical or theoretical approach to probability and to a deﬁnition of probability that one ﬁnds in most secondary school books known as the classical or theoretical deﬁnition of probability: Suppose that an experiment can result in n equally likely outcomes, and that we are interested in the probability of event, E, occurring. If E occurs in precisely m out of the n outcomes (that is, we have “success” m out of n times), then the prob- ability of that event occurring is mn. Here are some examples. Example 13.1 If one tosses an ordinary die, there are six possible outcomes, each (we believe) equally likely, so the probability of getting a “2” face up when the die falls is 16, and the probability of getting a prime number (2, 3, or 5) is 3 or 21. 6 Example 13.2 In a regular hexagon ABCDEF, a point is chosen at random inside the hexagon. What is the probability that the point, P, chosen, is closer to the vertex A than to any other vertex? Solution: The hexagon is symmetric. Thus, the events that (1) P is closer to A than the other ver- tices, (2) P is closer to B than to any other of the vertices, (3) P is closer to C than to any of the other vertices. . . (6) P is closer to F than to the other vertices, are all equally probable. Hence the proba- bility is 1/6. Example 13.3 Three numbers x, y, and z are chosen at random from the interval [0, 10]. What is the probability that x y z? Solution: There are only 6 possibilities: xyz xzy yxz yzx zxy z y x:

13.2 Basic Ideas of Probability 657 There is no reason any one should occur over any other, so all possibilities are equally likely. The probability therefore is 1/6 that x y z. There is yet a third school of thought on probability known as the subjective approach to probability. There, the probability of an event is a measure of how likely we feel the result is, which, of course, is subjective. Certainly, this can vary from person to person. Often in business, executives make decisions based on what they consider the subjective probability that an event will happen. 13.2.2 Issues With the Approaches to Probability There are issues with all of the preceding approaches to probability. For example, with the fre- quency approach, suppose that we want to determine the probability of an event happening. How many times must we perform the experiment before we are sure that the ratio of successes to the number of trials gives us a number close to the “real” probability of the event happening? Are we ever sure we are close? Do we know that the limit exists in the deﬁnition of probability according to the frequency deﬁnition, equation (13.1)? What if we performed the experiment say a million times and then repeated it another million times? Would the ratio of the number of successful outcomes to all outcomes be the same or even close to each other in each of the million trials? If not, how can we be sure our deﬁnition means anything? Suppose that it is not pos- sible to perform an experiment many times, (for example, dropping an atomic bomb on a country), how can we determine the probability of an event related to this? The classical deﬁnition also has problems. For one, it can only be applied to situations with a ﬁnite number of outcomes. What if there are inﬁnitely many outcomes as will occur in later exam- ples? Also, if we ﬁgure out the classical probability of an event to be 1/6, will the frequency approach to the experiment also yield the same 1/6? That is, are the approaches compatible? When we use the classical approach, we are assuming that the outcomes are equally likely. Perhaps this is not the case and we are building on a false foundation. (Indeed there are real exam- ples in atomic physics where everything points to the outcomes being equally likely, but something in nature makes them not equally likely. (See Feller, 1968, pp. 40–41.) Of course, the subjective approach to probability has even more problems. Can we conclude anything about probabilities using the subjective approach? If they are subjective, what does a probability of 50% mean in practice? What does this tell us to expect? If one person assesses the probability of an event to be 0.50 and another 0.90, is either right, and aren’t the conclusions each makes likely to be different? Given the difﬁculties with each of the three approaches to probability, mathematicians for whom precision is paramount, tried to develop a way to put probability on a solid mathematical foundation, which we will discuss in the next section. Student Learning Opportunities 1* Suppose we have a coin with heads on both sides. What is the probability that when we toss the coin we get a head? A tail? 2* Four numbers x, y, z, and w are chosen at random from the interval [0, 10]. According to the classical approach, what is the probability that x y z w?

658 Chapter 13 Data Analysis and Probability 3 (C) You pose the following problem to your class: You are given two boxes of lollipops, one has 7 greens and 3 reds. The other box has 70 greens and 30 reds. Each box is shaken and the lollipops are mixed up. You want to get a red lollipop, but you are only allowed to pick out one candy without looking. From which box would you choose? One of your students, Lenny claims that he would deﬁnitely pick from the second box since it has more red lollipops in it and he therefore has a better chance of picking a red lollipop. What is your response? Is Lenny correct? Explain. 4 (C) You told your students that you were going to ﬂip two nickels and if one landed on heads and the other on tails that you would bring in ice cream for them. Your students insisted that it was a bad deal since there was only a probability of 1/3 that this would happen. They justiﬁed it by saying that either the nickels both landed on heads, both landed on tails, or each landed differently. What is your reply? 5* A pair of dice is tossed. According to the classical approach, (a)* what is the probability of getting a sum of 7? (b) what is the probability of getting a 1 on one of the dice if the sum rolled is 7? 6 If a pair of dice is rolled, what is the probability of a double? What is the probability of rolling a sum of 11 or more? 7* By listing all possibilities and using the classical approach, ﬁnd the probability of getting exactly two heads when 3 coins are tossed simultaneously. 8* A number is picked at random from the integers 1 to 36 inclusive. (a) What is the probability that it is a perfect square? (b) What is the probability that it is divisible by 3? (c) What is the probability that it is divisible by 2 or 3 or both? Explain. 9 (C) Your student claims that she tossed a coin 20 times and got all heads. She asks you if this means her coin is unfair? How do you respond? 10* If you roll 2 dice, what is the most likely sum you will get? According to the classical approach to probability, what is the probability of getting that sum? 11* A pair of dice is thrown. What is the probability that at least one of them shows an even number? What is the probability that both show numbers less than 3? 12* Suppose we have two bags, one containing blue tiles numbered 5À9 and the other containing red tiles numbered 5À9. Suppose we pick one tile from each bag. What is the probability that their product will be odd? Explain. 13* In the game of craps seen in almost all casinos, a pair of dice is thrown by a player. If you make a “pass” bet, and a 7 or 11 comes up you win whatever you bet. If a 2, 3, or 12 comes up, you lose. All other outcomes result in no win or loss. Find the probability of your winning and the probability of a player’s losing on this type of bet. 14 (C) Comment on the following argument made by one of your students: We deﬁned the prob- ability of an event to be m where there are n equally likely events, and m of them represent n “success” with this event. But doesn’t the word “likely” mean probable? So isn’t the deﬁnition of probability circular in the sense that we are deﬁning a concept in terms of itself? 15 (C) Your students are questioning the fact that the probability of getting a head is 1/2 when a coin is tossed as is the probability of getting a tail. They point out that since there might be a

13.3 The Set Theoretic Approach to Probability 659 chance that the coin will land on its side that the probability of heads might not really be 1/2. How do you respond? 16 (C) Give two examples of where it might be better to use a frequency approach to estimate the probability of an event happening, and two examples of where it might be better to use the classical approach to estimate the probability of an event happening. 17 Explain what is wrong with each of the following: (a) A die has 5 of its faces painted red and one side painted blue. The die is tossed. The prob- ability of the die landing with a red face up is 1 since it either falls with the red face up or 2 with the blue face up. Since there are only two choices and one of them results in red, the probability of getting a red is 21. (b) There are 3 prisoners, A, B, and C in a jail and two are going to be released. A wants to know if he is one of those to be released. He ﬁgures if two are released, then they will be A and B, B and C, or A and C and thus, he has a 2/3 chance of being released. He is thinking of asking the warden to name one of the other persons who will be released, ﬁg- uring that if the warden says B then he has a 50% chance of being the other one being released. This lowers his chances from 2/3. So he decides not to ask. (c) If there is a 50% chance that something will go wrong, then by Murphy’s Law, we can expect this to happen about 90% of the time. (Murphy’s law says that what can go wrong will go wrong.) 13.3 The Set Theoretic Approach to Probability LAUNCH In a crime at a University in upstate New York, it is determined that the perpetrator is a student who was wearing Cornell sweatpants and a New York sweatshirt. A student is arrested who was wearing both of these items of clothing on the evening of the crime. The defense provides evidence that shows the probability that a randomly selected student in upstate New York is wearing Cornell sweatpants is 1/10, and the probability that a randomly selected student in Cornell is wearing a New York sweatshirt is 1/5. The prosecutor concludes that the probability that a student is wearing both Cornell sweatpants and a New York sweatshirt is (1/10)(1/5) = 1/50 = 2%, which is large enough to cause reasonable doubt for the jury. Problem: Do you think the prosecutor’s claim about the probability is true or false? In three sen- tences or less, justify your conclusions. Assume the 1/10 and 1/5 are accurate probabilities. Problems that depend on probabilistic analysis, such as that shown in the launch question, are surprisingly part of our daily lives. Very often, probabilistic statements are derived from data that have implications for such things as: criminal justice, decisions we make about how to dress

660 Chapter 13 Data Analysis and Probability each day, decisions buyers make about the number and types of items to purchase for their company to sell, and costs insurance companies charge depending on certain given conditions. To derive the probabilistic statements from the data, a set theoretic approach is used, which is described in this section. To put probability on ﬁrm mathematical ground, the mathematician Kolmogorov suggested the following set theoretic approach, which has gained large favor among mathematicians and is seen in secondary school courses. It is felt by mathematicians worldwide that any viable approach to probability must satisfy the axioms presented in this section. First, we need some background. When we perform an experiment, the set of all outcomes is known as the sample space for this experiment. We denote the sample space by the letter S. Thus, if we toss a die (our “experiment”) and want to know what number of dots will face up when the die lands, our sample space, S = {1, 2, 3, 4, 5, 6}. If we toss a pair of dice (yet another “experiment”), say a red die and a blue die, and wish to know how the dice fell, our sample space might be S = {(1, 1), (1, 2), (1, 3), . . ., (1, 6), (2, 1), (2, 2), (2, 3), . . ., (2, 6), . . ., (6, 1), (6, 2), (6, 3), . . ., (6, 6)} which consists of 36 ordered pairs of numbers where the ﬁrst number in each ordered pair is the number that faced up on the red die and the second number is the number that faced up on the blue die. Of course, if we were not interested in what came up on the individ- ual dies but were interested in the total sum of dots that came up when the dice were tossed, then our sample space would be S = {2, 3, 4, . . . 11, 12} since these are the only possible sums one can get. In summary, the sample space depends on what it is that we are interested in recording when we perform our experiment. An event, according to this theoretical approach, is deﬁned to be a subset of the sample space. Thus, if we tossed a die, and we were interested in the event, E, that a prime number came up, then E could be described by the subset E = {2, 3, 5} of the sample space S = {1, 2, 3, 4, 5, 6}. The event, F, that an even number came up could be described by the subset F = {2, 4, 6} of the sample space S = {1, 2, 3, 4, 5, 6}. Every event is made up of singleton events. For example, the event that a an even number came up on a die, that is, the event F = {2, 4, 6} is made up of the three single events {2}, {4}, and {6}. If any of these singleton events occur, F has occurred. That is, if say a 2 came up on a toss, then the event that “an even number came up” occurred. When two events, A and B, cannot occur at the same time, the events are called mutually exclusive. Thus, the event A, that tossing a single die results in an even number, and the event B, that the toss of the die results in an odd number, are mutually exclusive, since you can’t get a toss that is both an even number and an odd number at the same time. On the other hand, if we pick a card from a deck and, if we let C be the event that the card is a picture card and D be the event that the card is from the suit of diamonds, then C and D are not mutually exclusive, since drawing a jack of diamonds means that we have achieved both C and D simultaneously. It is easy to describe the words “mutually exclusive” in terms of sets. We already observed two paragraphs ago that whenever a singleton in the event set A occurs, then event A has occurred. Thus, for two events A and B to be mutually exclusive, there cannot be any singletons that occur in both, or else they would both occur. That is, A \\ B = f. Of course, if A \\ B ¼6 ϕ, then there is a common element and A and B can occur simultaneously. Thus, they would not be mutually exclu- sive. In short, two events, A and B, are mutually exclusive if and only if A \\ B = ϕ. There is more. Since any common element in both sets A and B means that A and B can occur simultaneously, it means that the set of common elements to A and B is the set of outcomes that result in both events occurring. Thus, the event that A and B occur simultaneously is described by

13.3 The Set Theoretic Approach to Probability 661 the set A \\ B. The advantage of this approach is that we can now use results about sets, and we will, to show that certain probability statements are true. In a similar manner, A [ B consists of all singletons there are in A or B or both. If we pick any such singleton in A [ B, then A or B or both events have occurred. In summary, the event that both A and B have occurred is the set A \\ B, and the event that A or B or both have occurred is the set A [ B. We now present Kolmogorov’s theoretical axioms for probability. You will notice that he inten- tionally excluded providing the deﬁnition of probability to avoid the issues discussed earlier. Thus, we can think of probability as an undeﬁned term about which we have an intuitive idea. This is no different from the agreement that we make in geometry to leave certain words like “point” and “line” undeﬁned and to work with our intuitive notion of what they are. Kolmogorov’s three axioms for probability: If E is an event and P(E) is the probability of the event occurring, then 1. 0 P(E) 1. 2. Finite Additivity: If A and B are mutually exclusive events, that is, if A \\ B = ϕ, then P(A [ B) = P(A) + P(B). (In words, if the events A and B are mutually exclusive, then the probability that A or B occurs is the sum of the probabilities that A occurs and that B occurs.) 3. P(S) = 1 where S is the sample space. (Recall the sample space is the set of all outcomes.) A note about Axiom 2: While (2) is the axiom one ordinarily sees in secondary school books, the correct version of (2) is (2)0 Countable Additivity If A1, A2, A3, and so on, are mutually exclusive events, that is, no two of them can occur at the same time, then P(A1 [A2 [ A3 . . .) = P(A1) + P(A2) + . . . . It is this rule that leads to some of the surprising results we get later on. This rule is also used quite a bit in the proofs of some of the major results in probability. Axioms (1)–(3) were motivated by what seemed to be logical consequences of the frequency and classical deﬁnitions. Let us illustrate with the classical deﬁnition of probability. Since probabil- ities were measured as the ratio of times that “success” occurred to the total number of times we performed these experiments, and these ratios were between 0 and 1, the probability of an event should be between 0 and 1. Hence, Axiom (1) holds. Let us illustrate Axiom (2). Suppose we toss a die and we let A be the event when we get a number of dots facing up which is a perfect square, and B be the event when we get a number which is prime. The event A occurs when the die falls on either 1 or 4. Thus, “success” occurs 2 times out of our 6 possibilities. Thus, PðAÞ ¼ 2/6. The event B occurs when either a 2, 3, or 5 shows up. Thus, we have success with 3 outcomes out of the six possible outcomes. So PðBÞ ¼ 36. The events A and B are mutually exclu- sive since they can’t occur at the same time. Finally, the event that A or B occurs can happen when we have any of the following outcomes: 1, 4, 2, 3, 5. So the outcomes associated with the event A or B occurs 5 out of 6 times. Thus, PðA or BÞ ¼ 56. We see that P(A or B) = P(A) + P(B), and it is exam- ples like this that motivated Axiom 2. Using the frequency approach, we can also deduce rule 2. Here is how. Suppose that A and B are any two mutually exclusive events and that we perform an experiment n times and that A occurs mA times and B occurs mB times. Then the event A or B occurs mA + mB times out of the n times.

662 Chapter 13 Data Analysis and Probability Now P(A) is deﬁned by PðAÞ ¼ lim number of successes ¼ lim mA ð13:2Þ n!1 total number of trials n!1 n and P(B) is deﬁned by PðBÞ ¼ lim number of successes ¼ lim mB : ð13:3Þ n n!1 total number of trials n!1 Finally, P(A or B) is deﬁned by PðA [ BÞ ¼ lim number of successes ¼ lim mA þ mB : ð13:4Þ total number of trials n n!1 n!1 (The number of successes is mA + mB, since the events A and B have no overlap.) But (assuming all the limits exist) this last limit is the same is nli!m1mnA þ mB ð13:5Þ n ¼ lim mA þ lim mB ð13:6Þ n n n!1 n!1 ¼ PðAÞ þ PðBÞ: ð13:7Þ From (13.4)À(13.7) we get that PðA [ BÞ ¼ PðAÞ þ PðBÞ: Since S represents the set of all outcomes, Axiom (3) is saying that the probability that some outcome occurs in an experiment is 1. Of course, this seems like an obvious axiom to include. 13.3.1 Some Elementary Results in Probability Suppose we toss a die and A is the event that an odd number comes up. Then the complementary event is the event that occurs when A does not happen (i.e., when an odd number doesn’t come up). Thus, the complementary event in this case is the event A0 that an even number comes up. If B is the event that a prime number comes up, then B0 is the event that a prime number doesn’t come up. If E is any event, then the complementary event, E0 and E are mutually exclusive, that is, cannot occur at the same time, and E [ E0 = S since every outcome in S, the sample space, is either in E or not in E (making it in E0). From this it follows that: Theorem 13.4 If E is an event and E0 is the complementary event, then P(E0) = 1 À P(E). Proof. Since E [ E0 = S, it follows from Axiom (3) that PðE [ E0Þ ¼ PðSÞ ¼ 1: But, from Axiom (2), since E and E0 are mutually exclusive, this simpliﬁes to PðEÞ þ PðE0 Þ ¼ 1 Solving for P(E0) in this equation, we get P(E0) = 1 À P(E). &

13.3 The Set Theoretic Approach to Probability 663 Thus, if the probability of winning a contest is 34, then the probability of not winning the contest is 1 À 3 or 41. If the probability of getting a disease is 0.1, then the probability of not 4 getting the disease is 1 À 0.01, or 0.9. If A and B are events, then we denote by A À A \\ B the event that A occurs, but not both A and B occur. We need the following: Theorem 13.5 P(A À A \\ B) = P(A)À P(A \\ B). Proof. A = (A À A \\ B) [ (A \\ B), as the Venn diagram in Figure 13.2 shows. AB A– A ∩ B A∩B Figure 13.2 Thus, PðAÞ ¼ PððA À A \\ BÞ [ ðA \\ BÞÞ ¼ PðA À A \\ BÞ þ PðA \\ BÞ; since (A À A \\ B) and (A \\ B) are mutually exclusive. Subtracting P(A \\ B) from both sides, we get that PðAÞ À PðA \\ BÞ ¼ PðA À A \\ BÞ which we can rewrite as P(A À A \\ B) = P(A) À P (A \\ B) by just switching the entries on both sides. And this, is what we were trying to prove. & There is a corollary of this that we will use in the section on geometric probability that will lead to a surprising result. That corollary in words says that, if whenever the event C occurs, the event A occurs, A0s probability of occurring is greater than or equal to that of C occurring. This makes perfect sense. Here it is in symbols. Corollary 13.6 If C is a subset of A, then P(C) P(A). Proof. If C is a subset of A, then A À C = A À A \\ C since A \\ C, would be the same as C. (Draw a Venn diagram to convince yourself.) So by the theorem, PðA À CÞ ¼ PðAÞ À PðA \\ CÞ ¼ PðAÞ À PðCÞ: Since by Axiom (1) P(AÀC) ! 0, we have that P(A)ÀP(C) ! 0 or P(C) P(A). & This brings us to an important result.

664 Chapter 13 Data Analysis and Probability Theorem 13.7 If E and F are two (not necessarily exclusive) events, then P(E [ F) = P(E) + P(F) À P(E \\ F). Proof. E [ F = (E À E \\ F) [ F. (See the Venn diagram in Figure 13.3.) EF The shaded area is E – E ∩ F Figure 13.3 Thus, PðE [ FÞ ¼ PðE À E \\ FÞ þ PðFÞ ¼ PðEÞ À PðE \\ FÞ þ PðFÞ ðby the last theorem and since E À E \\ F and F are mutually exclusive:Þ ¼ PðEÞ þ PðFÞ À PðE \\ FÞ ðby rearranging the last lineÞ: & Here is a typical secondary school problem that makes use of this theorem. Example 13.8 A card is picked from an ordinary deck of playing cards. What is the probability that the card picked is a face card or a diamond? Solution: There are 12 face cards and 13 diamonds, and 3 cards which are both face cards and diamonds (jack, queen, and king of diamonds). So, if F is the event of obtaining a face card, and D is the event of obtaining a diamond, then P(F or D) = P(F [ D) ¼ PðFÞ þ PðDÞ À PðF \\ DÞ ¼ 12 þ 13 À 3 ¼ 5222. 52 52 52 Here is another example. Example 13.9 In the small town of Cherokee, there is an 80% chance that a household subscribes to the local newspaper, a 50% chance that a household will subscribe to the metropolitan newspaper from a nearby town, and a 40% chance a household subscribes to both. If a household is selected at random from this town: (a) What is the probability that it will subscribe to at least one of the two newspapers? (b) What is the probability that it will subscribe to exactly one of the two newspapers? Solution: If L stands for the event that a household subscribes to the local paper and M is the event that a person subscribes to the metropolitan paper, then we are given that P(L) = 0.80, P(M) = 0.50, and that P(L \\ M) = 0.40. In part (a) we seek P(L [ M). We know that PðL [ MÞ ¼ PðLÞ þ PðMÞ À PðL \\ MÞ ¼ 0:80 þ 0:50 À 0:40 ¼ 0:90:

13.3 The Set Theoretic Approach to Probability 665 To solve part (b) we ﬁrst notice that the event, L [ M, means that a household subscribes to at least one paper. This can be represented by, the union of the event, E, that it subscribed to exactly one paper, and the event L \\ M, that it subscribe to both papers. And since E and L \\ M are mutually exclusive, we have that PðL [ MÞ ¼ PðEÞ þ PðL \\ MÞ: Substituting the values we know in this equation, we get 0:9 ¼ PðEÞ þ 0:4: From which it follows that P(E), the probability that a household subscribes to exactly one paper, is 0.5. Student Learning Opportunities 1 (C) You asked your students whether they thought there was a higher percentage of people who were 55 or older and had had a heart attack or people who only had had a heart attack. They chose the ﬁrst group. Were they correct? Why or why not? 2 (C) You ask one of your students, Julie, to come to the front of the room and randomly pick a card from an ordinary deck of playing cards. Now, you ask the class what the probability is that Julie picked a card that is either red or has an even number (10 or less). Ricky gives the follow- ing answer: Since half the deck has red cards, the probability that it is red is 2562. Also, since there are only 5 possible even numbers in the deck (2, 4, 6, 8, 10) and there are 4 suits, that makes 4 times 5, or 20 cards with even numbers. Therefore, the probability of picking a card with an even number is 5220. So, the probability is 26 þ 20 ¼ 4562. Is Ricky correct? Why or why not? 52 52 3 A construction ﬁrm has bid on two projects. The probability of getting the ﬁrst project is 0.35 and the probability of getting the second project is 0.40. The probability that the construction ﬁrm gets both projects is 0.22. Find: (a)* The probability that the construction ﬁrm gets one or the other or both projects. (b) The probability that the construction ﬁrm doesn’t get the ﬁrst project. (c) The probability that the construction ﬁrm gets the ﬁrst project but not both. (d)* The probability that the construction ﬁrm gets neither project. (e) Which of the 4 probabilities is the greatest, and why does this make sense? 4 Ed, the painter, has two paint jobs for this week that must be completed. He estimates that there is a 95% chance he can ﬁnish the ﬁrst job on time, a 70% chance he can ﬁnish the second job on time, and a 99% chance that he can ﬁnish one or the other on time. Assuming that these subjective probabilities are correct, and that the Kolmogorov axioms apply, what is the probability that: (a) He will ﬁnish both on time? (b) He will ﬁnish neither on time? (c) He will ﬁnish just one of the projects on time? (d) Which of the 3 probabilities is the greatest, and why does this make sense?

666 Chapter 13 Data Analysis and Probability 5 Show that if the sample space in an experiment is S = {a1, a2, a3, . . .,an}, that P({a1}) + P({a2}) + . . . + P({an}) = 1. Show also that it follows from this that if all the singleton events are equally likely, that each has probability 1 and that any event with m elements in it has probability mn . n Explain why this means that the Kolmogorov axioms imply the classical approach. 6 (C) One of your students has calculated the probability of three mutually exclusive events A, B, and C. She has arrived at the solution that P(A) = 0.5 and P(B) = 0.3 and P(C) = 0.4. She asks you if she is correct. Can you tell? How? 7* An unfair die has the probability that P(1) = P(2) = x, that P(3) = P(4) = 2x, and that Pð5Þ ¼ Pð6Þ ¼ 2x. Find x and then ﬁnd the probability that a number divisible by 3 comes up on a roll of this particular die. 8 Show, using a Venn diagram, that if E and F are sets, then E À F (the set of those things in E but not in F) is the same as E À E \\ F. From this result conclude that P(E À F) = P(E) À P(E \\ F). 9 In the newspaper example, 13.9, show that the event E can be described as (L À L \\ M) [ (M À L \\ M). Use this and the theorems in this section to compute P(E) another way than was shown in the example. 10 Show that if A = ϕ (where ϕ is the empty set) then P(A) = 0. [Hint: A = A [ ϕ.] 11 (C) We indicated that it is reasonable to assume that the probability of getting heads facing up when a coin is tossed is taken as 1/2, since it has been shown experimentally many times that when a coin is, in fact, tossed repeatedly, the ratio of the number of heads to the total number of ﬂips is close to 50%. But, your astute student argues, “Well, 49.98% is close to 50% and if we ﬂipped a coin millions and millions of times, we really don’t expect that we will get exactly half of the ﬂips heads, so how do we know that the probability of getting heads isn’t 49.98%? Furthermore, if we take the probability to be 50% when in fact it is less, aren’t we opening our- selves up to accumulated error if we compute using a wrong probability?” What is your answer? 13.4 Elementary Counting LAUNCH You have rented a room for a family birthday party that you are hosting in which there are 20 people in attendance (including you). You have been told that you must vacate the room at exactly 3 PM since the room will be needed at that time for another event. Knowing that each of your family members must hug each other goodbye, and knowing that it takes each family member approxi- mately 6 seconds per hug, at approximately what time should you announce to your family members that they must begin their good-bye hugs?

13.4 Elementary Counting 667 One of the ﬁrst things you learned how to do as a very young child was to count. However, as you discovered in secondary school, simple counting is often not sufﬁcient when it comes to prob- lems such as the one you encountered in the launch. For problems such as this, straightforward counting takes far too long, so short cuts are needed that require a far more sophisticated approach. Furthermore, since probability problems often require knowing the number of elements in a sample space, accurate counting methods are essential. That is why such topics as permutations and combinations are included in the secondary school curriculum. We begin with the counting principle after a small introductory example. Example 13.10 Let us return to the family birthday party presented in the launch. Suppose that each child had a choice of picking one entrée, one side dish, and one dessert. There are 4 entrees (pizza (P), hamburger (H), chicken ﬁngers (C), Linguini (L)), 2 side dishes (French fries (F), salad (S)), and 3 des- serts (birthday cake (B), ice cream (I), melon (M)). How many different meals are possible consisting of one entrée, one side dish, and one dessert? Solution: Solving this problem involves a lot of counting and clearly requires some orga- nized thought. Using the letters given, one might create an organized list that looks like the fol- lowing. (Note that the ﬁrst letter is the entrée, the second is the side dish, and the third is the dessert.) Thus, PFB, for example, means a meal consisiting of pizza, french fries, and birthday cake. PFB, PFI, PFM PSB, PSI, PSM HFB, HFI, HFM HSB, HSI, HSM CFB, CFI, CFM CSB, CSI, CSM LFB, LFI, LFM LSB, LSI, LSM Á ÁAs we can see, this list shows that there are 24 possible meals. Also, notice that 24 = 4 2 3. That is, the answer is the product of the number of choices for entrees, the number of choices for sides, and the number of choices for desserts. Another way of representing this situation is with a tree diagram. In Figure 13.4, we have drawn a tree diagram.

668 Chapter 13 Data Analysis and Probability F B I P M Start S B F H I S M B F I C M S B F I L M S B Figure 13.4 I M B I M B I M B I M Each path from the start represents a meal. Thus, the path, Start-P-F-B, represents the meal con- sisting of pizza, french fries, and birthday cake The tree diagram clearly shows the 24 possible Á Ámeals, and also seems to indicate why the answer for the number of meals is 4 2 3, since there are 4 choices for the entree, 2 for the sides, and 3 for the desserts, resulting in 24 paths. Although the organized list and tree diagrams are effective strategies to solve this problem, if the numbers were larger, these methods would be too cumbersome, to say the least. Listing the choices would be frustrating, and drawing the tree diagram would be impossible. Imagine if, as in most diners, there were 30 entrees, 15 sides, and 25 desserts to choose from. Then what? Our analysis of the tree diagram and the listing both seem to indicate that we would expect to get Á Á30 15 25, or 11, 250 possible different meals. This example illustrates a very powerful principle that we discuss next, known as the counting principle. Counting principle: Suppose we have k tasks T1, T2, T3, . . ., Tk.. If a task T1 can be done in n1 ways, and once T1 is done T2 can be done in n2 ways, and once these are done T3 can be done in n3 ways, and so on, then the number of ways of performing the tasks T1, T2, . . ., Tk in succession is given by n1n2 . . . nk.

13.4 Elementary Counting 669 So, in the preceding example, we can let task 1 be the task of choosing an entree, task 2 be the task of choosing a side dish, and task 3 be the task of choosing a dessert. Then task 1 can be done in n1 = 4 ways, task 2 can be done in n2 = 2 ways, and task 3 can be done in n3 = 3 ways. Thus, the total Á Ánumber of ways to perform tasks 1, 2, and 3, that is, the number of ways to form a meal, is 4 2 3. Example 13.11 How many code words can be formed from two letters of the alphabet followed by a single digit, if the letters chosen must be different? Solution: Let task 1 be the task of choosing the ﬁrst letter, task 2 the task of choosing the second letter, and task 3 the task of choosing the single digit. Task 1 can be done in any of 26 ways since we have 26 letters of our alphabet to choose from. Task 2 can then be done in 25 ways since we have 25 letters to choose from, since we have already chosen one letter. Finally, task 3 can be done in any of Á Á10 ways since we can pick any digit from 0 to 9. Thus, the number of code words is 26 25 10, or 6500. Example 13.12 A photographer wants to take a picture of 5 children in a family. She lines up 5 chairs in a row in which the children will sit. (a) How many different seating arrangements can she make with the 5 children? (b) Next the photographer wants to take some pictures using only 3 of the 5 children. So she removes two of the chairs. How many different seating arrangements can she make using only 3 of the 5 children? Solution (a): Let T1 be the task of putting a child in the ﬁrst seat. Since the photographer has 5 children from which to choose, this task can be done in 5 ways. Let T2 be the task of ﬁlling the second seat. Since 4 children are left to be placed in the second seat, this task can be done in 4 ways. Letting T3, T4, and T5 represent the tasks of ﬁlling the third, fourth, and ﬁfth seat respectively, these tasks can be done in 3, 2, and 1 way respectively. Thus, by the counting principle, we get that Á Á Á Áthe total number of seating arrangements is 5 4 3 2 1, or 120. (b) Now we have only 3 seats. Proceeding as we did earlier, we let T1, T2, and T3 represent the tasks of ﬁlling the ﬁrst, second, and third seat. These can be done in 5, 4, and 3 ways respectively, so Á Áby the counting principle, now the number of seating arrangements is 5 4 3, or 60. An arrangement of objects in which the position counts is called a permutation. In part (a) of the example, we asked for the number of different seating arrangements of the 5 children. Each dif- ferent seating arrangement of the children represents a different permutation of the children in the seats. In general, when we arrange r distinguishable objects taken from a set of n objects, the number of permutations we get is known as nPr. Using the counting principle and arguing as earlier, we can compute nPr as follows: nPr ¼ nðn À 1Þðn À 2Þ . . . :ðn À r þ 1Þ: ð13:8Þ Thus, in part (a) of the example, we were arranging r = 5 children from a set of n = 5 children in Á Á Á Áseats, and this can be done in 5P5 ways. By equation (13.8) this evaluates to 5 4 3 2 1, which as most of you probably know is abbreviated as 5! (ﬁve factorial). Notice that the ﬁnal 1 in this product is n À r + 1, since n and r are both 5. In part (b) of the example, we were arranging r = 3 children taken from a set of n = 5 children. This can be done in 5P3 ways. Again, by equation (13.8), this evaluates to Á Á5 4 3 ways. Notice the ﬁnal 3 in the product is n À r + 1, since n = 5 and r = 3.

670 Chapter 13 Data Analysis and Probability Another, more concise way of representing nPr is nPr ¼ ðn n! rÞ! : ð13:9Þ À Thus, in part (b) of our example, the number of different seating arrangements that can be made using only 3 of the 5 children could have been computed as 5P3 ¼ ð5 5! À 3Þ! 5! 5Á4Á3Á2Á1 2! 2Á1 Á Áor¼ or 5 4 3, which is the same number of arrangements that we got before. Let us revisit the previous problem with a slight change. Suppose that from the 5 children we just want to choose 3 of them to go shopping with. How is this different from the previous example? Here, we are only looking for the number of different groupings of 3 people we get. But, the order in which we pick these 3 children to shop with us is irrelevant. It is the same 3 people regardless of who we choose ﬁrst, who we choose second and who we choose third. But if we were photograph- ing them, the position in which we photograph them would determine the picture we get There would be 3! = 6 possible permutations (orders) that we could photograph them in (from left to right). (For example if the people chosen were Maria (M), Jose (J) and Anna (A) these six orders would be MJA, MAJ, JMA, MJA, AJM and AMJ.) This is true for each set of 3 children selected from the 5 children. So in this case, the number of ways we can select 3 children to go shopping from 5 children is 1/6 the number of permutations of this group of children, or, 53P!3. This new way of counting, where position is irrelevant, is an example of combinations. The same idea works for choosing r items from n items where the order is irrelevant. Namely, if we have n objects, and we wish to choose a set of r of them, the number of sets of Àr Áobjects, where the position of the r objects is irrelevant, is given by nCr, alternately denoted by n , where either one of these means r n n! r ¼ nCr ¼ nPr ¼ ðn À rÞ! ¼ n! : r! r! r!ðn À rÞ! Since nCr ¼ nPr we see that nCr is always less than or equal to nPr. Here are some typical sec- r! , ondary school problems: Example 13.13 The QC Mathletes is a club with 10 students. (a) Suppose we wish to choose 4 Mathletes to ﬁll the positions of president, vice president, secretary, and treasurer. In how many ways can this be done? Is this a permutation or a combination? Why? (b) Suppose we wish to choose 4 of them to compete against Franklin High School in a math com- petition. How many such sets of 4 people can we choose? Is this a permutation or a combination? How does this differ from part (a)? Solution: (a) In this part we are assigning people to positions. How we assign them matters. So, Jack being president and Jill being vice president, is different from Jill being president and Jack being vice president. Therefore, this is a permutation question, and we can use the formula (13.8) to compute this, where n = 10 and r = 4. We get 10P4 = (10)(9) . . . (10 À 4 + 1) = (10)(9) (8)(7), or we can use formula (13.9) and get 10! ¼ 10! ¼ 10 Á 9 Á 8 Á ::: Á 1 ¼ 10 Á 9 Á 8 Á 7 ¼ ð10 À 4Þ! 6! 6 Á 5 Á 4 Á ::: Á 1 5040 ways. Wow! That is a lot of ways to assign the four positions to students.

13.4 Elementary Counting 671 (b) In this problem there is a slight change from the previous one in that now, we do not care about the order in which we select the 4 people, we just want 4 people. So this is a combination problem where we wish to choose 4 people from 10. This can be done in ÀÁ ¼ 10! 4Þ! ¼ 10! ¼ 10 4!ð10 À 4!6! 4 10 Á 9 Á 8 Á 7 ¼ 210 ways. This is a lot less than the number of permutations. 1Á2Á3Á4 Example 13.14 You visit an antique dealer and decide to buy 5 of the 14 antiques available. Exactly 2 of the 14 items are a pair, and the dealer will not sell them separately. So you either buy them both, or buy neither. How many sets of 5 antiques can you buy? Solution: You either buy the pair that must be sold together or you don’t. If you buy the pair, then you need to choose 3 more antiques from the remaining 12 antiques. If you don’t buy the pair, then all 5 of your antiques must be chosen from the 12 antiques remaining, not counting the pair. Thus, the total number of sets of 5 antiques you can buy is 12 12 3 þ 5 ¼ 1012: Example 13.15 We wish to choose a team consisting of 5 boys and 3 girls. The boys are to be chosen from a set of 10 boys and the girls from a set of 6. (a) How many such teams can we form? What is the probability that if we choose a team of 8 people at random from this group of 10 boys and 6 girls that the team will consist of 5 boys and 3 girls? Solution: (a) Let task 1 be thÀe Átask of choosing the boys and task 2 the task of choÀosÁing the girls. Then task 1 can be done in 10 or 252 ways, and task two can be done in any of 6 or 20 ways. 5 3 Thus, the number of teams we can choose consisting of 5 boys and 3 girls is therefore 10 6 5 Á 3 ¼ 5040; and that is a lot of teams! À Á (b) The number of teams consisting of 8 people chosen from the 10 boys and 6 girls is 16 ¼ 12; 870. Thus, the probability of getting a team consisting of 5 boys and 3 girls when a 8 team of 8 is drawn at random is 5040 ¼ 0:39161, which is not very likely. 12 870 Student Learning Opportunities 1 (a)* How many license plates can one form using the digits from 1 to 9, if the license plates only consist of six digits and the digits can be repeated? (b)* How many license plates can be formed from any three letters followed by any three single digits from 1 to 9 if repetition is allowed?

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Dina Widiastuti

The Mathematics That Every Secondary School Math Teacher Needs to Know

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The Mathematics That Every Secondary School Math Teacher Needs to Know

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