Trigonometry lecture notes

lecture notes on spherical trigonometry and lecture notes in plane trigonometry. how trigonometry is used in architecture, how does trigonometry apply to engineering
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Chapter 0 Trigonometry f. Gr. r p y wvo- v triangle + -tH pa measurement. - Oxford English Dictionary In this chapter we will look at some results in geometry that set the stage for a study of trigonometry. 1 What is new about trigonometry? Two of the most basic figures studied in geometry are the triangle and the circle. Tngonometry will tell us more than we learned in geometry about each of these figures. For example, in geometry we learn that if we know the lengths of the three sides of a triangle, then the measures of its angles are completely 1 determined (and, in fact, almost everything else about the triangle is de­ termined). But, except for a few very special triangles, geometry does not tell us how to compute the measures of the angles, given the measures of the sides. Example 1 The measures of the sides of a triangle are 6, 6, and 6 cen­ timeters. What are the measures of its angles? 1 It is sometimes said that the lengths of three sides determine a triangle, but one must be careful in thinking this way. Given three arbitrary lengths, one may or may not be able to form a triangle (they form a triangle if and only if the sum of any two of them is greater than the third). But if one can form a triangle, then the angles of that triangle are indeed determined. 2 Trigonometry Solution. The triangle has three equal sides, so its three angles are also equal. Since the sum of the angles is 180", the degree-measure of each angle is 180/3 = 60°. Geometry allows us to know this without actually measuring the angles, or even drawing the triangle. o Example 2 The measures of the sides of a triangle are 5, 6, and 7 cen­ timeters. What are the measures of its angles? Solution. We cannot find these angle measures using geometry. The best we can do is to draw the triangle, and measure the angles with a pro­ tractor. But how will we know how accurately we have measured? We will answer this question in Chapter 3. o Example 3 Two sides of a triangle have length 3 and 4 centimeters, and the angle between them is 90°. What are the measures of the third side, and of the other two angles? Solution. Geometry tells us that if we know two sides and an included angle of a triangle, then we ought to be able to find the rest of its measure­ ments. In this case, we can use the Pythagorean Theorem (see page 7) to tell us that the third side of the triangle has measure 5. But geometry will not tell us the measures of the angles. We will learn how to find them in Chapter2. o Exercise Using a protractor, measure the angles of the triangle below as · accurately as you can. Do your measurements add up to 180°? Let us now turn our attention to circles. Example 4 In a certain circle, a central angle of 20° cuts off an arc that is 5 inches long. In the same circle, how long is the arc cut off by a central angle of 40°? 1. Whut is new uhout tri,Zollomctty' 3 Solution. We can divide the 40°angle into two angles of 20°. Each of these angles cuts off an arc of length 5", so the arc cut off by the 40°angle is 5 + 5 = 10 inches long. ? 5" That is, if we double the central angle, we also double the length of the arc it intercepts. 0 Example 5 In a certain circle, a central angle of 20° determines a chord that is 7 inches long. In the same circle, how long is the chord determined by a central angle of 40°? Solution. As with Example 4, we can try to divide the 40° angle into two 20° angles: ? 7" However, it is not so easy to relate the length of the chord determined by the 40° angle to the lengths of the chords of the 20° angles. Having doubled the angle, we certainly have not doubled the chord. o Exercises 1. In a circle, suppose we draw any central angle at all, then draw a second central angle which is larger than the first. Will the arc of the second central angle always be longer than the arc of the first? Will the chord of the second central angle also be larger than the chord of the first? 2. What theorem in geometry guarantees us that the chord of a 40° of a 20° angle? angle is less than double the chord 4 Trigonometry 3. Suppose we draw any central angle, then double it. Will the chord of the double angle always be less than twice the chord of the original central angle? Trigonometry and geometry tell us that any two equal arcs in the same circle have equal chords; that is, if we know the measurement of the arc, then the length of the chord is determined. But, except in special circum­ stances, geometry does not give us enough tools to calculate the length of the chord knowing the measure of the arc. Example 6 In a circle of radius 7, how long is the chord of an arc of 90°? Solution. If we draw radii to the endpoints of the chord we need, we will have an isosceles right triangle: Then we can use the Pythagorean Theorem to find the length of the chord. 2 2 2 If this length is x, then 7 + 7 = x , so that x = .J98 = 7 ../i. o · Example 7 In a circle of radius 7, how long is the chord of an arc of 38°? Solution. Geometry does not give us the tools to solve this problem. We can draw a triangle, as we did in Example 6: But we cannot find the third side of this triangle using only geometry. How­ ever, this example does illustrate the close connection between meaure­ ments in a triangle and measurements in a circle. o 1. What is new about tri,go11onwt1y' 5 Exercises 1. What theorem from geometry guarantees us that the triangle in the diagram for Example 7 is completely determined? 2. Note that the triangle in Example 7 is isosceles. Calculate the mea­ sure of the two missing angles. Trigonometry will help us solve all these kinds of problems. However, trigonometry is more than just an extension of geometry. Applications of trigonometry abound in many branches of science. Example 8 Look at any pendulum as it swings. If you look closely, you will see that the weight travels very slowly at either end of its path, and picks up speed as it gets towards the middle. It travels fastest during the middle of its journey. o Example 9 The graph below shows the time of sunrise (corrected for daylight savings) at a certain latitude for Wednesdays in the year 1995. The data points have been joined by a smooth curve to make a continuous graph over the entire year. 7.001 i 5.00-1 '.) . . § T ..... I 3.oo T 1.00 4+-++-+-++ 0 50 100 250 300 350 150 200 Jan Mar Jun Sep Dec We expect this curve to be essentially the same year after year. How­ ever, neither geometry nor algebra can give us a formula for this curve. In Chapter 8 we will show how trigonometry allows us to describe it mathe­ matically. Trigonometry allows us to investigate any periodic phenomenon - any physical motion or change that repeats itself. o 6 Trigonomerry 2 Right triangles We will start our study of trigonometry with triangles, and for a while we will consider only right triangles. Once we have understood right triangles, we will know a lot about other triangles as well. Suppose you wanted to use e-mail to describe a triangle to your friend in another city. You know from geometry that this usually requires three pieces of information (three sides; two sides and the included angle; and so on). For a right triangle, we need only two pieces of information, since we already know that one angle measures 90°. In choosing our two pieces of information, we must include at least one side, so there are four cases to discuss: a) the lengths of the two legs; b) the lengths of one leg and the hypotenuse; c) the length of one leg and the measure of one acute angle; d) the length of the hypoteluse and the measure of one acute angl. (b) (a) ___________ _d (d) Suppose we want to know the lengths of all the sides of the triangle. For cases (a) and (b) we need only algebra and geometry. For cases (c) and (d), however, algebraic expressions do not (usually) suffice. These cases will introduce us to trigonometry, in Chapter 1. 3. The Pythagorean thcorr.m 7 3 The Pythagorean theorem We look first at the chief geometric tool which allows us to solve cases (a) and (b) above. This tool is the famous Pythagorean Theorem. We can separate the Pythagorean theorem into two statements: Statement 1: If a and b are the lengths of the legs of a right triangle, and c 2 2 2 is the length of its hypotenuse, then a + b = c . 2 2 2 Statement II: If the positive numbers a, b, and c satisfy a + b = c , then a triangle with these side lengths has a right angle opposite the side with 2 length c. These two statements are converses of each other. They look similar, but a careful reading will show that they say completely different things about triangles. In the first statement, we know something about an angle of a triangle (that it is a right angle) and can conclude that a certain relation­ ship holds among the sides. In the second statement, we know something about the sides of the triangle, and conclude something about the angles (that one of them is a right angle). The Pythagorean theorem will allow us to reconstruct a triangle, given two legs or a leg and the hypotenuse. This is because we can find, using this information, the lengths of all three sides of the triangle. As we know from geometry, this completely determines the triangle. Example 10 In the English university town of Oxford, there are some­ times lawns occupying rectangular lots near the intersection of two roads (see diagram). B 9 meters A 12 meters c 2 In fact, we can make a stronger statement than statement II: 2 2 Statement n': If the positive numbers a, b, and c satisfy a + b2 = c , then there exists a triangle with sides a, b, and c, and this triangle has a right angle opposite the side with length c. 2 2 2 This statement includes, for example, the fact that if a + b = c , then a+ b c. Trigonometry In such cases, professors (as well as small animals) are allowed to cut across the lawn, while students must walk around it. If the dimensions of the lawn are as shown in the diagram, how much further must the students walk than the professors in going from point A to point B? Solution. Triangle ABC is a right triangle, so statement I of the Pythag­ orean theorem applies: 2 2 2 AB = AC + BC = 122 + 92 = 144 + 81 = 225 . So AB = 15 meters, which is how far the professor walks. On the other hand, the students must walk the distance A(; + C B = 12 + 9 = 21. This is 6 meters longer than the professor's walk, or 40% longer. o Example 11 Show that a triangle with sides 3, 4, and 5 is a right tnangle Solution. We can apply statement II to see if it is a right triangle. In 2 2 2 fact, 5 = 25 = 3 + 4 , so the angle opposite the side of length 5 is a right angle. Notice that we cannot use statement I of the Pythagorean theorem to solve this problem. o Exercises The following exercises concern the Pythagorean theorem. In solving each problem, be sure you understand which of the two statements of this theorem you are using. 1. Two legs of a right triangle measure 10 and 24 units. Find the length of the hypotenuse in the same units. 2. The hypotenuse of a right triangle has length 41 units, and one leg measures 9 units. Find the measure of the other leg. 3. Show that a triangle with sides 5, 12, and 13 is a right triangle. 4. One leg of a right triangle has length 1 unit, and the hypotenuse has length 3 units. What is the length of the other leg of the triangle? 5. The hypotenuse of an isosceles right triangle has length 1. Find the length of one of the legs of this triangle. 4. Our best li"icnds (lllllon,g rigllt triunglcs) 9 6. In a right triangle with a 30oangle, the hypotenuse has length 1. Find the lengths of the other two legs. Hint: Look at the diagram in the footnote on page 11. 7. Two points, A and B, are given in the plane. Describe the set of 2 2 2 points X such that AX + BX = AB . (Answer: A circle with its center at the midpoint of AB.) 8. Two points, A and B, are given in the plane. Describe the set of 2 2 points for which AX - BX is constant. 4 Our best friends (among right triangles) There are a few right triangles which have a very pleasant property: their sides are all integers. We have already met the nicest of all (because its sides are small integers): the triangle with sides 3 units, 4 units and 5 units. But there are others. Exercises 1. Show that a triangle with sides 6, 8, and 10 units is a right triangle. 2. Look at the exercises to section 3. These exercises use three more right triangles, all of whose sides are integers. Make a list of them. (Later, in Chapter 7, we will discover a way to find many more such right triangles.) 3. The legs of a right triangle are 8 and 15 units. Find the length of the hypotenuse. 4. We have used right triangles with the following sides: II Leg I Leg I Hypotenuse II 4 5 3 6 8 10 9 12 15 By continuing this pattern, find three more right triangles with inte­ ger sides. 10 'J'rigonomet1y 5. We have seen that a triangle with sides 5, 12, and 13 is a right tri­ angle. Can you find a right triangle, with integer sides, whose short­ est side has length 10? length 15? 6. Exercises 4 and 5 suggest that we can construct one integer-side right triangle from another by multiplying each side by the same number (since the new triangle is similar to the old, it is still a right tri­ angle). We can also reverse the process, dividing each side by the same number. Although we won't always get integers, we will al­ ways get rational numbers. Show that a triangle with sides 3/5, 4/5, and 1 is a right triangle. 7. Using the technique from Exercise 6, start with a 3-4-5 triangle and find a right triangle with rational sides whose shorter leg' is 1. Then find a right triangle whose longer leg is 1. 8. Start with a 5-12-13 right triangle, and find a right triangle with ra­ tional sides whose hypotenuse is 1. Then find one whose shorter leg is 1. Finally, find a right triangle whose longer leg is 1. 9. Note that the right triangles with sides equal to 5, 12, 13 and 9, 12, 15 both have a leg equal to 12. Using this fact, find the area of a triangle with sides 13, 14, and 15. 10. (a) Find the area of a triangle with sides 25, 39, 56 (b) Find the area of a triangle with sides 25, 39, 16. 5 Our next best friends (among right triangles) In the previous section, we explored right triangles with nice sides. We will now look at some triangles which have nice angles. For example, the two acute angles of the right triangle might be equal. Then the triangle is isosceles, and its acute angles are each 45o. Or, we could take one acute angle to be double the other. Then the triangle has acute angles of 30 and 60°. But nobody is perfect. It turns out that the triangles with nice angles never have nice sides. For example, in the case of the 45° right triangle, we have two equal legs, and a hypotenuse that is longer: ·5. Our next best friends (llmong right triungles) II 1 If we suppose the legs are each 1 unit long, then the hypotenuse, measured in the same units, is about 1.414213562373 units long, not a very nice number. For a 30°right triangle, if the shorter leg is 1, the hypotenuse is a nice 3 length : it is 2. But the longer leg is not a nice length. It is approximately 1.732 (you can remember this number because its digits form the year in which George Washington was born- and the composer Joseph Haydn). It also turns out that triangles with nice sides never have nice angles. If we want an example of some theorem or definition, we will look at how the statement applies to our friendly triangles. Exercises 1. Find the length of each leg of an isosceles right triangle whose hy­ potenuse has length 1. Challenge: Find the length, correct to nine decimal places without using your calculator (but using information contained. in the text above). 2. Using the Pythagorean theorem, find the hypotenuse of an isosceles right triangle whose legs are each three units long. 3 If you don't remember the proof, just take two copies of such a triangle, and place them back-to-back: You will find that they form an equilateral triangle. The side opposite the 30°angle is half of one side of this equilateral triangle, and therefore half of the hypotenuse. 12 'J'rigonometly 3. The shorter leg of a 30-60-90°triangle is 5 units. Using the Pythago­ rean theorem (and the facts about a 30°-60°-90° triangle referred to above), find the lengths of the other two sides of the triangle. 4. In each of the diagrams below, find the value of x and y: X (c) 1 1 X X y A X X 4 X (e) (f) y p c B s AC==BC PQRS is a square 6 Some standard notation A triangle has six elements ("parts"): three sides and three angles. We will agree to use capital letters, or small Greek letters, to denote the measures of the angles of the triangle (the same letters with which we denote the vertices of the angles). To denote the lengths of the sides of the triangle, we will use the small letter corresponding to the name of the angle opposite this side. Some examples are given below: A A B b y B a c c Appendix I. C/LJssif'ying triungles 13 Appendix I. Classifying triangles Because the angles of any triangle add up to 180°, a triangle can be classi­ fied as acute (having three acute angles), right (having one right angle), or obtuse (having one obtuse angle). We know from geometry that the lengths of the sides of a triangle determine its angles. How can we tell from these side lengths whether the triangle is acute, right, or obtuse? Statement II of the Pythagorean theorem gives us a partial answer: If 2 2 2 the side lengths a, b, c satisfy the relationship a + b = c , then the triangle is a right triangle. But what if this relationship is not satisfied? We can tell a bit more if we think of a right triangle that is "hinged" at its right angle, and whose hypotenuse can stretch (as if made of rubber). The diagrams below show such a triangle. Sides a and b are of fixed length, and the angle between them is "hinged." B B a C b A c A b As you can see, if we start with a right triangle, and "close down" the hinge, then the right angle becomes acute. When this happens, the third 2 2 2 side (labeled c) gets smaller. In the right triangle, c = a + b , so we can see that: 2 2 2 Statement III: If angle C of :::,.ABC is acute, then c a + b . In the same way, if we open the hinge up, angle C becomes obtuse, and the third side gets longer: B B a A A b c c b Sowe see that 2 2 2 Statement IV: If angle C of :::,.ABC is obtuse, then c a + b • 14 'f'rigonomelry Exercise Write the converse of statements III and IV above. While the converses of most statements require a separate proof, for these particular cases, the converses follow from the original statements. 2 2 2 For example, if, in .6ABC, c a +b , then angle C cannot be right (this would contradict statement II of the Pythagorean Theorem) and cannot be . obtuse (this would contradict statement IV above). So angle C must be acute, which is what the converse of statement III says. Statements III and IV, together with their converses, allow us to decide whether a triangle is acute, right, or obtuse, just by knowing the lengths of its sides. Some examples follow: 1. Is a triangle with side lengths 2, 3, and 4 acute, right or obtuse'? 2 2 2 Solution. Since 4 = 16 2 + 3 = 4 + 9 = 13, the triangle is obtuse, with the obtuse angle opposite the side of length 4. 2 2 2 2 Question: Why didn't we need to compare 3 with 2 + 4 , o: 2 2 2 with 3 + 4 ? 2. Is a triangle with sides 4, 5, 6 acute, right, or obtuse? 2 Solution. We need only check the relationship between 6 and 2 2 2 2 2 4 +5 . Since 6 = 36 4 +5 = 41, the triangle is acute. 3. Is the triangle with side lengths 1, 2, and 3 acute, right, or obtuse? 2 2 2 Solution. We see that 3 = 9 2 + 1 = 5, so it looks like the triangle is obtuse. Question: This conclusion is incorrect. Why? Exercise If a triangle is constructed with the side lengths given below, tell wheth­ er it will be acute, right, or obtuse. a) 6, 7, 8 b) 6, 8, 10 c) 6,8,9 d) 6, 8, 11 e) 5, 12, 12 f) 5, 12, 14 g) 5, 12, 17 II. Proof of the Pythagorean theorem There are many proofs of this classic theorem. Our proof follows the Greek tradition, in which the squares of lengths are interpreted as areas. We first recall statement I from the text: Appendix II. Proof' of t/w l'ytlwgorean tlleorem 15 If a and b are the lengths of the legs of a right triangle, and c is the length 2 2 2 of its hypotenuse, then a + b = c • Let us start with any right triangle. The lengths of its legs are a and b, and the length of its hypotenuse is c: b We draw a square (outside the triangle), on each side of the triangle: c2 We must show that the sum of the areas of the smaller squares equals the area of the larger square: = + Or: 16 Trigonometry I The diagram below gives the essence of the proof. If we cut off two copies of the original triangle from the first figure, and paste them in the correct niches, we get a square with side c: We fill in some details of the proof below. We started with an oddly shaped hexagon, created by placing two squares together. To get the shaded triangle, we lay off a line segment equal to b, starting on the lower left-hand comer. Then we draw a diagonal line. This will leave us with a copy of the original triangle in the comer of the hexagon: a (Notice that the piece remaining along the bottom side of the hexagon has length a, since the whole bottom side had length a+ b.) Triangle ABC is congruent to the one we started with, because it has the same two legs, and the same right angle. Therefore hypotenuse A B will have length c. Next we cut out the copy of our original triangle, and fit it into the niche created in our diagram: Appendix II. Proof of tlw Jytlwgorcun t11corcm 17 B D / A A The right angle inside the triangle fits onto the right angle outside the hexagon (at D), and the leg of length a fits onto segment BD, which also has length a. Connecting A to E, we form another triangle congruent to the original (we have already seen that AF =a, and EF = b because each was a side of one of the original squares). B B F A A This new copy of the triangle will fit nicely in the niche created at the top of the diagram: Why will it fit? The longer leg, of length b, is certainly equal to the upper side of the original hexagon. And the right angles at G must fit together. But why does G H fit with the other leg of the triangle, which is of length a? 18 'Ji"igonometry Let us look again at the first copy of our original triangle. If we had placed of side b, it would have looked like this: it alongside the square r I ' I b I I a• I I I I I But in fact we draw it sitting on top of the smaller square, so it was pushed up vertically by an amount equal to the side of this square, which is a: H (1 , I I I I I I I ib I I a I I I I .__ __ _.. _________________ , I So the amount that it protrudes above point G must be equal to a. This is the length of G H, which must then fit with the smaller leg of the second copy of our triangle. One final piece remains: why is the final figure a square? Certainly, it has four sides, all equal to length c. But why are its angles all right angles? Bk'v c Let us look, for example, at vertex B. Angle C B D was originally a right angle (it was an angle of the smaller square). We took a piece of it away when we cut off our triangle, and put the same piece back when we pasted the triangle back in a different position. So the new angle, which is one Appendix II. Proof of tile l'ytlw,gorea/1 theorem 19 in our new figure, is still a right angle. Similar arguments hold for other vertices in our figure, so it must be a square. In fact all the pieces of our puzzle fit together, and we have transformed the figure consisting of squares with sides a and b into a square with side c. Since we have not changed the area of the figure, it must be true that a az+bz=cz. Finally, we prove statement II of the text: 2 2 2 If the positive numbers a, b, and c satisfy a + b = c , then a triangle with these side lengths has a right angle opposite the side with length c. We prove this statement in two parts. First we show that the numbers a, b, and care sides of some triangle, then we show that the triangle we've created is a right triangle. Geometry tells us that three numbers can be the sides of a triangle if and only if the sum of the smallest two of them is greater than the largest. But can we tell which of our numbers is the largest? We can, if we remember 2 2 2 2 2 that for positive numbers, p q implies that p q. Since c = a + b , 2 2 2 and b 0, we see that c a , so c a. In the same way, we see that c b. Now we must show that a+ b c. Again, we examine the squares of 2 2 2 2 2 2 our numbers. We find that (a+ b ) = a + 2ab + b a + b = c (since 2ab is a positive number). So a + b c and segments of lengths a, b, and c form a triangle. What kind of triangle is it? Let us draw a picture: a b Does this triangle contain a right angle? We can test to see if it does by copying parts of it into a new triangle. Let us draw a new triangle with sides a and b, and a right angle between them:

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