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.11.1 Fundamental vocabulary terms Measuring angles The magnitude of an angle does not depend on the length of its sides; it only depends on the relative direction of the two sides. E.g. the adjacent edges of a postcard are at an angle of 90– with respect to each other. But so is the Empire State Building in New York City with respect to Fifth Ave. and 34th Street (NOTE: its SA... maybe this should be the baobab tree with respect to the Limpopo river (i kid... i kid)). So we can’t use notions of length to measure angles. How do we measure angles then? We begin by ﬂnding a way to enumerate angles. What is the smallest angle you can draw? Two lines subtending almost no angle. Two coincident line segments pointing in the same direction subtend an angle of 0–, e.g., lines AB and AC in the ﬂgure below. Now if we keep one line ﬂxed and move the other while still pivoted at the common vertex, we can obtain any other angle. Two line segments with a common vertex and facing in opposite direction are said to form an angle of 180–, e.g., XY and XZ in the ﬂgure below. ZXY is a straight line. 0– A B Z C 180– X Y The choice of a measure of 180– for the angle subtended by the segments of a straight line is a matter of historical convention. Once we have decided what the measure of an angle formed by a straight line is, we have also ﬂxed the measure of all the other angles. This is because we would like the angles to obey some desirable properties. E.g. if we have an angle a– between two lines AB and AC, and another angle b– between AB and AD, we would like 88 the angle between AC and AD to be (a + b)–. This makes the measurement of angles intuitive and conforms to our notion of measurement of length, weight, etc. Similarly, a line segment that deﬂnes a direction exactly half way between AB and AC should create an angle of (a=2)–. D B B D b– a– A C A C a– (a=2)– An angle of 90– is termed a right angle. A right angle is half the measure of the angle subtended by a straight line
(180–). An angle twice the measure of a straight line is 360–. An angle measuring 360– looks identical to an angle of 0–, except for the labelling. All angles after 360– also look like we have seen them before. Angles that measure more than 360– are largely for mathematical convenience to maintain continuity in our enumeration of angles. C Right Angle 90– A B 360– A B C We deﬂne some other terms at this point. These are simply labels for angles in particular ranges. Acute angle: An angle 0– and < 90–. ‚ Obtuse angle: An angle > 90– and < 180–. Straight angle: An angle measuring 180–. Reex angle: An angle > 180– and < 360–. † † † † C acute C A B A obtuse C B reex A B Once we can number or measure angles, we can also start comparing them. E.g. all right angles are 90–, hence equal. An obtuse angle is larger than an acute angle, etc. An alternative measure of angles is used on a compass. E.g. if North(N) is 0–, North-East(NE) is 45–, NNE is 22:5–, etc. 89 Special Angle Pairs In the previous section, we classiﬂed angles based on measurement. In this section we’ll examine some interesting properties of angles formed by a pair of intersecting lines. First we consider a single straight line, AB. There’s a point on the line AB called X. The measure of angle AXB is 180– as deﬂned in the previous section. Now let us draw another straight line intersecting the ﬂrst. Without loss of generality, let the point of intersection be X. We call the four angles formed with X as the vertex a, b, c and d. At this point, we introduce some deﬂnitions for convenience Vertical angles Deﬂnition: The angles formed by two intersecting straight lines that share a vertex but do not share any sides are called vertical angles. E.g., a and c in the ﬂgure above are vertical angles. b and d are also vertical angles. Adjacent Angles Deﬂnition: Two angles that share a common vertex and a common side are called adjacent
angles. E.g., (a;b) and (c;d) are adjacent angles. Linear pairs Deﬂnition: The adjacent angles formed by two intersecting straight lines are said to form a linear pair. E.g. (a;b), (b;c), (c;d) and (d;a) all form linear pairs. Since the non common sides of a linear pair are part of the same straight line, the total angle formed by the linear pair is 180– by deﬂnition. E.g. a + b = 180–, etc. What can we say about the vertical angles? Looking at ﬂgure above, it seems like the vertical angles are equal to one another. We can prove the following result. Theorem: The vertical angles formed by intersection of two straight lines are equal. Proof: Since a and b form a linear pair, a + b = 180–. Similiarly, b and c form a linear pair, so, b + c = 180–. Thus, a + b = b + c. Since the angle b contributes equally to both sides of the equation, it can be cancelled out leaving, a = c. The proof for the pair (b;d) is identical. 90 This proof conﬂrms our intuition that vertical angles are equal in magnitude. This result will later be used when proving properties for parallel lines. We end this section with some more deﬂnitions. Supplementary and Complementary pairs Deﬂnition: Two angles are called supplementary if their sum equals 180–. E.g. angles that constitute a linear pair are supplementary. Deﬂnition: Two angles are called complementary if their sum equals 90–. Note that in order to be labelled supplementary or complementary, the two angles being considered need not be adjacent. E.g. x and y in the ﬂgure below are supplementary, but they are not adjacent and thus do not form a linear pair. C D x y x + y = 180– A B E F 6.11.2 Parallel lines intersected by transversal lines Two lines are said to intersect if there is a point that lies on both lines. Informally, two lines intersect if they meet at some point when extended indeﬂnitely in either direction. E.g. at a tra–c intersection, two or more streets intersect; the
middle of the intersection is the common point between the streets. It is possible that two lines that lie on the same plane never intersect even when extended to inﬂnity in either direction. Such lines are termed parallel lines. E.g. the tracks of a straight railway line are parallel lines. We wouldn’t want the tracks to intersect as that would be catastrophic for the train! A section of the Australian National Railways Trans-Australian line is perhaps one of the longest pairs of man-made parallel lines. Longest Railroad Straight (Source: www.guinnessworldrecords.com) The Australian National Railways Trans-Australian line over the Nullarbor Plain, is 478 km. (297 miles) dead straight, from Mile 496, between Nurina and Loongana, Western Australia, to Mile 793, between Ooldea and Watson, South Australia. A transversal of two or more lines is a line that intersects these lines. E.g. in the ﬂgure below, AB and CD are two lines and EF is a transversal. We are interested in the properties of the angles formed by these intersecting lines, so we’ll introduce some deﬂnitions for various angle pairs. 91 Deﬂnitions: † † † † † Interior angles: When two lines are intersected by a transversal, the angles that lie between the two lines are called interior angles. E.g. in the ﬂgure above, 1, 2, 3 and 4 are interior angles. Exterior angles: When two lines are intersected by a transversal, the angles formed that lie outside the two lines are called exterior angles. E.g. 5, 6, 7 and 8 are exterior angles. Alternate interior angles: When two lines are intersected by a transversal, the interior angles that lie on opposite sides of the transversal are termed alternate interior angles. E.g. in the above example, 1 and 3 are are a pair of alternate interior angles. 2 and 4 are also alternate interior angles. Interior angles on the same side: As the name suggests, these are interior angles that lie on the same side of the transversal. E.g. (1,4) and (2,3). Corresponding angles: The angles on the same side of the transversal and the same side of the two lines are called corresponding
angles. E.g. (1;5), (4;8) and (3;7), etc., are pairs of corresponding angles. In order to prove relationships between the angles deﬂned above, we will assume the following postulate regarding parallel lines. Euclid’s Parallel Line Postulate: Postulate: If a straight line falling on two straight lines makes the two interior angles on the same side less than two right angles (180–), the two straight lines, if produced indeﬂnitely, meet on that side on which the angles are less than two right angles. The above is one of the fundamental postulates of Euclidean geometry and has no proof based on the other postulates. Now we’ll use the above postulate to prove some other properties. Theorem 1: If two parallel lines are intersected by a transversal, the sum of interior angles on the same side of the transversal is two right angles (180–). Proof: Consider parallel lines AB and CD intersected by the transversal EF in the ﬂgure above. Suppose that the sum of the interior angles is less than 180– on one side of the transveral, e.g. 1 + 4 < 180–. Then Euclid’s 92 ¡ ¡ ¡ (1) and (3) = 180– (4). So (2+3) = 360– Parallel Line Postulate implies that the AB and CD meet on that side of the transversal and are not parallel. This contradicts the assumption that the lines are parallel. Now suppose that the sum of the interior angles 1 and 4 is greater than (1+4). 180–. Now, (2) = 180– Since (1 + 4) > 180–, (2 + 3) < 180–. Thus the parallel line postulate implies that the lines will meet on that side of the transveral and are not parallel. Thus both pairs of interior angles on the same side need to sum up to 180– for the lines to be parallel. Theorem 2: If two parallel lines are intersected by a transversal, the alternate interior angles are equal. Proof: In the ﬂgure above, using Theorem 1, (1 + 4) = 180– Also, since AB is a straight line, 1 and 4 are supplementary. (4 + 3) = 180– Thus, 1 = 3. Similarly
, 2 = 4. Theorem 3: If two parallel lines are intersected by a transversal, the corresponding angles are equal. Proof: Again using Theorem 1, in the ﬂgure above, (1 + 4) = 180– Also, since EF is a straight line, (4 + 5) = 180– So 1 = 5, etc. Theorem 4: The sum of the three angles in a triangle is 180. Proof: Consider triangle ABC shown in the ﬂgure below. The three angles are denoted 1, 2 and 3. We have to show that 1 + 2 + 3 = 180–. Consider a straight line DE through point A that is parallel to BC. We denote the angles between DE and the sides of the triangle as 4 and 5 Since DE is a straight line, 1 + 4 + 5 = 180–. Now DE is parallel to BC, and 2 and 4 are alternate interior angles (AB is the transversal). So 2 = 4. Similarly, 3 = 5. So substituting these in the ﬂrst equation, 1 + 2 + 3 = 180–. The following theorems help in determining when two lines are parallel to each other. Theorem 5: If two lines are intersected by a transversal such that any pair of interior angles on the same side is supplementary, then the two 93 lines are parallel. Proof: We’ll prove that if two lines are not parallel, the interior angles on the same side are not supplementary. We’ll prove this by contradiction. Assume that two non-parallel distinct lines are intersected by a transversal such that interior angles 1 and 4 are supplementary. 1 + 4 = 180– (Eq. (i)) Since the lines are not parallel, they have to intersect at some point Z. Since the two lines are distinct, they have to form a non-zero angle at their point of intersection. Z 9 X 1 4 Y XY Z is a triangle. So 1 + 4 + 9 = 180–, using Theorem 4. But using Eq. (i), 1 + 4 = 180–, so 9 = 0–. This contradicts the fact that distinct intersecting lines create a non-zero angle at their point of intersection. So our original assumption is not supportable and the interior angles 1 and 4 cannot be supplementary. Theorem 6: If two lines are intersected by a transversal such that a pair
of alternate interior angles are equal, the lines are parallel. Proof: Left as an exercise. Theorem 7: If two lines are intersected by a transversal such that a pair of alternate corresponding angles are equal, the lines are parallel. Proof: Left as an exercise. Theorem 8: Prove that if a line AB is parallel to CD, and AB is parallel to EF, then CD is parallel to EF. Proof: Left as an exercise. (Hint: We can prove this in two steps: 1. Prove that if two lines are parallel, then a line that intersects one also intersects the other. 2. Use the equivalence of corresponding angles to get the result.) 94 Sides Name 5 6 7 8 10 15 pentagon hexagon heptagon octagon decagon pentadecagon Table 6.1: Table of some polygons and their number of sides. 95 Chapter 7 Trigonometry 7.1 Syllabus 7.1.1 Triangles The syllabus requires: { similarity of triangles is the basis of trig functions (NOTE: perhaps this is best left in geometry) { solve problems in 2D by constructing and interpreting geometric and trig models including scale drawings, maps and building plans 7.1.2 Trigonometric Formul‰ The syllabus requires: { some history from various cultures { derive reduction formul‰ for trig ratios { recognise equivalence of trig expressions by reduction { solve 2D problems by establishing sin/cos/area rules (NOTE: perhaps just do the rules here, and do the problems in section 8.5) { use trig for height and distances (NOTE: SH isn’t this already done in 7.1.1?) 7.2 Radian and Degree Measure You should be familiar with the idea of measuring angles from geometry but have you ever stopped to think why there are 360 degrees in a circle? 96 The reason is purely historical 1. There are, in fact, many diﬁerent ways of measuring angles. The two most commonly used are degrees (the one you have been using up to now) and radians. arc le The radian measure of an angle is deﬂned as the ratio of the arc length subtending the angle to the radius of the circle. A = arclength radius (7.1) We know from geometry that the circumference of a circle is found using the equation c = 2…r. If we divide through by the radius of the circle
, r, we ﬂnd that the radian angle subtended by the complete circumference, (or in other words the number of radians in a full circle) is 2…r r = 2…. This means that 2… radians is the same as 360–. With this in mind we can easily work out how to convert between degrees and radians. Deﬂnition: (rad) = (–) 2… 360 £ or (–) = (rad) 360 2… £ Using these formulae we can express common angles in radians. It is worth learning these as questions may be asked using either degrees, radians or a mixture of both. Degrees Radians 30– … 6 45– … 4 60– … 3 90– … 2 180– … 270– 3… 2 360– 2… 7.2.1 The unit of radians You may be wondering what the unit of radians is. The answer is that it doesn’t have one. This is because a radian is the ratio of two lengths: the arc length divided by the radius. Now, both of these will have the 1There are 360 degrees in a circle because the ancient Babylonians had a number system with base 60. A base is the number you count up to before you get an extra digit. The number system that we use everyday is called the decimal system (the base is 10), but computers use the binary system (the base is 2). 360 = 6 £ 60 so for them it make sense to have 360 degrees in a circle. 97 same unit, so when you divide them, the units simply disappear! This is what is known as a dimensionless quantity. Sometimes we write radians (or simply rad ) after the number to emphasise that we are using radians, but this is not necessary. In general, if an angle is expressed in terms of … it is meant to be in radians. Be careful though. If the question does not explicitly say whether the angle is measured in degrees or radians you need to use common sense to decide which to use. 7.3 Deﬂnition of the Trigonometric Functions 7.3.1 Trigonometry of a Right Angled Triangle Consider a right-angled triangle. We deﬂne b a c sin = cos = tan = a b c b a c (7.2) (7.3) (7.4) These are abbreviations for sine,
cosine and tangent. These functions, known as trigonometric functions, relate the lengths of the sides of a triangle to its interior angles. How to remember the deﬂnitions Diﬁerent people have diﬁerent ways of remembering these ratios. One way involves deﬂning opposite to be side of the triangle opposite to the angle, hypotenuse to be the side opposite to the right-angle (just like we use the term in geometry) and adjacent to be the side next to the angle, which is not the hypotenuse. This is illustrated in the following picture, where we 98 show the adjacent, opposite and hypotenuse for the angle. e t i s o p p o hypotenuse adjacent So, using these deﬂnitions we have: sin = cos = tan = opposite hypotenuse adjacent hypotenuse opposite adjacent (7.5) (7.6) (7.7) There is a mnemonic to remember these: Sine S O Opposite H Hypotenuse C Cos A Adjacent H Hypotenuse T Tan O Opposite A Adjacent Another mnemonic that is perhaps easier to remember goes as follows: Silly Old Hens Sin = Cackle And Howl Cos = Till Old Age Tan = Opposite Hypotenuse Adjacent Hypotenuse Opposite Adjacent CAUTION! The deﬂnitions of opposite, adjacent and hypotenuse only make sense when you are working with right-angled triangles! Always check to make sure your triangle has a right angle before you use them, otherwise you will get the wrong answer. We will ﬂnd ways of working with the trigonometry of non right-angled triangles later in the chapter. By using the appropriate triangles it is possible to work out the following values of the sine, cosine and tangent functions for a number of common angles. 99 cos sin tan 0– 1 0 0 30– p3 2 1 2 1 p3 45– 1 p2 1 p2 1 60– 1 2 p3 2 p3 90– 0 1 ¡ 180– 1 ¡ 0 0 These values are useful to remember as they often occur in questions. They are also a good way of helping us to visualise the graphs of the sine, cosine and tangent functions. 7.3.2 Trigonometric Graphs Sine and Cosine Graphs Let
us look back at our values for sin { 0– 0 30– 1 2 45– 1 p2 60– p3 2 90– 1 180– 0 sin As you can see, the function sin has a value of 0 at = 0–. Its value then smoothly increases until = 90– when its value is 1. We then know that it later decreases to 0 when = 180–. Putting all this together we can start to picture the full extent of the sine graph. The sine graph is shown in ﬂgure 7.1. 360 ¡ 180 ¡ 1 1 ¡ 180 360 Figure 7.1: The sine graph. Let us now look back at the values of cosine{ cos 0– 1 30– p3 2 45– 1 p2 60– 1 2 90– 0 180– 1 ¡ If you look carefully you will notice that the cosine of an angle is the same as the sine of the angle 90–. Take for example, ¡ cos 60– = 1 2 = sin 30– = sin (90– 60–) ¡ 100 This tells us that in order to create the cosine graph all we need to do is to shift the sine graph 90– to the left2. The cosine graph is shown in ﬂgure 7.2. 360 ¡ 180 ¡ 90– shift 180 360 1 1 ¡ Figure 7.2: The cosine graph (in black) with the sine graph (in gray). Tangent graph Now that we have the sine and cosine graphs there is an easy way to visualise the tangent graph. Let us look back at our deﬂnitions of sin and cos in a right angled triangle. sin cos = opposite hypotenuse adjacent hypotenuse = opposite adjacent = tan This is the ﬂrst of an important set of equations called trigonometric identities. An identity is an equation which holds true for any value which is put into it. In this case we have shown that tan = sin cos for any value of. So we know that for values of for which sin = 0, we must also have tan = 0. Also, if cos = 0 our value of tan = 0 is undeﬂned as we cannot divide by 0. The complete graph3 is shown in ﬂgure 7.3. 2You may have noticed that the transformation we are using is in fact a translation of 90–
followed by a reection in the y axis due to a negative sign in front of the. However, because cosine is an even function (i.e. symmetric about the y axis) this reection doesn’t really matter! 3The dotted lines in the tangent graph are known as asymptotes and the graph is said to display asymptotic behaviour. This means that as approaches 90–, tan approaches inﬂnity. In other words, there is no deﬂned value of the function at the asymptote values. Another 1 graph which displays asymptotic behaviour is y = x whose asymptotes are the x and y axes themselves. 101 360 ¡ 180 ¡ 4 2 2 ¡ 4 ¡ 180 360 Figure 7.3: The tangent graph. 7.3.3 Secant, Cosecant, Cotangent and their graphs In the sections that follow it will often be useful to deﬂne the reciprocal functions of sine, cosine and tangent. We shall deﬂne them as follows{ csc = sec = cot = 1 sin 1 cos 1 tan = = = hypotenuse opposite hypotenuse adjacent adjacent opposite (7.8) (7.9) (7.10) j j j cos j and The graphs of these functions are shown in ﬂgures 7.4{7.6. There are a number of points worth noting about these graphs. Firstly, since sin j and are always less than or equal to 1 their reciprocal functions j must always be greater than or equal to 1. Secondly csc sec j notice that the sectant graph can be obtained from the cosecant graph by performing a 90– shift, just like we did with sine and cosine. Notice also that these graphs have asymptotes whenever their reciprocal function is 0. One important feature of all these trigonometric functions is that they are periodic with a period of 360–. This is most easily understood by looking back at a circle. j 102 360 ¡ 180 ¡ ¡ 180 360 Figure 7.4: The cosecant graph. 360 ¡ 180 ¡ ¡ 180 360 Figure 7.5: The sectant graph. 103 360 ¡ 180 ¡ 4 2 2 ¡ 4 ¡ 180 360 Figure 7.6: The cotangent graph. Imagine that we are measuring the
angle on this circle. Now let us add 360– to our angle so that our line sweeps all the way around the circle and ends up back where it started as indicated in the diagram. There is no way of knowing whether we have swept around the circle in this way as everything ends up exactly where it started. In other words, if we add 360– to an angle we eﬁectively have the same angle we started with. Since our diagram is the same after the rotation the values of our trigonometric functions also remain unchanged. This is the reason that all trigonometric functions have a period of 360– { adding 360– degrees to an angle does nothing more than sweep it all the way round a circle back to where it 360–. began, so all of our functions must have the same value for and § 7.3.4 Inverse trigonometric functions Like all functions the trigonometric functions have inverses. These funcopposite hypotenuse in the case of inverse sine) and give tions take in ratios (such as out the angle it corresponds to. However, due to the periodicity of the 104 trigonometric functions there are many possible angles for one ratio. For example, both sin 30– and sin 150– give values of 1 2 can be either 30–, 150– or any one of an inﬂnite amount of other possibilities. 1 and Two notations are commonly used for the inverse functions { sin¡ 1 and arc versions can be used for any of the six arcsin. Both the ¡ trigonometric functions. However, you must be careful not to confuse the reciprocal functions (csc, sec and cot) with the inverse functions (arcsin, arccos and arctan). They are diﬁerent functions with diﬁerent meanings and will give diﬁerent answers4. 2 so sin¡ 1 1 7.4 Trigonometric Rules and Identities 7.4.1 Translation and Reection We found earlier that all trigonometric functions are periodic, with a period of 360–. We can express this more formally by writing{ sin ( § 360–) = sin This identity states that the sine of an angle is unchanged if we add or subtract 360–. Another way to think of this is as a translation of the sine graph by 360– to the right or left. 360– shift 1 360 ¡ 180 ¡ 1 �
� 180 360 As you can see, if we shift the whole graph by 360– left or right it will end up back on top of itself. The sine of an angle is therefore completely unchanged. Identities of this form can be very useful. We shall consider a few such identities here using the ideas of chapter (NOTE: Add in correct ref for transformations in geometry chapter). n First let us consider reecting the sin graph in the x and y axes. We know that if we reect in the x axis we will get the graph of sin. Figure 7.7 shows the original sine graph (gray) and its reection in the x axis. Just by looking at the graph we can see that reecting sine in the y axis would give the same result as reecting in the x axis did5. Mathematically, a ¡ 4Remember { you can have inverse reciprocal functions such as arccsc and sec¡1. 5If you have a mirror you can check this by putting it along the y axis. 105 360 ¡ 180 ¡ 1 1 ¡ 180 360 Figure 7.7: The sine graph reected in the x axis. reection in the y axis gives us the function sin ( second identity{ ¡ ). This gives us a sin ( ) = ¡ sin ¡ We could also obtain the black function in ﬂgure 7.7 by translating the gray sine graph 180– to the right or left. This tells us that{ sin ( ) = ¡ sin = sin( 180–) § ¡ Let us now look at the cosine function. Again we can use the fact that cosine is periodic with period 360– to give{ cos ( § 360–) = cos This time our reections are a little more complicated. Firstly let’s reect the cosine function in the y axis to generate cos ( ). Since the cosine graph is symmetric about the y axis the reection will not change our graph. This tells us that { ¡ cos ( ) = cos ¡ Reecting in the x axis we obtain ﬂgure 7.8. The ﬂrst point of interest 360 ¡ 180 ¡ 1 1 ¡ 180 360 Figure 7.8: The cosine graph reected in the x axis. is that unlike with sine we get diﬁerent graphs by reecting in the x and 106 y axes.
We can however still consider the black graph as a translation of 180– to the left or right. As before this gives us an identity{ cos = cos( 180–) § ¡ One ﬂnal and very important class of translation identity are those that convert sine into cosine and vice versa. We have already seen one of these when we looked at the graphs of sine and cosine{ cos = sin (90– ) ¡ This can be written in many ways by using the identities we have already proven. One more sensible version is{ cos ( ¡ 90–) = sin It is now easier to see that the sine graph is just the cosine graph moved 90– to the right. We shall prove that these two identities are the same in the worked example at the end of this section. There are many more identities such as these for sine and cosine as well as tangent and the reciprocal functions. One way to ﬂnd such identities is to uses the addition and subtraction formulae which we will derive in section 7.4.6. Always remember to check if your expressions can be simpliﬂed using these identities. 7.4.2 Pythagorean Identities Consider a right angled triangle{ r x (0;0) y Let us use our deﬂnitions of sine and cosine in a right{angled triangle on the angle. sin = opposite hypotenuse = y r and cos = adjacent hypotenuse = x r y = r sin and x = r cos ) Using Pythagorases theorem we can say that{ (7.11) (7.12) (7.13) x2 + y2 = r2 107 Substituting in our expressions for x and y{ (r cos )2 + (r sin )2 = r2 r2 cos2 + r2 sin2 = r2 ) If we divide both sides through by r2 we arrive at{ cos2 + sin2 = 1 This is known as a Pythagorean identity. We can express the identity in terms of the other trigonometric functions by dividing through by either sin2 or cos2 on both sides{ sin2 sin2 sin2 cos2 + + cos2 sin2 cos2 cos2 = = 1 sin2 ) 1 cos2 ) 1 + cot2 = csc2 tan2 + 1 = sec2 These 3 versions of the Pythagorean Identity can be used in
any question. They do not need to be used in a triangle they will work for any angle in any situation. Deﬂnition: Pythagorean Identities{ cos2 + sin2 = 1 1 + cot2 = csc2 tan2 + 1 = sec2 7.4.3 Sine Rule So far we have only dealt with the trigonometry of right angled triangles where we are able to use our deﬂnitions of sin, cos and tan. There are some rules which we can derive that hold true for any triangle. One of these is the sine rule. Consider a scalene triangle (i.e. one with all sides diﬁerent lengths and all angles diﬁerent){ a B b A p c 108 Note that the order in which the angles of the triangle are labelled is important. The three sides are labelled a, b, and c in any order. Then we deﬂne angle A as the angle opposite side a and so on. We can split it into two right-angled triangles by choosing a perpendicular, p, to one side (in this case side c) which passes through the opposite vertex, as shown. You may want to draw a few triangles and convince yourself that we can always do this regardless of the shape of the triangle. Now, since we have right angled triangles we can use our old deﬂnition of sin = opposite hypotenuse to ﬂnd{ sin A = p b and sin B = p a or, rearranging slightly{ p = b sin A and p = a sin B If we now set these two equations for p equal to each other (to eliminate the p) and rearrange again (by dividing through both sides by ab) we get the following{ p = b sin A = a sin B sin A a = sin B b ) (7.14) We could have chosen our perpendicular to go through any side of our triangle so let us repeat the proof with the perpendicular through a. b C p0 c a B Now we have{ p0 = b sin C = c sin B sin B b = sin C c ) (7.15) Check this result for yourself to make sure you understand where it came from! Putting equations 7.14 and 7.15 together we obtain the sine rule{ Deﬂnition: Sine Rule { sin A a = sin B b = sin C c 109 Many books now
tell us that we have not proved the sine rule fully. Consider the following triangle{ b p a A B c This type of triangle which has all of its angles smaller than 90– is called an acute triangle. We have already proved the sine rule for acute triangles. If we ‘fold’ this acute triangle along the perpendicular we obtain the following{ p b a A A A0 B c This kind of triangle, which contains an angle greater than 90–, is called an obtuse triangle. Notice that we now have no way of ﬂnding a perpendicular through side c which goes through a vertex of the new triangle. Instead, the perpendicular must be drawn outside of the triangle. This is the reason that some books say that our proof is incomplete { the perpendicular which we used for acute triangles does not always exist in obtuse ones! We can either repeat our proof from the start for obtuse triangles(which is messy!) or we can use our knowledge of trigonometry to prove these other books wrong. Notice that the lengths of a, b and p have not changed, neither has the value of angle B. The only change is that we have replaced the old angle, A, with a new angle, A0. If we look at A and A0 we can see that they lie on a straight line so from our knowledge of geometry we can say that A + A0 = 180– or alternatively that A0 = 180– A. We have already shown that{ ¡ sin (180 ¡ ) = sin 110 so we can say that{ sin A0 = sin (180 A) = sin A ¡ We now know that although angle A has changed, its sine has stayed the same. Since we are only interested in the sine of the angle our equation must still be valid! Any obtuse triangle can be formed by ‘folding’ an acute triangle in this way so the sine rule must be true for any triangle we choose. 7.4.4 Cosine Rule Let us return to our scalene triangle We have deﬂned a perpendicular, p, just as we did before. This perpendicular divides side c into two parts. One part lies between vertex A and the perpendicular. We shall call its length x. The other part between the perpendicular and vertex B must, therefore, have a length of c x. Using the usual deﬂnitions of sine and cosine on the left had section
of the triangle we ﬂnd{ ¡ sin A = p b p = b sin A and and cos A = x b x = b cos A ) Now we will use Pythagorases theorem on the right hand side of the triangle. a2 = p2 + (c = p2 + c2 x)2 2cx + x2 ¡ ¡ Substituting in p = b sin A and x = b cos A a2 = b2 sin2 A + c2 = b2(sin2 A + cos2 A) + c2 ¡ 2cb cos A + b2 cos2 A 2bc cos A ¡ (7.16) (7.17) (7.18) (7.19) Using the Pythagorean identity sin2 A + cos2 A = 1 we get the cosine rule. 111 Deﬂnition: Cosine rule { a2 = b2 + c2 2bc cos A ¡ 7.4.5 Area Rule One last triangle rule is an extension of the area formula for a triangle. height base From geometry{ area = 1 2 £ base £ height where height is measured perpendicular to the base. From trigonometry{ b height c sin = opposite adjacent = height b height = b sin ) Calling the base c we can write{ area = 1 2 £ base £ height = 1 2 bc sin where must be the angle between sides b and c. 112 Deﬂnition: Area rule { area = 1 2 bc sin 7.4.6 Addition and Subtraction Formulae Let us return to our scalene triangle complete with a perpendicular as we had before. ` p 1 x h ` ¡ 90– y The perpendicular, p, divides the top angle into two pieces, and `. We can use the fact that the internal angles of a triangle sum to 180– to ﬂnd that the bottom right angle must be 90– `. If we deﬂne one side of the triangle to be of legnth 1 for simplicity we can easily see from the left hand triangle that{ ¡ sin = cos = = sin p = cos From the right hand triangle we can see that{ cos ` = sin = p h ) y h ) h = p cos ` y = h sin ` = p sin ` cos ` Substituting in our expression for p we get{ y = cos sin ` cos ` (7.
20) (7.21) (7.22) (7.23) Putting these expressions togeher we ﬂnd that the base of the triangle, x + y, has a length{ x + y = sin + cos sin ` cos ` 113 + ` 1 90– sin + cos sin ` cos ` h ` ¡ We can now use the sine rule on the two angles that we know{ sin( + `) sin + cos sin ` cos ` = sin(90– 1 `) ¡ As we learnt in section 7.4.1, sin (90– cross multiplying we get{ ¡ ) = cos. Using this identity and sin( + `) sin + cos sin ` cos ` = cos ` 1 sin ( + `) = sin cos ` + cos sin ` ) (7.24) (7.25) This is the sine addition formula. If we take this formula and replace ) (remember, we can only do this with identities as they must with (90– be true for any value of their variables) we get{ ¡ sin (90– + `) = sin(90– ¡ As before we can use the identities ) cos ` + cos(90– ) sin ` ¡ ¡ ¡ ¡ Using these identities we can obtain the cosine subtraction formula{ sin (90– cos (90– `) = cos ` `) = sin ` (7.26) (7.27) sin (90– ( ¡ ¡ `)) = cos( ¡ `) = sin(90– ) cos ` + cos(90– ¡ = cos cos ` + sin sin ` ¡ ) sin ` (7.28) (7.29) We are missing two identities, the sine subtraction and cosine addition ` in the sine formulae. To obtain the sine subtraction we replace ` with addition formula. Since we know that sin( `) = cos ` from section 7.4.1 we ﬂnd that ¡ sin ` and cos( `) = ¡ ¡ ¡ sin ( ¡ `) = sin cos( ¡ = sin cos ` `) + cos sin( cos sin ` ¡ `) ¡ (7.30) (7.31) Proof of the cosine addition formula can be done in the same way, starting with the subtraction formula. cos ( + `) = cos
cos( `) + sin sin( `) ¡ = cos cos ` ¡ sin sin ` ¡ (7.32) (7.33) 114 Similar equations can be found for the tangemt function. Proof of these is left to the reader6. Deﬂnition: Addition and Subtraction Formulae{ ¡ ¡ sin ( + `) = sin cos ` + cos sin ` cos sin ` sin ( `) = sin cos ` sin sin ` cos ( + `) = cos cos ` ¡ cos ( ¡ tan ( + `) = `) = cos cos ` + sin sin ` tan ` + tan tan tan ` 1 ¡ tan ` tan ¡ 1 + tan tan ` `) = tan ( ¡ (7.34) (7.35) (7.36) (7.37) (7.38) (7.39) 7.4.7 Double and Triple Angle Formulae Let us remind ourselves of the addition formulae for sine and cosine{ cos ( + `) = cos cos ` sin sin ` sin ( + `) = sin cos ` + cos sin ` ¡ (7.40) (7.41) If we set ` to be equal to we get the following equations{ cos ( + ) = cos (2) = cos cos sin2 (7.42) sin ( + ) = sin (2) = sin cos + cos sin = 2 cos sin (7.43) sin sin = cos2 ¡ ¡ These are known as the double angle formulae. By using the phythagorean identity cos2 + sin2 = 1 we can substitute into the cosine double angle formula for cos2 or sin2 to get diﬁerent forms. We can do the same for the tangent function. Deﬂnition: Double angle formulae{ cos (2) = cos2 sin2 ¡ = 2 cos2 1 ¡ 2 sin2 = 1 sin (2) = 2 cos sin ¡ tan (2) = 2 tan tan2 1 ¡ (7.44) (7.45) (7.46) (7.47) (7.48) 6HINT{Remember, tan = sin cos so you can use the sine and cosine addition formulae to ﬂnd the tangent one. To get it into the same form as
in the deﬂnition you need to divide everything through by a factor. The fact there is a 1 in the denominator should give you a clue as to what the factor is! Once you have the addition forumula you need to remember that tan(¡) = ¡ tan to ﬂnd the subtraction formula. 115 Now that we have the double angle formulae it is easy to ﬂnd higher order multiple angle formulae. We shall derive the cosine triple angle formulae here. We start by taking the cosine addition faormula and setting ` = 2{ cos ( + 2) = cos (3) = cos cos (2) sin sin (2) ¡ We now substitute in the double angle formulae for cos (2) and sin (2). We have a choice of forms for the cos (2) formula. We shall choose cos (2) = cos2 sin2. ¡ cos (3) = cos cos (2) sin sin (2) ¡ sin2 ) 3 sin2 cos ¡ ¡ = cos (cos2 = cos3 ¡ sin (2 cos sin ) (7.49) (7.50) (7.51) The corresponding sine triple angle formula is{ sin (3) = 3 cos2 sin sin3 ¡ 7.4.8 Half Angle Formulae We can rearrange the double angle formulae to ﬂnd the half angle formulae. We shall start by rearranging the cosine double angle formula of the form cos (2) = 2 cos2 1. ¡ 2 cos2 ) ¡ 1 = cos (2) 2 cos2 = 1 + cos (2) 1 + cos (2) 2 1 + cos (2) 2 cos2 = cos = §r ) ) (7.52) (7.53) (7.54) (7.55) Another way to write this is to halve both of the angles (we can do this because it is an identity, so must be valid for any angle) { cos 2 = §r 1 + cos 2 Using the same method to rearrange the identity cos (2) = 1 obtain{ ¡ 2 sin2 we §r We can now use the ratio identity to ﬂnd the tangent half angle formula{ sin = 2 1 ¡ cos 2 tan 2 = sin 2 cos 2 = § q q 1
cos 2 ¡ 1+cos 2 = §r 1 cos 1 + cos ¡ As with all identities the half angle formulae can be expressed in a number of ways. Some of these will be proven in the worked example and more 116 given in the summary of identities at the end of the chapter. Deﬂnition: Half Angle Formulae cos sin tan 2 2 2 = = = §r §r §r 1 1 + cos 2 cos 2 1 cos 1 + cos ¡ ¡ (7.56) (7.57) (7.58) ‘Product to Sum’ and ‘Sum to Product’ Identi- 7.4.9 ties For completeness we include a brief comment on the ‘product to sum’ and ‘sum to product’ identities. They can be derived from the addition and subtraction formulae. We shall only derive one of each type here as their derivations are broadly similar. We start by proving a product to sum identity. This identity us an experssion linking the product of two cosine functions (cos cos `) to a sum of cosine functions. The derivation is as follows{ cos ( + `) + cos ( ¡ `) = (cos cos ` = 2 cos cos ` ¡ (7.59) sin sin `) + (cos cos ` + sin sin `) (7.60) cos cos ` = ) 1 2 [cos ( + `) + cos ( `)] ¡ Sum to product identities are messier to prove. Here we prove the identity linking the sum of two cosines by exchange of variables. We substitute `0 into the product to sum identity above (the = 0 + `0 and ` = 0 primes just prevent us getting confused, we shall drop them later). ¡ (7.61) (7.62) 2 cos(0 + `0) cos(0 ¡ `0) = cos ((0 + `0) + (0 = cos(20) + cos(2`0) ¡ `0)) + cos ((0 + `0) (0 ¡ (7.63) `0)) ¡ (7.64) As always with identities we can divide all our variables by 2 for convenience (since it must be true for any angle) and drop out the primes to give{ cos + cos ` = 2 cos 0 + `0 2 ¶ cos 0
`0 ¡ 2 ¶ 7.4.10 Solving Trigonometric Identities A standard type of question in an exam is of the form \show that sin(2) tan = 2 cos2 ". As well as being important in examinations being able to prove 117 38:7– 100m Figure 7.9: Determining the height of a building using trigonometry. identities is a key mathematical skill. Most identities can be proven by using the standard identities we have already learnt earlier in this section. There are three ways that the two sides of an identity can diﬁer{ 1. The functions are diﬁerent e.g. sin cot = cos 2. The operations are diﬁerent e.g. sin cos3 = sin cos (1 3. The angles are diﬁerent e.g. sin = sin(2) 2 cos sin2 ) ¡ Of course, real identities (and even the examples above) contain a mixture of these three diﬁerences, but they can be solved by dealing with each of the diﬁerences, one at a time. 7.5 Application of Trigonometry Trigonometry is very important in many areas of every day life. In this section we shall learn to use trigonometry to solve problems which would otherwise require very complicated solutions. 7.5.1 Height and Depth One simple task is to ﬂnd the height of a building using trigonometry. We could just use tape measure lowered from the roof but this is impractical (and dangerous) for tall buildings. It is much more sensible to measure a distance along the ground and use trigonometry to ﬂnd the height of the building. Figure 7.9 shows a building whose height we do not know. We have walked 100m away from the building and measured the angle up to top. This angle is found to be 38:7–. We call this angle the angle of elevation. As you can see from ﬂgure 7.9 we now have a right angled triangle, one side of which 118 A 127– B 255– C Figure 7.10: Two lighthouses, A and B, and a boat, C. is the height of the building, which also includes our 100m distance and the angle of elevation. Using the standard deﬂnition of tangent{ tan 38:7– = = opposite adjacent height 100 ) height = 100 =
80m £ tan 38:7– (7.65) (7.66) (7.67) (7.68) 7.5.2 Maps and Plans Maps and plans are usually scale drawings. This means that they are an enlagement (usually with a negative scale factor so that they are smaller than the original) so all angles are unchanged. We can use this to make use of maps and plans by adding information from the real world. Let us imagine that there is a coastline with two lighthouses, one either side of a beach. This is shown in ﬂgure 7.10. The two lighthouses are 0:67km apart and one is exactly due east of the other. Let us suppose that no boat may get closer that 200m from the lighthouses in case it runs aground. How can the lighthouses tell how close the boat is? Both lighhouses take bearings to the boat (remember { a bearing is an angle measured clockwise from north). These bearings are shown on the map in ﬂgure 7.10. We can see that the two lighthouses and the boat form a triangle. Since we know the distance between the lighthouses and we have two angles we can use trigonometry to ﬂnd the remaining two sides of the triangle, the distance of the boat from the two lighthouses. Figure 7.11 shows this triangle more clearly. We need to know the legnths of the two sides AC and BC. We can choose to use either sine or cosine rule to ﬂnd our missing legnths. We shall use both here. Using the sine rule{ sin 119 (7.69) A 0:67km B 37– 15– 128– C Figure 7.11: Two lighthouses, A and B, and a boat, C. 7.6 Trigonometric Equations Trigonomeric equations often look very simple. Consider solving the equation sin = 0:7. We can take the inverse sine of both sides to ﬂnd that 1(0:7) = 1(0:7). If we put this into a calculator we ﬂnd that sin¡ = sin¡ 44:42–. This is true, however, it does not tell the whole story. As you 360 ¡ 180 ¡ 1 1 ¡ 180 360 Figure 7.12: The sine graph. The dotted
line represents sin = 0:7. ¡ can see from ﬂgure 7.12, there are four possible angles with a sine of 0:7 between 360– and 360–. If we were to extend the range of the sine graph to inﬂnity we would in fact see that there are an inﬂnite number of solutions to this equation! This di–culty (which is caused by the periodicity of the sine function) makes solving trigonometric equations much harder than they may seem to be. Any problem on trigonometric equations will require two pieces of information to solve. The ﬂrst is the equation itself and the second is the range in which your answers must lie. The hard part is making sure you ﬂnd all of the possible answers within the range. Your calculator will always give 90– and 90– for you the smallest answer (i.e. the one that lies between tangent and sine and one between 0– and 180– for cosine). Bearing this in mind we can already solve trigonometric equations within these ranges. ¡ 120 Worked Example 7 : Question: Find the values of x for which sin 3x = 0:5 if it is given that 0 < x < 90–. Answer: Because we are told that x is an acute angle, we can simply apply an inverse trigonometric function to both sides. sin x = 0:5 x = arcsin 0:5 x = 30– ) ) (7.70) (7.71) (7.72) (7.73) We can, of course, solve trigonometric equations in any range by drawing the graph. Worked Example 8 : Question: For what values of x does sin x = 0:5, when x < 360–? Answer: 360– < ¡ Step 1 : Draw the graph We take look at the graph of sin x = 0:5 on the interval [-360, 360]. We want to know when the y value of the graph is 0.5, so we draw in a line at y=0.5. 360 ¡ 180 ¡ 1 1 ¡ 180 360 Step 2 : Notice that this line touches the graph four times. This means that there are four solutions to the equation. Step 3 : Read oﬁ those x values from the graph as x = and 150–. 210–,30– 330–, ¡
¡ 121 1 0 1 ¡ 1st 2nd 3rd 4th 90 180 270 360 – 180 +VE +VE -VE -VE – 90 2nd +VE 1st +VE 3rd -VE 4th -VE – 270 – 0 /360 – Figure 7.13: The graph and unit circle showing the sign of the sine function. 1 90 180 270 360 360 ¡ 270 ¡ 180 ¡ 90 ¡ 1 ¡ This method can be time consuming and inexact. We shall now look at how to solve these problems algebraically. 7.6.1 Solution using CAST diagrams The Sign of the Trigonometric Function The ﬂrst step to ﬂnding the trigonometry of any angle is to determine the sign of the ratio for a given angle. We shall do this for the sine function ﬂrst. In ﬂgure 7.13 we have split the sine graph unto four quadrants, each 90– wide. We call them quadrants because they correspond to the four quadrants of the unit circle. We notice from ﬂgure 7.13 that the sine graph is positive in the 1st and 2nd quadrants and negative in the 3rd and 4th. Figure 7.14 shows similar graphs for cosine and tangent. All of this can be summed up in two ways. Table 7.1 shows which trrigonometric fuctions are positive and which are negative in each quadrant. A more convenient way of writing this is to note that all fuctions are positive in the 1st quadrant, only sine is positive in the 2nd, only tangent in the 3rd and only 122 1 0 1 ¡ 1st 2nd 3rd 4th 90 180 270 360 +VE -VE -VE +VE ¡ 1st 2nd 3rd 4th 90 180 270 360 +VE -VE +VE -VE Figure 7.14: Graphs showing the sign of the cosine and tangent functions. 2nd 1st sin +VE +VE -VE cos +VE -VE +VE tan +VE 3rd 4th -VE -VE -VE +VE -VE Table 7.1: The signs of the three basic trigonometric functions in each quadrant. 123 cosine in the 4th. We express this using the CAST digram (ﬂgure 7.15). This diagram is known as a CAST
diagram as the letter, taken anticlock- 180– 90– S T A C 270– 0–/360– S T A C Figure 7.15: The two forms of the CAST diagram. wise from the bottom right, read C-A-S-T. The letter in each quadrant tells us which trigonometric functions are positive in that quadrant. The ‘A’ in the 1st quadrant stands for all (meaning sine, cosine and tangent are all positive in this quadrant). ‘S’, ‘C’ and ‘T’,of course, stand for sine, cosine and tangent. The diagram is shown in two forms. The version on the left shows the CAST diagram including the unit circle. This version is useful for equations which lie in large or negative ranges. The simpler version on the right is useful for ranges between 0– and 360–. Magnitude of the trigonometric functions Now that we know which quadrants our solutions lie in we need to know which angles in these quadrants satisfy our equation. Calculators give us the smallest possible answer (sometimes negative) which satisﬂes the equation. For example, if we wish to solve sin = 0:3 we can apply the inverse sine function to both sides of the equation to ﬂnd{ = arcsin 0:3 = 17:46– However, we know that this is just one of inﬂnitely many possible answers. We get the rest of the answers by ﬂnding relationships between this small angle,, and answers in other quadrants. To do this we need to condider the modulus7 of the sine graph. = As you can see in ﬂgure 7.16 there is a solution to the equation 0:3 in every quadrant. The 1st quadrant solution is of course 17:46– as our calculator told us. The 2nd quadrant solution can be seen to be 180–. Another way to see this is to look at the identity sin ¡ j j 7This means we plot only the magnitude of the function. This is the same as reecting sin = sin(180– ) ¡ negative sections in the x axis. 124 1 0.3 0 0 1st 2nd 3rd 4th 90 180 270 360 Figure 7.16: The modulus of the sine graph. proved in section 7
.4. Using the same logic the 3rd quadrant solution can be seen to be (180– + ) and the 4th quadrant solution (360– ). It is now left to the reader to show, using similar graphs for cosine and tangent, that these relationships are true for all three of the trigonometric functions. These rules can be expressed in a simpler way. If we deﬂne the solution lying between 0– and 90– as `{ ¡ Deﬂnition: { If we are in the 1st or 3rd quadrants our solution is the lower boundary of the quadrant plus `. { If we are in the 2nd or 4th quadrants our solution is the upper boundary of the quadrant minus `. 7.6.2 Solution Using Periodicity ¡ Up until now we have only solved trigonometric equations where the argument (the bit after the function, e.g. the in cos or the (2x 7) in tan(2x 7)) has been. If there is anything more complicated than this we need to be a little more careful. 360–. We want soLet us try to solve tan(2x) = 2:5 in the range 0– lutions for positive tangent so using our CAST diagram we know to look in the 1st and 3rd quadrants. Our calulator tells us that arctan(2:5) = 68:2–. This is our ﬂrst quadrant solution for 2x. Our 3rd quadrant lies between 180– and 270– so our solution is 180– + 68:2–. Putting this together{ ¡ • • x 2x = 68:2– x = 34:1– ) or or 248:2– 124:1– 125 x • • Notice that we did not divide by the 2 until we had found our answers. Now try to put x = 214:1 into the equation. This gives 2x = 428:2 and we ﬂnd that tan(428:2) = 2:5! This solution, x = 214:1, lies within the range 0– 360– so we should have included it in our answer. Why did we not ﬂnd this solution before? The answer is that when we halved our solutions for 2x to ﬂnd x we also 360– so, after halved our range. We looked for solutions for 0–
• halving, our ﬂnal answer gave us solutions in the range 0– 180–. There are two ways of dealing with this. We could redo the problem looking the the range 0– 720–. This will work but there is a simpler method. We know that all the trigonometric functions are periodic with a period of 360–. This means we can add (or subtract) a factor of 360n– (where n is an integer) our solution to ﬂnd another equally valid solution. Let us try this with tan(2x) = 2:5. If n = 0 we regain our original answers{ 2x 2x • • • • • x 2x = 68:2– or 248:2– Adding 360– (n = 1) to our solutions for 2x we ﬂnd the next two solutions{ 2x = 68:2– + 360– or = 428:2– ) or x = 214:1– ) 248:2– + 360– or 608:2– 304:1– 7.6.3 Linear Triginometric Equations Just like with regular equation solving without trigonometric functions the equations can become a lot more complicated. You should solve these just like normal equations and once you have a signal trigonometric ratio isolated, then you follow the strategy outlined in the previous section.(ADD AN EXAMPLE HERE) 7.6.4 Quadratic and Higher Order Trigonometric Equations The simplest quadratic trigonometric equation is of the form{ sin2 x 2 = 1:5 ¡ ¡ This type of equation can be easily solved by rearranging to get a more familiar linear equation{ sin2 x = 0:5 sin x = ) p0:5 § (7.74) (7.75) This gives two linear trigonometric equations. The solutions to either of these equations will satisfy the original quadratic. (ADD AN EXAMPLE HERE) The next level of complexity comes when you need to solve a trinomial 126 which contains trig functions. Here you can make you life a lot easier if you use temporary variables. Consider solving{ tan2 (2x + 1) + 3 tan (2x + 1) + 2 = 0 Here you should notice that tan(2x+1) occurs twice in the equation, hence we let y = tan(2x + 1) and rewrite: y2 + 3y + 2 = 0
That should look rather more familiar so that you can immediately write down the factorised form and the solutions: (y + 1)(y + 2) = 0 ¡ Next one just substitutes back for the temporary variable: ) y = 1 OR y = 2 ¡ tan (2x + 1) = 1 ¡ or tan (2x + 1) = 2 ¡ And then we are left with two linear trigonometric equations. Be careful: sometimes one of the two solutions will be outside the range of the In that case you need to discard that solution. trigonometric function. For example sonsicer the same equation with cosines instead of tangents{ cos2 (2x + 1) + 3 cos (2x + 1) + 2 = 0 Using the same method we ﬂnd that{ cos (2x + 1) = 1 ¡ or cos (2x + 1) = 2 ¡ 1 and The second solution cannot be valid as cosine must lie between 1. We must, therefore, reject the second equation. Only solutions to the ﬂrst equation will be valid. ¡ 7.6.5 More Complex Trigonometric Equations Here are two examples on the level of the hardest trig equations you are likely to encounter. They require using everything that you have learnt in this chapter. If you can solve these, you should be able to solve anything! (ADD AN EXAMPLE HERE) 127 7.7 Summary of the Trigonomertic Rules and Identities Pythagorean Identities Reciprocal Identities Ratio Identities cos2 + sin2 = 1 1 + cot2 = csc2 tan2 + 1 = sec2 csc = 1 sin sec = 1 cos tan = 1 cot tan = sin cos cot = cos sin Odd/Even Identities Periodicity Identities Cofunction Identities ) = sin sin( ¡ ¡ cos( ) = cos ¡ tan ) = tan( ¡ ¡ cot ) = cot( ¡ ¡ ) = csc( csc ¡ ¡ ) = sec sec( ¡ sin( cos( tan( cot( csc( sec( § § § § § § 360–) = sin 360–) = cos 180–) = tan 180–) = cot 360–) = csc 360–) = sec sin(90– cos(90– tan(90– cot(90– csc
(90– sec(90– ¡ ¡ ¡ ¡ ¡ ¡ ) = cos ) = sin ) = cot ) = tan ) = sec ) = csc Double Angle Identities Addition/Subtraction Identities Half Angle Identities sin(2) = 2 sin cos sin ( + `) = sin cos ` + cos sin ` sin ( ¡ `) = sin cos ` cos sin ` ¡ cos (2) = cos2 ¡ cos (2) = 2 cos2 cos (2) = 1 sin2 1 ¡ 2 sin2 ¡ tan (2) = 2 tan tan2 1 ¡ cos ( + `) = cos cos ` cos ( sin sin ` `) = cos cos ` + sin sin ` ¡ ¡ tan ( + `) = tan `+tan tan tan ` tan tan ( ¡ 1+tan tan ` 1 `) = tan ` ¡ ¡ sin 2 = cos 2 = 1 cos 2 ¡ 1+cos 2 §q §q tan 2 = 1 cos 1+cos ¡ §q Sine Rule Area Rule Cosine Rule sin A a = sin B b = sin C c Area = 1 Area = 1 Area = 1 2 bc cos A 2 ac cos B 2 ab cos C a2 = b2 + c2 b2 = a2 + c2 c2 = a2 + b2 2bc cos A 2ac cos B 2ab cos C ¡ ¡ ¡ 128 cos cos ` = 1 Product to Sum Identities 2 [cos( + `) + cos( `) 2 [cos( 2 [sin( + `) + sin( ¡ ¡ sin sin ` = 1 sin cos ` = 1 cos( + `)] `)] ¡ `)] sin + sin ` = 2 sin ¡ sin ¡ sin ` = 2 cos cos + cos ` = 2 cos Sum to Product Identities ‡ ‡ · +` 2 +` 2 +` 2 · +` 2 · cos sin ` ¡ 2 · ` · ` ¡ 2 ‡ ‡ cos ‡ sin ¡ sin ¡ cos ¡ cos ` = 129 Chapter 8 Solving Equations 8.1 Linear Equations The syllabus requires: { solve linear equations 8.1.1 Introduction Let’s imagine you have a friend called Joseph. He picked up your test results from the Biology class and now he refuses to tell what you scored
, or what he scored! Obviously you are trying everything to get him to tell you, and he decides to tease you and makes you work it out for yourself. He says the following: \I have 2 marks more than you and the sum of both our marks is equal to 14. How much did we get?" Now if the numbers are simple like in the example, you might be able to work it out in your head. Can you? But to make it easier, you can use a linear equation! This is how it works: We use a placeholder for your amount and that placeholder is x. So: Y ou = x Then we need a placeholder for Joseph Joseph = y BUT the trick is that we have some information about Joseph’s mark, which is that Joseph has 2 more than you. We need to use that, so how 130 about: Joseph = you + 2 Or Joseph = (x + 2) Or y = (x + 2) Now we need to use the last bit of information we have and that is: Y ou + Joseph = 14 Or using placeholders x + y = 14 Or substituting y x + (x + 2) = 14 What we have here is the actual linear equation. You already know what an equation is but what does linear mean? Linear means the highest power of the unknown variable, usually called x, is one. 8.1.2 Solving Linear equations - the basics To ﬂnd out what your test result is we need to now simplify this equation until we only have the x on the one side of the equal sign and a value on the other side. There are a few rules on how to simplify these equations to get a value for x. They can be organized into 3 groups. Once we have worked through them and we are sure about them, then we can attempt to ﬂnd out what the answer to our problem is. So here they are: Rule one - Addition or subtraction You are allowed to subtract or add any amount as long as you do it on both sides of the equal sign: Example 18.1) (8.2) (8.3) ) ) x + 5 x = ¡ 11 ¡ 131 Example 2 15 Rule two - Multiply or divide The same principle applies for multiplication and division: Example 1: Example 2: 2x = 9 2x = 20 4 = 5 4 £ ) ) Rule three - Fractions (8.4) (8.5)
(8.6) (8.7) (8.8) (8.9) (8.10) (8.11) (8.12) If x is multiplied by a fraction we need to divide both sides of the equal sign with that fraction to get x alone. We do that by ipping the fraction around and then multiplying both sides with it Example 1: )x( 2 3 14 3 x = ) = 7( 2 3 ) ) ) (8.13) (8.14) (8.15) These are the basic rules to apply when simplifying a linear equation. But most linear equations will require a few combinations of these before x is 132 sitting alone on the one side of the equal sign. That means we might have to make use of the rules above a number of times one after the other. Let’s do a few examples where we will use multiple steps to solve the equation: 8.1.3 Solving linear equations - Combining the basics in a few steps TIP: Start with eliminating the terms without x, that way you avoid having to calculate too many fractions Example 1: Example 2 = 62 7 + 5x = 62 7 + 5x ¡ 5x = 55 5x 55 5 5 x = 11 = 55 = 5x + 3 4 3 4 = 5x + 3 4 ¡ 3 4 ) = 5x = 5x 55 54( ¡ 1 4 217 4 217 4 £ 1 5 = 5x 1 5 £ Doing that is the the same as dividing by 5 217 20 ) = x (8.16) (8.17) (8.18) (8.19) (8.20) (8.21) (8.22) (8.23) (8.24) (8.25) (8.26) TIP: Start by moving all the terms with x to the one side and all the terms without x to the opposite side of the equal sign. Remember we can do that by changing the sign of the term 133 Example 3: 5x = 3x + 45 ) ) ) ) ) ) ) ) Example 4: Example 5: 3x = 45 5x ¡ 2x = 45 2x 45 2 2 1 2 x = 22 = 23x 12 = 6 + 2x 2x 12 = 6 2x = 6 + 12 ¡ ¡ 23x 23x ¡ ¡ 21x = 18 x = 21 18 12 6x
+ 34x = 2x ¡ 6x + 34x = 2x 6x + 34x ¡ 2x = 24 64 ¡ 24 ¡ 64 ¡ 24 ¡ 64 ¡ ¡ ¡ 12 12 ) ¡ ) ¡ ¡ 100 ) ) ) 26x = x = x = ¡ ¡ ¡ 100 26 50 13 (8.27) (8.28) (8.29) (8.30) (8.31) (8.32) (8.33) (8.34) (8.35) (8.36) (8.37) (8.38) (8.39) (8.40) (8.41) (8.42) We simpliﬂed the answer - but this is not necessarily a required step TIP: If there are parentheses (brackets) in the equation, start by removing them - multiply with the coe–cient Example 6(3x 4) = 8 ¡ 9x + 12 = 8 9x = 8 12 9x = 4 9 ¡ 4 ¡ (8.43) (8.44) (8.45) (8.46) (8.47) see the term is now positive - do you remember why? 134 Example 7: lets start with the parentheses - don’t forget the minus! 6x + 3x = 4 5(2x 3) ¡ ¡ next we move the like terms to their own sides 6x + 3x = 4 10x + 15 ¡ ) 6x + 3x + 10x = 4 + 15 19x = 19 x = 1 ) ) ) Example 8: (8.48) (8.49) (8.50) (8.51) (8.52) 8(3x 14) 34 = 2(4x 22) 5(3 + 2x) (8.53) ¡ Looks like a big one? Lets take it step by step ¡ ¡ ¡ 24x 112 ¡ ¡ ) 34 = 8x 44 ¡ ¡ 15 ¡ 10x (8.54) that’s all the brackets gone ) ) 24x 24x ¡ ¡ 34 112 ¡ ¡ 8x + 10x = and now solve! 8x + 10x = 15 44 15 + 112 + 34 ¡ ¡ 44 ¡ ¡ ) )
26x = 87 x = 87 26 (8.55) (8.56) (8.57) (8.58) And that is it for our examples. This covers all the types of linear equations you can be expected to solve. It’s the best to always keep your priority of steps in mind and then just simply do them one by one. If you are unsure about your answer you can just substitute it into your original equation and see if you get the same value for both sides: Example : 5(x 3) = 5 ¡ 15 = 5 5x 5x = 5 + 15 ¡ ) ) ) ) 5x = 20 x = 4 135 (8.59) (8.60) (8.61) (8.62) (8.63) Test: 3) = 5 5(4 ¡ 5(1) = 5 5 = 5 ) ) (8.64) (8.65) (8.66) and there we see it works! Now lets get back to our original problem of your test results! The linear equation was: x + (x + 2) = 14 x + x + 2 = 14 2 x + x = 14 ¡ 2x = 12 x = 6 ) ) ) ) (8.67) (8.68) (8.69) (8.70) (8.71) You scored 6 for your Biology test and Joseph scored 6 +2 = 8! 8.2 Quadratic Equations The syllabus requires: { solve quadratic equations by factorisation, completing the square and quadratic formula { identify ‘‘not real’’ numbers and how they occur. (see 2.1) 8.2.1 The Quadratic Function (NOTE: these notes have just been copied and pasted from the older structure notes and have not been written to the syllabus. it needs a serious edit and the worked example methodology needs redone (we now do inline examples and analogies with worked examples and exercises at the end of the chapter.)) A quadratic or parabolic function is a function of the form f (x) = ax2 + bx + c. (NOTE: very quick notes by sam for simple quadratics... this would be best as a decision tree. make the student appreciate that its basically trial and error to get the answer, but you can do some detective work ﬂrst to eliminate most possibilities
{ write the problem in the form ax2 + bx + c = 0 (with a positive) { write down two brackets, with an x in each, leaving room for a num- ber on each side ( x )( x ) (8.72) 136 { write out your options (in a table at the side) for multiplying two numbers together to give a. these numbers should go in front of the xs in your brackets. { if c is positive, then the other two numbers you need are either both positive or both negative. they are both negative if b is negative, and both positive id b is positive. if c is negative, it means only one of your numbers is negative, the other one beng positive. { your two numbers should multiply to give c, so write out your options if each number is multiplied by the (in a table oﬁ to the side). number in front of the x in the other bracket, then added together it gives you b. so try diﬁerent combinations of the numbers you have written). if it doesn’t work, go back to the 3rd step and try a diﬁerent combination of numbers to give you a. { once you get an answer, multiply out your brackets again just to make sure it works (sanity check). damn thats long winded!) Worked Example: Q: Draw a graph of the quadratic function y = x2 A: First let us set up a table of x and y values: 6. x ¡ ¡ x : y : -5 24 -4 14 -3 6 -2 0 -1 -4 0 -6 1 -6 2 -4 3 0 4 6 5 14 The graph of this function is shown in ﬂgure 8.1. Notice that the function 3). This shows that the x-intercepts can also be written as y = (x + 2)(x (where y = 0) are x = 2 and x = 3, which agrees with the graph. The y-intercept (where x = 0) is at y = ¡ ¡ 6. ¡ 8.2.2 Writing a quadratic function in the form f (x) = a(x p)2 + q. ¡ Consider the general form of the quadratic function y = ax2 + bx + c. Now adding and subtracting the same factor b2 4a from this expression does not
change anything. Therefore y = ax2 + bx + b2 4a ¡ b2 4a + c Taking out a factor of a then gives y = a(x2 + b a + ( b 2a )2) + c b2 4a ¡ (8.73) (8.74) 137 25 20 15 10 Figure 8.1: Graph of y = x2 x 6 ¡ ¡ The expression in brackets is then a perfect square so that y = a(x + ( b 2a ))2 + c ¡ = a(x ( ¡ ¡ b 2a ))2 + (c b2 4a ¡ b2 4a ) which can be written in the form y = a(x p)2 + q ¡ (8.75) (8.76) (8.77) where p = b 2a and q = c b2 4a. ¡ ¡ ¡ p)2 is a perfect square and therefore always positive, (x p)2 is Since (x at a minimum of 0 when x = p. This means that y is minimum (if a0) or maximum (if a < 0) when x = p and y = q. This point (p;q) is therefore called the turning point. ¡ In Now notice that the quadratic function is symmetric about x = p. v) = av2 + q for any real number v. This other words f (p + v) = f (p means that the part of the quadratic to the right of the vertical line x = p looks like the part to the left of x = p ipped about this line. Therefore we call the line x = p the axis of symmetry of the parabola. ¡ Worked Example: Q: Consider the quadratic function f (x) = into the form f (x) = a(x axis of symmetry. Plot a graph of f (x) showing all the intercepts. 5. Put this function p)2 + q and thus ﬂnd the turning point and ¡ ¡ ¡ x2 +6x 138 f (p v) ¡ v v f (p + v) x = p Figure 8.2: Graph of TODO (NOTE: original BMP (qf-gen1.bmp) missing, i made this up on the y) A: First we shall write the
quadratic in the form f (x) = a(x f (x x2 + 6x (x2 (x2 (x2 (x2 ¡ ¡ ¡ ¡ 6x + 5) 6x + 9 6x + 9) + 4 3)2 + 4 ¡ 9 + 5) p)2 + q. ¡ (8.78) (8.79) (8.80) (8.81) (8.82) ¡ Therefore the turning point is (3,4) and the axis of symmetry is x = 3 (in other words p = 3 and q = 4). ¡ Now to plot the graph we need to know the intercepts. The y-intercept 5. The x-intercepts can be found by solving the equation is y = f (0) = f (x) = 0 as follows: ¡ x2 + 6x 5 = 0 ¡ ¡ 6x + 5 = 0 ¡ x2 (x 1)(x x = 1 or x = 5 ¡ ¡ 5) = 0 ) ) ) (8.83) (8.84) (8.85) (8.86) Note that the technique of writing x2 5) is called factorisation. We shall learn more about this in the following chapter. For now just check that this is true by multiplying out the brackets. 6x + 5 = (x 1)(x ¡ ¡ ¡ Thus the graph of the quadratic function f (x) = x2 + 6x ¡ 5 is ¡ 8.2.3 What is a Quadratic Equation? An equation of the form ax2 + bx + c = 0 is called a quadratic equation. Solving this equation for x is the same as ﬂnding the roots (x-intercepts) of the quadratic function f (x) = ax2 + bx + c. 139 4 3 2 1 x1 1 2 3 4 x2 5 Figure 8.3: Graph of TODO (NOTE: original BMP (qf-eg2.bmp) missing, i made this up on the y) 8.2.4 Factorisation We have already seen examples of solving for the roots of a quadratic function by writing this function as the multiple of two brackets. For example, x2 2 or x = 3. This is called
factorising the quadratic function. 3) = 0 means that either x = 6 = (x + 2)(x ¡ ¡ ¡ ¡ x Knowing how to factorise a quadratic takes some practice, but here some general ideas which are useful. { First divide the entire equation by any common factor of the coe–cients, so as to obtain an equation of the form ax2 + bx + c = 0 where a, b and c have no common factors. For example, 2x2 + 4x + 2 = 0 can be written as x2 + 2x + 1 = 0 by dividing by 2. { Now, if ax2 + bx + c = (rx + s)(ux + v), then sv = c and ru = a. Therefore, by ﬂnding all the factors of a and c, one can try all the combinations and see if there is one which gives the correct result for b = su + rv. { Once writing the equation in the form (rx + s)(ux + v) = 0, it then follows that the two solutions are x = s r and x = u v. ¡ ¡ Worked Examples: Example 1: Q: Solve the equation x2 + 3x 4. ¡ A: Since a = 1, if this equation can be factorised, it must have the form x2 + 3x 4 = (x + s)(x + v) = x2 + (s + v)x + sv (8.87) ¡ 4, we know that s = 1, v = 4 or s = 1, 4 (excluding the options which just involve interchanging s and v, Now, as sv = v = which makes no diﬁerence to the ﬂnal answer). 2, v = 2 or s = ¡ ¡ ¡ ¡ 140 Also s + v = 3, so s = quadratic equation can be written as ¡ 1 and v = 4 is the correct combination. Thus the x2 + 3x 4 = (x ¡ ¡ 1)(x + 4) = 0 (8.88) Therefore the solutions are x = 1 and x = the roots of the quadratic function f (x) = 4. Example 2: Q: Find ¡ 2x2 + 4x ¡ ¡ 2. 2x2 + 4x 2 = 0. �
� 2. This gives A: We must ﬂnd the solutions to the equation f (x) = ¡ First we divide both sides of the equation be a factor of the equation ¡ Now, let us assume that x2 ¡ 2x + 1 = 0 (8.89) x2 ¡ 2x + 1 = (x + s)(x + v) = x2 + (s + v)x + sv (8.90) Then sv = 1 and therefore either s = v = 1 or s = v = s + v = 2, it follows that s = v = 1 and thus 1. Since ¡ ¡ ¡ ¡ x2 ¡ 2x + 1 = (x ¡ 1) = (x ¡ ¡ 1)2 = 0 (8.91) The only solution is therefore x = 1. Example 3: Q: Solve the equation 2x2 12 = 0. 5x A: This equation has no common factors, but still has a = 2. Therefore, we must look for a factorisation in the form ¡ 1)(x 2x2 5x 12 = (2x + s)(x + v) = 2x2 + (s + 2v)x + sv (8.92) ¡ ¡ We see that sv = considered below. Note: ¡ 12 and s + 2v = 5. All the options for s and v are Since we now have the factor of 2x in the ﬂrst ¡ s 2 -2 3 -3 4 -4 6 -6 v -6 6 -4 4 -3 3 -2 2 s + 2v -10 10 -5 5 -2 2 2 -2 bracket, it does make a diﬁerence, in this case, whether we interchange 6, v = 2 give the s and v values. For example, s = 2, v = ¡ diﬁerent solutions. We must therefore consider both options. 6 and s = ¡ We can see that the combination s = 3 and v = Therefore one can check that 4 gives s + 2v = ¡ 5. ¡ 2x2 ¡ 5x + 12 = (2x + 3)(x 4) = 0 (8.93) Therefore the solutions are x = ¡ ¡ 3 2 and x = 4. 141 8.2.5 Completing the Square It is not always
possible to factorize a quadratic function. We shall now derive a general formula, which gives the solutions to any quadratic equation. Consider a general quadratic equation ax2 + bx + c = 0. Adding and subtracting b2 equation. Thus 4a from the left-hand side does not change the ax2 + bx + b2 4a ¡ b2 4a + c = 0 (8.94) Taking out a factor of a from the 1st 3 terms gives a(x2 + b a a(x2 + ) b2 4a2 ) ¡ b2 4a + c = 0 + ( b 2a )2) = b2 4a ¡ c + b a (8.95) (8.96) We can now see that the term in brackets is the perfect square (x + b and therefore 2a )2 a(x + )2 = b 2a b2 4a ¡ c (8.97) Now dividing by a and taking the square root of either side gives the expression x + b 2a = §r b2 4a2 ¡ c a Finally, solving for x implies that x = b 2a § r b2 4a2 ¡ c a ¡ = b 2a § r ¡ b2 4ac ¡ 4a2 and taking the square root of 4a2 to obtain 2a gives b x = ¡ § 4ac pb2 2a ¡ (8.98) (8.99) (8.100) These are the solutions to the quadratic equation. Notice that there are two solutions in general, but these may not always exists (depending on the sign of the expression b2 4ac under the square root). ¡ Worked Examples: Example 1: Q: Solve for the roots of the function f (x) = 2x2 + 3x 7. ¡ A: One should ﬂrst try to factorise this expression, but in this case it turns out that this is not possible. Therefore we must make use of the general formula as follows: 142 4(2)( 7) ¡ b x = ¡ § (3) = ¡ 4ac pb2 2a ¡ § p (3)2 2(2) ¡ 3 = ¡ 3 = ¡ p56 § 4 2p14 § 4 (8.101) (8.
102) (8.103) (8.104) Therefore the two roots of the quadratic function are x = ¡ 3+2p14 4 and 3 2p14 ¡ 4 ¡. Example 2: Q: Solve for the solutions to the quadratic equation x2 5x + 8. ¡ A: Again it is not possible to factorise this equation. The general formula shows that b x = ¡ § 4ac pb2 2a ¡ 5)2 ( ¡ 2(1) p ( = ¡ ¡ 5(1)(8) ¡ (8.105) (8.106) (8.107) (8.108) Since the expression under the square root is negative these are not real solutions (p 7 is not a real number). Therefore there are no real solutions to the quadratic equation x2 5x + 8. This means that the quadratic function f (x) = x2 5x + 8 has no x-intercepts, but the entire function ¡ lies above the x-axis. ¡ ¡ Note to self: maybe add quadratic example about distance, velocity and acceleration... object falling under action of gravity (giving formula for distance as a function of time)?. 8.2.6 Theory of Quadratic Equations What is the Discriminant of a Quadratic Equation? Consider a general quadratic function of the form f (x) = ax2 + bx + c. The discriminant is deﬂned as ¢ = b2 4ac. This is the expression under the square root in the formula for the roots of this function. We have already seen that whether the roots exist or not depends on whether this factor ¢ is negative or positive. ¡ 143 The Nature of the Roots Real Roots: Consider ¢ case there are solutions to the equation f (x) = 0 given by the formula 0 for some quadratic function f (x) = ax2 + bx + c. In this ‚ b x = ¡ § pb2 2a 4ac b = ¡ ¡ p¢ § 2a (8.109) since the square roots exists (the expression under the square root is nonnegative.) These are the roots of the function f (x). There are various possibilities: Equal Roots: If ¢ = 0, then the roots are equal and, from the formula
, these are given by x = b 2a ¡ (8.110) Unequal Roots: There will be 2 unequal roots if ¢ > 0. The roots of f (x) are rational if ¢ is a perfect square (a number which is the square of a rational number), since, in this case, p¢ is rational. Otherwise, if ¢ is not a perfect square, then the roots are irrational. Imaginary Roots: If ¢ < 0, then the solution to f (x) = ax2 + bx + c = 0 contains the square root of a negative number and therefore there are no real solutions. We therefore say that the roots of f (x) are imaginary (the function f (x) does not intersect the x-axis). Summary of Cases: { Real Roots (¢ 0) ‚ Equal Roots (¢ = 0) Unequal Roots (¢ > 0) ⁄ ⁄ Rational Roots (¢ a perfect square) Irrational Roots (¢ not a perfect square) ¢ ¢ { Imaginary Roots (¢ < 0) Note to self: maybe add pictures showing these cases graphically? 144 Worked Examples: Example 1: Q: Consider the function f (x) = 2x2 + 5x equation f (x) = 0, discuss the nature of the roots of f (x). A: We need to calculate and classify ¢ = b2 for the roots. ¡ ¡ 11. Without solving the 4ac according to the cases ¢ = (5)2 ¡ 4(2)( 11) = 25 + 88 = 113 ¡ (8.111) Now ¢ is positive, so the roots are real and unequal. Also, since 113 is not a perfect square, the roots are irrational. Example 2: Q: Consider the quadratic function f (x) = x2 + bx + (2b some constant. Classify the roots of this function as far as possible. ¡ 5), where b is A: Let us calculate the discriminant ¢ = b2 4(1)(2b 5) = b2 8b + 20 (8.112) ¡ We shall now use a useful trick, which is to write the above expression as a perfect square plus a number. ¡ ¡ ¢ = b2 8b + 20 = (b2 8b + 16) + 4 = (b 4)2 + 4 (
8.113) ¡ 0 because this is a perfect square. Therefore we know that ¡ ¡ Now (b ¢ ¡ 4 > 0. 4)2 ‚ ‚ We can thus say that f (x) has real unequal roots. We do not know whether ¢ is a perfect square, since we do not know that value of the constant b, and therefore we cannot say whether the roots are rational or irrational. 8.3 Cubic Equations The syllabus requires: { (grade 12) solve cubic equations using factor theorem ‘‘and other techniques’’ (NOTE: SH: i want to hit whoever wrote this syllabus) 8.4 Exponential Equations The syllabus requires: { solve exponential equations { (grade 12) switch between log and exp form of an equation 145 8.5 Trigonometric Equations The syllabus requires: { solve trig equations (NOTE: this implies that this section comes after trig, which is also probably after geometry) 8.6 Simultaneous Equations The syllabus requires: { solve simultaneous equations algebraically and graphically 8.7 Inequalities The syllabus requires: { solve linear inequalities in 1 and 2 variables and illustrate graphically { solve quadratic inequalities in 1 variable and illustrate graphically 8.7.1 Linear Inequalities Let us say that we are given a general inequality as follows: 0; ‚ ax + by + c ax + by + c > 0; ax + by + c < 0 (8.114) Now there are many possible values of x and y, for which this may be true (these will depend on the values of the constants a, b and c). The set of all the (x;y) values which satisfy this inequality is called the solution set. ax + by + c or • 0 We shall now see how to draw the solution set on a graph. Let’s consider the following example. Worked Example 1: Q: Find the solution set of the inequality 2x + y ¡ A: First we solve for y by writing the inequality as 3 0. ‚ y 2x + 3 ‚ ¡ (8.115) 2x+3 is a straight line and the points (x;y), which Now the function y = satisfy the inequality are therefore all the points above the line. These can be drawn as ¡ The shaded section on the graph, which shows the solution set, is called the
feasible region. Now sometimes x and y must satisfy more than one inequality. In this case, we consider each inequality separately and then the feasible region is where the feasible regions of each inequality overlap. 146 3 2 1 3 2 1 2 3 4 Figure 8.4: Graph of y = 2x + 3. Points satisfying y ¡ 2x + 3 are shaded ‚ ¡ Worked Example 2: Q: Graphically represent the solution set for the following inequalities: ¡ 2x + 10 • ¡ ¡ A: Solving for y gives 1 y y < 2x + 5 ‚ y x + 10 • ¡ (8.116) (8.117) (8.118) (8.119) (8.120) (8.121) Now we draw the solution set to each of the inequalities separately and ﬂnd the region where these overlap as shown below. We draw the line y = 2x + 5 as a dashed line because the inequality is < and not (the line is not included in the feasible region). • 8.7.2 What is a Quadratic Inequality A quadratic inequality is an inequality of the form ax2 + bx + c > 0, ax2 + bx + c 0, ax2 + bx + c < 0 or ax2 + bx + c 0. ‚ • 8.7.3 Solving Quadratic Inequalities Solving a quadratic inequality corresponds to working out in what region a quadratic function lies above or below the x-axis. Here are some examples showing how this is done. 147 10 10 Figure 8.5: Graph of TODO Worked Examples: Example 1: Q: Find all the solutions to the inequality x2 A: Consider the function f (x) = x2 f (x) ‚ 5x + 6. We need to ﬂnd out where 0; in other words, where the function f (x) lies above/on the x-axis. 5x + 6 ¡ ¡ 0. We shall ﬂrst work out where f (x) intersects the x-axis by solving the equation which can be factorised to give x2 ¡ 5x + 6 = 0 (x ¡ 3)(x ¡ 2) = 0 (8.122) (8.123) The x-intercepts are therefore x1 = 2 and x2 =
3. We can see from ﬂgure 8.6 that f (x) is above/on the x-axis when x or x 2. 3 ‚ Therefore the solution to the quadratic inequality is or in interval notation ( [3; ;2] ). x : x f ‚ 3 or x 2 g • ¡1 [ 1 Note: The x-intercepts are included in this solution, since the f (x) inequality includes the solution f (x) = 0. 0 • ‚ • Example 2: Q: Solve the quadratic inequality Let f (x) = A: quadratic equation ¡ x2 ¡ x2 ¡ ¡ 3x + 5 > 0. 3x + 5. The x-intercepts are solutions to the 148 8 6 4 2 x1 2 x2 3 1 Figure 8.6: Graph of f (x) = x2 5x + 6 ¡ x2 ¡ x2 + 3x 3x + 8.124) (8.125) which has solutions (using the formula for the roots of a quadratic function (NOTE: reference that equation)) given by 4(1)( 5) ¡ 3 x = ¡ § (3)2 ¡ 2(1) p p29 3 = ¡ § 2 3 x1 = ¡ x2 = ¡ p29 ¡ 2 3 + p29 2 (8.126) (8.127) (8.128) (8.129) The graph of f (x) is shown in ﬂgure 8.7. The points x1 and x2 (where the function f (x) cuts the x axis) are labelled. 3 Now f (x) > 0 (the function is above the x-axis) when ¡ p29 ¡ 2 < x < 3+p29 2 ¡. Therefore the solution to the inequality is 3 or in interval notation ( ¡ 3+p29 2 Note: The x-intercepts are not included in the solution because the > sign has been used and therefore f (x) = 0 does not deﬂne a solution to the inequality. < x < ¡ p29 ¡ 2, g ; ¡ ). 3 x : ¡ f 3+p29 2 p29 ¡ 2 Example 3: Q: Solve the inequality 4x2 4x + 1 0
. • ¡ 149!! 6 4 2 x1 4 ¡ 3 ¡ 2 ¡ 1 ¡ x2 1 Figure 8.7: Graph of f (x) = x2 ¡ ¡ 3x + 5 1:5 1:0 0:5 0:5 1:0 Figure 8.8: Graph of f (x) = 4x2 4x + 1 ¡ A: Let f (x) = 4x2 f (x) = (2x ¡ 4x + 1. Factorising this quadratic function gives 1)2, which shows that f (x) = 0 only when x = 1 2. The function f (x) lies below/on the x-axis only at the x-intercept. Therefore the only solution to the inequality is x = 1 2. ¡ 8.8 Intersections The syllabus requires: { find solutions of 2 lines and interpret the common solution as the intersection 150 " " " # # { find solutions of linear and quadratic, interpret the common solution(s) as intersections 151 Chapter 9 Working with Data 9.1 Statistics The syllabus requires: { organise univariate numerical data to determine measure of central tendency; mean, median, mode and when each is appropriate. measure of dispersion; ranger, percentiles, quantities, interquartile, semi-interquartile range { represent data effectively, choosing from; bar and compound bar graphs, histograms, frequency polygons, pie charts, dot plots, line and broken line graphs, stem and leaf plots, box and whisker diagrams { (grade 12) variance, standard deviation { (grade 12) draw a suitable random sample from a populations, and understand the importance of sample size in predicting the mean and standard deviation { (grade 12) identifies data which is normally distributed about a mean by investigating appropriate histograms and frequency polygons 9.2 Function Fitting The syllabus requires: { represent bivariate data as a scatter plot and suggest what function (linear, quadratic, exp) would best describe it { tell the difference between symmetric and skewed data and make relevant deductions { (grade 12) use appropriate technology to calculate linear regression line which best fits a given set of bivariate data 152 9.3 Probability The syllabus requires: { Venn diagrams to solve probability problems. must know P (s) = 1qquadP (AorB) = P (A) + P (B) P (
AandB) ¡ { identify dependent and independent events and calculate the prob of 2 independent events occurring by applying the product rule for independent events P (AandB) = P (A):P (B) { identify mutually exclusive events. calculate prob of the events occuring by applying additive rule for mutually exclusive events P (AorB) = P (A) + P (B) { identify complementary events P (notA) = 1 { use prob models for comparing experimental results with theory; P (A) ¡ need many trials to get comparable results... coin example flipping a { comparing experimental results with each other { potential sources of bias, error in measurement, potential uses and misuses of stats and charts (NOTE: if SA people could get some popular TV adverts as examples, that would be good) { converts this theory into a project (NOTE: i don’t know how much of this project stuff we should do in the book. just ignore its existence) possibly 9.4 Permutations and Tree Diagrams The syllabus requires: { tree diagrams and other methods of listing all options to generalise counting principle (successive choices) { calculate the probability of compound events which are not independent { assess the odds in a variety of games of chance, lotteries, raffles { (grade 12) use investigate and solve problems involving the number of arrangements (permutations) of a number of discrete objects (when order matters) m! (m different items), m items selected from n { (grade 12) investigate and solve problems involving the number of possible solutions when order is not important (combinations) of m items from n where all are different or distinguishable { (grade 12) uses permutations and combinations to correctly calculate the probability of specified events occurring { determines the odds of various games of chance and the probability of events which depend on combinations and permutations 153 9.5 Finance The syllabus requires: { use simple and compound interest for relevant problems (hire purchase and inflation) { effective and nominal interest { understand fluctuating foreign exchange rates and their effect on local prices, travelling prices, imports and exports { solve straight line (simple) depreciation and depreciation on a reducing budget (compound depreciation) { (grade 12) apply geometric series to solve problems (future values of annuities, bond repayments, sinking fund contributions including the difference in time taken to pay when the monthly payment is changed) { (grade 12) critically analyse investment and loan options and make informed decisions
to the best options (pyramid and micro lenders schemes) 9.6 Worked Examples TODO 9.7 Exercises TODO 154 Part II Old Maths 155 Chapter 10 Worked Examples 10.1 Exponential Numbers (NOTE: All of these worked examples need to be updated to use the FHSST internal environments and also to use the new rules (i changed the originals). They are for the Exponential Numbers section) Worked Example 9 : Manipulating Exponential Num- bers Question: Simplify the expression 42:33 63 2 = 22 and 6 = 2:3 it follows that A: Noting that 4 = 2 £ 42:33 63 = = (22)2:32 (2:3)3 24:33 23:33 3 ¡ = 24 = 21 = 2 (10.1) (10.2) (10.3) (10.4) (10.5) Example 2: Q: Simplify ( 5 A: First note that 20 = 2 2 )2:20. 2 £ £ 5 = 22:5. Therefore 156 ( 5 2 )2:20 = 52 22 :22:5 = 52+1 = 53 = 5 £ = 125 5 5 £ (10.6) (10.7) (10.8) (10.9) (10.10) Example 3: Q: Solve for the variable x in the equation 2x+1 = 2x + 8. A: ¡ 2x+1 = 2x + 8 2x = 8 2x:2 2x(2 1) = 8 ¡ 2x = 8 = 2 x = 3 £ 2 (10.11) (10.12) (10.13) (10.14) (10.15) 2 = 23 £ ) ) ) ) It is also possible to talk about zero, negative and even fractional exponents. We shall assume that laws 1{5 are also true in these cases. This gives us 3 more laws. a0 = a1 1 = ¡ a1 a1 = a a = 1 Law 7 Since n = 0 ¡ ¡ 1 an n, it follows from laws 2 and 6 that n = a¡ a¡ n = a0 ¡ n = a0 an = 1 an (10.16) (10.17) This deﬂnes what is meant by a negative exponent. Note to self: add aside lay-out to
following paragraph Aside: A fraction to the power of a negative exponent is the same as the inverse fraction to the power of the corresponding positive exponent. Therefore ( a b )¡ n = 1 a b = a an bn = ( b a )n (10.18) 157 Example 1: Q: Simplify the expression ( 1 A: 2 )¡ 3. ( 1 2 )¡ 3 = 1 ( 1 2 )3 = 1 1 23 = 23 = 8 (10.19) Example 2: Q: Simplify (27) 1 3 : A: Now 27 = 3 £ q 3 £ 2 9. 3 = 33 and 9 = 3 3 = 32, so £ (27) 1 3 : q 2 9 = (33) 1 3 : 2 r 32 2 32 ) 1 2 = 33: 1 3 :( 1 2 1 2 = 31:2 (32) 3:p2 3 = p2 = Example 3: Q: Simplify (an + an) A: 1 2 (an + an) 1 2 = (2an) 1 2 1 2 1 2 :(an) = (2) = p2 an: 1 = p2 a n 2 2 (10.20) (10.21) (10.22) (10.23) (10.24) (10.25) (10.26) (10.27) (10.28) Example 4: Q: Find a solution to the equation ax b are constants. A: m n ¡ b = 0, where a and ) ) ) ) ) b = 0 m n ax ax ¡ m n = b b a xm = (x m n = x m n )n = ( x = (xm) 1 m = (( )n b a b )nn ( r 158 (10.29) (10.30) (10.31) (10.32) (10.33) (10.34) Therefore x = m 0. q ( b a )n is a solution to the equation ax m n b = ¡ (NOTE: surds) Worked Example: Q: Simplify the expression p32 2. A: p32 2 = p32 p4 32 4 = r = p8 = p4:2 = p4:p2 = 2p2 (10.35) (10.36) (10.37) (10.38) (10.39) (10.
40) Worked Example: Q: Which of the numbers 3p100 and p20 is bigger? (You may not use a calculator to answer this question.) A: The two numbers must ﬂrst be converted into like surds. Since we have a cube root and a square root, we must ﬂrst ﬂnd the lowest common multiple of 2 and 3 which is 6. We then convert each of the surds into 6th roots as follows: 3p100 = 3 q p1002 = 3:2p1002 = 6p10000 (10.41) p20 = 2 3p203 = 2:3p203 = 6p8000 q (10.42) Now, since 100000 is bigger than 8000, it follows that 6p10000 is bigger than 6p8000. Therefore 3p100 is bigger than p20. Worked Examples: Example 1: Q: Rationalize the denominator of the fraction 1 p6. A: The denominator can be changed into the rational number 6 by multiplying the numerator and denominator by p6. Therefore 1 p6 = 1 p6 £ p6 p6 = p6 6 159 (10.43) Example 2: Q: Rationalize the denominator of p2 (p3+p8). A: Multiplying the numerator and denominator by (p3 p8) gives ¡ p2 (p3+p8) = = = = = (p3 (p3 p8) p8) ¡ ¡ p2 (p3 + p8) £ p8) p2(p3 3 p2p3 ¡ 8 ¡ p2p8 ¡ 5 ¡ p16 ¡ 5 p6 ¡ 5 p6 ¡ 4 (10.44) (10.45) (10.46) (10.47) (10.48) 1.1.4 d) Equations Here we shall solve some equations involving surds. Worked Examples: Example 1: Q: Find a solution to the following equation x + 3 ¡ p6x + 13 = 0 (10.49) A: First the square root must be moved to the right-hand side of the equation, and everything else to the left. x + 3 = p6x + 13 (10.50) Now we square both sides of the equation. (x + 3)2 = 6x + 13
x2 + 6x + 9 = 6x + 13 x2 = 4 x = 2 or x = 2 ¡ ) ) ) (10.51) (10.52) (10.53) (10.54) When we square both sides of the equation, it is possible that we introduce extra solutions, which may not actually satisfy the original equation. This is the reason one should always check the answers by substituting these into the original equation. x = 2: 160 p6x + 13 = 2 + 3 6(2) + 13 = 5 p25 = 5 ¡ ¡ 5 = 0 (10.552: x+3 p6x + 13 = 2+3 6( 2) + 13 = ¡ ¡ ¡p Therefore, in this case, both the solutions x = 2 and x = to the original equation ¡ ¡ 1 ¡ p1 = 1 1 = 0 (10.56) ¡ 2 are solutions ¡ Note to self: Maybe the following example should only be included after the chapter on quadratics? But need non-trivial example where not all solutions valid. Example 2: Q: Solve the equation px + 7 = x ¡ 1 ¡ (10.57) As before, we ﬂrst move the surd to the right-hand side and the other terms to the left. x + 1 = px + 7 (10.58) Squaring both sides of the equation gives (x + 1)2 = x + 7 x2 + 2x + 1 = x + 7 x2 + x ¡ (x + 3)(x 2 or x = 2 ¡ ) ) ) ) (10.59) (10.60) (10.61) (10.62) (10.63) Again we must check these answers: x = -3: px + : 3 + 7 = p ¡ p4 = 10.64) ¡ px + 7 = 2 x ¡ p2 + 7 = 2 p9 = 2 3 = 1 ¡ ¡ ¡ ¡ (10.65) Therefore x = only solution. 3 does not satisfy the original equation, so x = 2 is the ¡ 161 6 Scientiﬂc Notation Example 1: The speed of light in a vacuum c is c = 2:9979 = 2:9979 £ £ 108 m s 100000000 m s (
10.66) (10.67) (10.68) = 299790000 m s Example 2: The approximate radius of the hydrogen atom (called the Bohr radius a0) is a0 = 5:3 = 5:3 = 5:3 £ £ £ 11 m 10¡ 1 1011 m 1 10000000000 0:00000000001 m m = 5:3 = 0:000000000053 m £ (10.69) (10.70) (10.71) (10.72) (10.73) Note to self: check the above number... I think a0 = 0:53”A?... Example 3: Q: The universe is known to be 13.7 billion years old. Convert this number into scientiﬂc notation. A: The age of the universe is 13700000000 years (since 1 billion is a thousand million or 1000000000). The decimal point must move 10 places to the left to convert this number into 1.37. Since 13700000000 is much bigger than 1.37, the power 10m must make the number 1.37 bigger and therefore m must be positive (so that 10m is greater than 1). Therefore m = 10 (the decimal point moves 10 places). Therefore 13700000000 = 1:37 1:37 1010 years. £ 1010. The age of the universe is thus £ Example 4: Q: The charge on an electron is e = 0:0000000000000000001602 C. What is this constant in scientiﬂc notation. A: The decimal point must be moved 19 places to the right to change this number into 1.602 (so a = 1:602). Now e is much smaller than 1 and thus 162 the exponent m must be negative, so that 10m is less than 1. Therefore m = 19. ¡ Therefore e = 1:602 10¡ 19 C. £ Note: As a general rule, if the decimal point must move to the left then the exponent m is positive and if the decimal point moves to the right then m is negative. patterns sequences Worked Example 10 : Calculating the nth Term of a Sequence ¡ is given by 2 + 4(n Question: Check that the formula for the nth term of the sequence 1), and calculate 2;6;10;14;18;::: g f the thousandth term. Answer: The sequence we are given starts at two, and each term is equal to the previous term
plus four. We can check whether the formula is valid by going through the ﬂrst few terms, and seeing whether the terms in the sequence correspond to the If we substitute n = 1 into the terms given by the formula. formula, we should get the ﬂrst term of the sequence, and this is indeed the case: 2 = 2 + 4(1 1). If we substitute n = 2, we ¡ get the second term, namely 6. We can continue in this way, substituting n = 3, n = 4, and so on - each time we get the expected value back. This means that the formula is indeed valid. To calculate the thousandth term, we must substitute 1000 into the formula. When we do so we get 2 + 4(1000 1) = 2 + 4(999) = 3998. ¡ Worked Example 11 : Calculating the Sequence When Given the Formula ¡ Question:The formula for the nth term of a sequence is given 1).Write down the ﬂrst four terms of the sequence, and by 2(2n describe the pattern that you observe. Answer:This question is similar to the one in the previous example. To get the ﬂrst term of the sequence, we must substitute 1 into the formula. Doing so gives us 2(21 1) = 2(20) = 2(1) = 2. To ﬂnd the second term we must substitute two into the sequence: 2(22 1) = 2(2) = 4. For the third and fourth terms we substitute three and four respectively, and so we ﬂnd that the ﬂrst four terms of the sequence are 2;4;8;16;:::. >From these four terms we can see that every term in the sequence is equal to the previous term multiplied by two. ¡ ¡ 163 Worked Example 12 : Calculating the Sequence When Given the Formula Question:The ﬂrst term of a sequence is 3, and the formula for the (n + 1)th term is given by an+1 = an + 2n. Write down the ﬂrst four terms of the sequence. Answer:We start by calculating the second term, since we already know the ﬂrst. To do this we need to substitute n = 1 into the formula, since the formula is for the (n + 1)th term, NOT
for the nth term, as was previously the case. So, substituting n = 1, we get that a1+1 = a1 + 2(1) = 3 + 2 = 5: To get the next term, we must substitute n = 2: a2+1 = a2 + 2(2) = 5 + 4 = 9. Lastly, we calculate that a4 = 9 + 2(3) = 15. So the ﬂrst four terms of the sequence are 3;5;9;15;:::. Worked Example 13 : Checking That a Given Se- quence is Arithmetic ¡ Question:Check that the sequence given by the formula an = 1) is an arithmetic sequence, and ﬂnd d for this se3 + 4(n quence. Answer:We must check to see that the diﬁerence between successive terms is a constant. There are two ways of doing this: we could write out the ﬂrst few terms of the sequence and check that they are evenly spread - i.e. they diﬁer by a constant amount, or we could do the calculation in general, using the formula directly. We will do the example using both these approaches alternately. ¡ ¡ 3 = 11 7 = 15 11 = 19 { First approach: It is easy to calculate the ﬂrst few terms using the formula. They are given by 3;7;11;15;19;:::. We can see that the diﬁerence between successive terms is always 4, since 7 15 = 4, so the sequence is indeed an arithmetic sequence, and d = 4 { Second approach: We know that the formula for the nth term is given by an = 3 + 4(n 1). From this it should be clear that the (n + 1)th term is given by an+1 = 3 + 4((n + 1) If we work out the diﬁerence between successive terms, we get that an+1 ¡ 1) = 4n we got using the previous method. ¡ 4n + 4 = 4 = d, which is the same answer that 1) = 3 + 4n. an = 3 + 4n 4(n ¡ ¡ ¡ ¡ ¡ ¡ ¡ 3 Worked Example 14 : Calculating the Formula for the nth term of a Sequence Question:Find a formula
for the nth term of the sequence 6; 17; 28; 39;:::. Which term in the sequence equals 688? 164 Answer:First we must check that the given sequence is in fact an arithmetical one, otherwise we can’t use the formula. This is easily seen, since the diﬁerence between successive terms is 11. Now we must use (??). We know that a1 = 6 and that If we substitute these into the formula, we see that d = 11. an = 6 + 11(n 1). Lastly, we must ﬂnd which term equals 688. Notice that in the ﬂrst worked example we were given an n (namely 1000), and asked to calculate an. Now we are given an an and asked to calculate n. We can do this easily by rearranging the formula: ¡ an = 688 = 6 + 11(n 6 = 11(n 6 ¡ 688 688 ¡ 11 = n 1) 1) ¡ ¡ 1 ¡ n = 6 688 ¡ 11 + 1 = 63 We conclude that the 63rd term will equal 688. geometric Worked Example 15 : Checking That a Given Se- quence is Geometric ¡ 1) is a geometric sequence, and ﬂnd r for this sequence. Question:Check that the sequence given by the formula an = 2(3n Answer:We must check to see that the ratio between successive terms is a constant. As in example four, there are two ways of doing this: we could write out the ﬂrst few terms of the sequence and check that successive terms diﬁer by a common factor, or we could do the calculation in general, using the formula directly. We will do the example using both these approaches alternately. { First approach: It is easy to calculate the ﬂrst few terms using the formula. They are given by 2;6;18;54;162;:::. We can see that the ratio between successive terms is always 3, since 6 6 = 54 54 = 3, so the sequence is indeed an arithmetic sequence, and r = 3 18 = 162 2 = 18 { Second approach: We know that the formula for the nth 1). From this it should be clear term is given by an = 2(3n ¡ that the (n + 1)th term is given by an+1 =
2(3n). If we work out the ratio between successive terms, we get that an+1 an = 2(3 n+1 = 3, which is the same answer that we got using the previous method. 2(3n¡1) = 3n ¡ n ) Worked Example 16 : Calculating the Formula for the nth term of a Sequence 165 Question:Find a formula for the nth term of the sequence 2;4;8;16;:::. Which term in the sequence equals 8192? Answer:First we must check that the given sequence is in fact a geometric one, otherwise we can’t use the formula. This is easily seen, since the ratio between successive terms is 2. Now we must use (??). We know that a1 = 2 and that r = 2. If 1). we substitute these into the formula, we see that an = 2(2n Lastly, we must ﬂnd which term equals 8192. We can do this easily by rearranging the formula: ¡ a n = 688 (688 688 = 6 + 11(n = = ¡ 6)=11 + 1 = 6 ¡ 6)=11 11(n n ¡ 63 ¡ 1 1) ¡ 1) n = (688 ¡ We conclude that the 13th term will equal 8192. 10.2 series Worked Example 17 : Calculating Sn Question:Calculate S4 for the series 2 + 4 + 8 + 16 + 32 + :::. Answer:Recall that S4 is the sum of the ﬂrst four terms of the series. This is given by 2 + 4 + 8 + 16 = 30. Worked Example 18 : Calculating a Series Question:Calculate ﬂrst ﬂve terms of the series which corresponds to the sequence an = 2 + 2(n Answer:First we calculate the ﬂrst ﬂve terms of the sequence. They are 2,4,6,8,10. To get the corresponding series, we simply need to put addition signs in between the terms: 2+4+6+8+10. 1). ¡ Worked Example 19 : Checking Sn for a given series 2 [4 + 2(n Question:For the series 2+4+6+8+:::, check that the formula for the sum of the ﬂrst n terms is given by Sn
= n 1)] Answer:Let us start by writing out the ﬂrst few terms of Sn. S1 equals the ﬂrst term of the series, so S1 = 2. S2 is the sum of the ﬂrst two terms, so S2 = 2 + 4 = 6. S3 is the sum of the ﬂrst three terms, namely 12. Continuing in this fashion, we can see that the ﬂrst few terms of Sn are 2;6;12;20;28;:::.What we need to determine is whether these correspond to the given formula for Sn, and indeed they do. We must simply note that when we substitute 1 into n 1)], we get 2, when we substitute 2, we get 6, when we substitute 3, we get 12, then 20, then 28, 2 [4 + 2(n ¡ ¡ 166 and so on. This means that the formula does indeed give us the sum up to n terms. Worked Example 20 : Calculating the Sequence that Corresponds to a Given Sn ¡ 4n. Find the sequence which corresponds to this series. Question:For a certain series, Sn is given by the formula Sn = 2n2 Answer:Let us start by writing out the ﬂrst few terms of Sn: 2;0;6;16;30;:::. Think carefully about what this means - S1 = ¡ 2 means that the ﬂrst term of the series is -2, S2 = 0 means ¡ that the sum of the ﬂrst two terms is 0. Therefore the second term must be 2, since when we add 2 to -2 we get 0. S3 = 6 means that the sum of the ﬂrst three terms of the series is 6, so, by similar reasoning, the third term must be 6. S4 = 16 means that the fourth term must be 10. Now one should start to see the pattern - the ﬂrst few terms of the sequence which corresponds to this series are 2;2;6;10;:::, so we are dealing ¡ with an arithmetical sequence that has a common diﬁerence of 4. Worked Example 21 : Calculating Sn for a Given Series Question:Calculate the value of the series 1 + 4 + 7 + 10 + 13 + ::: + 46
. Answer:We wish to ﬂnd S16 (since 46 is the 16th term of the series). Of course we can do this with a calculator, but there is a much quicker way. S16 = 1 + 4 + ::: + 43 + 46 S16 = 46 + 43 + ::: + 4 + 1 2S16 = 47 + 47 + ::: + 47 + 47 2S16 = 16 47 = 752 £ S16 = 752 2 = 376 Worked Example 22 : Using the Formula for Sn Question:Find the sum of all the integers between 1 and 100, i.e. ﬂnd 1+2+3+...+99+100. Answer:Since we are dealing with an arithmetic series,we can use the formula for Sn that we have derived. In order to use it we need to know which values to put in for a1, d, and n. It should Since the series starts at 1, we know that a1 = 1. 167 also be clear that d = 1 (since the common diﬁerence between successive terms is 1), and that n = 100 (since we are summing up to the hundredth term). Now it is just a question of plugging these values into the formula. S100 = n S100 = 100 S100 = 50(101) = 5050 2 [2a1 + (n ¡ 2 [2 + (100 1)d] 1)] ¡ Worked Example 23 : Calculating Sn for a Geomet- ric Series in General Question:Calculate a formula for Sn for any geometric series, given r and a1. Answer:This example is slightly diﬁerent from the case for arithmetic series. Try writing out the calculation for yourself to make sure that you understand all the steps. r £ Sn = a1 + a1r + a1r2 + ::: + a1rn Sn = a1r + a1r2 + ::: + a1rn ¡ 2 + a1rn 2 + a1rn ¡ 1 + a1rn 1 rSn ¡ Sn = Sn(r ¡ ¡ ¡ ¡ a1 + 0 + 0::: + 0 + a1rn 1) = a1rn a1 ¡ a1(rn r ¡ = 1, otherwise we would have This formula is only valid when r 0 in the denominator. When r = 1, Sn =
a1 +a1 +:::+a1 = na1. (10.74) Sn = ¡ 1 1) Worked Example 24 : Using the Formula for Sn Question:What is 2 + 4 + 8 + 16 + ::: + 32768? Answer:We are dealing with a geometric series, so we have to use equation (10.74). In order to use it we need to know which values to put in for a1, r, and n. Since the series starts at 2, we know that a1 = 2. It should also be clear that d = 2 (since the common ratio between successive terms is 2). It is not so clear what n should be, but we can work it out using the equation for the nth term of a geometric series. 32768 = 2(2n 1 an = a1rn ¡ 1) = 2n ¡ log 32768 = n log 2 log 32768 log 2 = 15 n = 168 6 Now that we have values for a1,n, and r, we can use the formula to work out S15. n Sn = a1(r r ¡ S15 = 2(215 1) 1) ¡ 1 ¡ 1 S15 = 2(215 ¡ 2 ¡ 1) = 65534 Worked Example 25 : Sigma Notation Question:Calculate the values of the following expressions: { { { { 7 k=2 k + 1 4 t=2 t2 4 t=2 2t 4 k=1 3 P P P Answer: P { We have to sum the expression k +1 from k = 2 until k = 7: (1+1)+(2+1)+(3+1)+(4+1)+(5+1)+(6+1)+(7+1) = 35. 4 t=2 t2 = 22 + 32 + 42 = 4 + 9 + 16 = 29 4 t=2 2t = 22 + 23 + 24 = 4 + 8 + 16 = 28 4 k= = 12 { { { P P P Worked Example 26 : Converging or diverging Question:Which of the following inﬂnite series do you think will converge, and which do you think will diverge: { { { { P P P 1k=1 k 1 1k=1 k 1 1k=1 k2 1k=1( ¡ 1)(k + 1) P Answer:
We don’t really have any systematic way of working this out yet, but we can easily guess by using our calculators. { Working out the ﬂrst few terms of the sum, we get S1 = 1, S2 = 1 + 2 = 3, S3 = 1 + 2 + 3 = 6, S4 = 1 + 2 + 3 + 4 = 10, S5 = 1 + 2 + 3 + 4 + 5 = 15. Clearly this is getting larger and larger, and it would be reasonable to guess that if we kept on adding terms, we would not get a ﬂnite number. So this series diverges. { As before, we can work out the ﬂrst few terms by hand: 6, S4 = 2 1 S1 = 1, S2 = 1 + 1 12, S5 = 2 17 60, and, using a calculator, S20 = 3:6. This is still not absolutely clear, so now we can write a computer program It turns out that to work out even higher values of Sn. 2, S3 = 169 S100 = 5:19, S1000 = 7:49, and S1000000 = 14:4. This series is in fact divergent. Even though it grows at a very slow rate, it never stops getting larger when we add more terms. { As we mentioned in the text above, this series never gets larger than 1, no matter how many terms we care to add. In fact, the more terms we add, the closer the series gets to 1 (try it on your calculator), so we say that the series converges to 1. 1 = 0, S3 = 1 { This example seems a bit strange at ﬂrst. Let us try to write out a few terms of Sn to give us an idea of what is happening: S1 = 1, S2 = 1 1+1 = 1, S4 = 0, S5 = 1. It seems that the values we get are oscillating between 0 and 1. Remember we said that a series converges if we get closer and closer to some number when we add more terms - and that is clearly not what is happening here. That means that this series is divergent, even though it never gets larger than 1. ¡ ¡ Worked Example 27 : Using the formula for S 1 Question:Calculate the sum of the inﬂnite series Answer:
We are asked to determine the sum of the inﬂnite series, which is 1 and 1, so we can use the formula. The formula between ¡ = a gives S 1 ¡ P 8 + :::. The common ratio is clearly 1 r = 1 = 2. 1 ¡ 1k=1( 1 2 )k 1 2 1 1 ¡ 10.3 functions Q: State the domain and range of the function y = x2 notation and interval notation. ¡ 4 in set-builder A: There is nothing to stop x from taking on the value of any real number, but, since x2 cannot be negative, we see that y 4. Thus the domain of the function is ‚ ¡ x : x † R g f or ) ; ( 1 ¡1 and the range is given by y : y f ‚ 4 and y † R g or [ ¡ 4; ) 1 (10.75) (10.76) Worked Example: Q: Plot a graph of the function f (x) = A: The x-intercept is x + 1. ¡ 170 x + 1 = 0 x = 1 ¡ ) The y-intercept is ¡ Therefore the graph is as follows: y = (0) + 1 = 1 (10.77) (10.78) (10.79 Figure 10.1: Graph of f (x) = 1 x ¡ Now we have seen that, in general, the constant b is the y-intercept, but what is a? The bigger a the faster the y-values change when the x values change. Therefore a is called the slope and shows how steep the straight line is. Worked Example: Example 1: Q: We are given a straight line graph f (x) = ax + b and two diﬁerent = x2. What is the slope of the line? points (x1;y1) and (x2;y2), where x1 6 A: Now we need to calculate the slope a and we know that y1 = ax1 + b and y2 = ax2 + b, since (x1;y1) and (x2;y2) are points on the straight line. This means that y2 ¡ y1 = (ax1 + b) = ax2 ¡ = a(x2 ¡ ¡ ax1 + b x1) (
ax2 + b) b ¡ (10.80) (10.81) (10.82) 171 $ $ Therefore the slope describes the change in y (sometimes called ¢y = x1) between any two y2 ¡ diﬁerent points on the line, i.e. y1) divided by the change in x (¢x = x2 ¡ a = ¢y y2 ¡ y1 x2 ¡ x1 ¢x = 0) because one cannot x1 6 We have used the fact that x1 6 = x2 (i.e. x2 ¡ = y2 gives an inﬂnite slope divide by zero. (The case of x1 = x2 but y1 6 and this describes a vertical line at constant x.)(NOTE: This section needs to work in the relevance of ﬂgure 10.2 more) (10.83) = y2 y1 ¢y ¢x x1 x2 Figure 10.2: Graph showing ¢x and ¢y for a line of the form y = ax + b Note: A straight line with a positive slope (a > 0) increases from left to right and a straight line with a negative slope (a < 0) increases from right to left. a > 0 a < 0 Figure 10.3: A straight line with positive slope (a > 0) and a straight line with negative slope (a < 0) Example 2: Q: Consider the straight line shown in the graph below: 172 % % Figure 10.4: Graph of straight line with y0 = 2 and x0 = 5 ¡ Find the equation of the straight line describing this graph. A: We need to ﬂnd f (x) = ax + b, so we need the a and b values for this line. We see that the y-intercept is y = 2. Since we have two points (the y-intercept (0,-2) and the x-intercept (5,0)), we can ﬂnd the slope using the equation from the previous example. Therefore 2 so b = ¡ ¡ a = y2 ¡ x2 ¡ y1 x1 0 = ¡ 5 2) ( ¡ 0 ¡ = 2 5 (10.84) So the equation of the straight line is f (x) = 2 5 x 2. ¡ Now
, for a general parabola of the form f (x) = ax2 + c where a is positive, the term ax2 is always positive so the function is at a minimum when x = 0. Therefore the arms of the parabola point upwards. Otherwise, if a is negative, then ax2 is always negative and thus the function is maximum at x = 0. This means that the arms of the parabola must point downwards. Worked Examples: Example 1: Q: Find the intercepts and thus plot a graph of the function f (x) = x2 + 4. ¡ A: The y-intercept is and the x-intercepts are y = (0)2 + 4 = 4 ¡ (10.85) 173 & a > 0 a < 0 Figure 10.5: A parabola of the form y = ax2 + c with positive curvature (a > 0) and a parabola with negative curvature (a < 0)(NOTE: is curvature the right word to use here?) x2 + 4 = 0 x2 = 4 2 x = § ¡ ) ) (10.86) (10.87) (10.88) Since a = 1 is negative, we know that the arms of the parabola must point downwards. We can now use the three intercepts to draw the graph of this function Figure 10.6: Graph of f (x) = x2 + 4 ¡ 174 ’ ’ ’ Example 2: Q: Plot a graph of the parabola f (x) = x2 + 1. A: The y-intercept is y = (0)2 + 1 = 1 (10.89) and the x-intercepts are x2 + 1 = 0 x2 = 1 ¡ ) (10.90) (10.91) (10.92) This is not possible if x is a real number, so there are no x-intercepts. The parabola must therefore be entirely above the x-axis. This agrees with the fact that we know the arms of the parabola point upwards because a = 1 is positive. Thus the graph is as follows: 5 4 3 2 1 0 Figure 10.7: Graph of the parabola f (x) = x2 + 1 1 ¡ 1 175 ( Example 2: Q: Plot a graph of the hyperbola xy = ¡
A: A table of x and y values is as follows: 1. x : y = 1 x : -4 1 4 -2 1 2 -4 1 2 -2 1 -1 2 - 1 2 4 - 1 4 This gives the graph shown below Figure 10.8: Graph of the hyperbola xy = 1 ¡ We can see that a hyperbola has no x or y-intercepts. However, there are two general forms for hyperbolae, depending on whether a is positive or negative. Both types of hyperbola are symmetric about the lines y = x and y = x (in other words, the part of the hyperbola on one side of the line is just the reection of the part on the other side). ¡ In the ﬂrst case (a > 0), the hyperbola intersects line y = x at two points. At these intersections xy = a and y = x (10.93) which implies that 176 ) ) ) ) ) ) ) ) ) ) a > 0 a < 0 Figure 10.9: A hyperbola of the form y = a x with positive coe–cient (a > 0) and straight line y = x. A hyperbola with negative curvature (a < 0) and straight line y = x. ¡ x(x) = x2 = a pa x = § ) (10.94) (10.95) and, as y = x, it follows that the points of intersection are ( and (pa;pa). ¡ pa; ¡ pa) In the second case (a < 0), the hyperbola intersects the line y = two intersection points occur when x. The ¡ xy = a and y = x ¡ and therefore x) = x2 = a ¡ x( ¡ x2 = x = a ¡ p § a ¡ ) ) and, since y = pa). ¡ x, the two intersection points are ( (10.96) (10.97) (10.98) (10.99) pa;pa) and (pa; ¡ ¡ Worked Examples: Example 1: Q: Draw a graph of the hyperbola xy = 9. A: First we note that, since a = 9 is positive, the hyperbola must be in the top right and bottom left quadrants. The points at which it intersects
the line y = x are (-3,-3) and (3,3). It is also clear that the points (-1,-9), (-9,-1), (1,9) and (9,1) are part of the hyperbola. Thus the graph is 177 Figure 10.10: Graph of the hyperbola xy = 9 1:5 1:0 0:5 0:5 1:0 1:5 1:5 ¡ 1:0 ¡ 0:5 ¡ 0:5 ¡ 1:0 ¡ 1:5 ¡ 1 2 ; 1 2 ) and Figure 10.11: Graph of a hyperbolic function turning at the points ( ( 1 2 ; 1 2 ) ¡ ¡ Example 2: Q: The graph of a hyperbolic function is shown below. 178 * * * * * * + + What is the function deﬂning this hyperbola? A: Since the hyperbola is in the top left and bottom right quadrants, we known that a < 0. Now the point ( 2 ) lies on the hyperbola, so 2 ; 1 1 ¡ a = xy = ( ¡ 1 2 )( 1 2 ) = 1 4 ¡ (10.100) which is negative as we originally worked out. Therefore the hyperbolic function is f (x) = 1 4x. ¡ Worked Example: Q: Plot graph of the relation x2 + y2 = 4. A: This is the equation of a circle centered at the origin with radius 2. The graph is as follows: 2 2 ¡ 2 ¡ 2 Figure 10.12: Graph of the circle x2 + y2 = 4 Worked Example: Q: Draw the semi-circles described by the equations y = 9 x2 and y = ¡ Also give the domains and ranges of these semi-circles. p ¡ x2 9 ¡p (10.101) A: These equations describe semi-circles with radius 3. The ﬂrst semicircle lies above the x-axis (since the positive square root is being considered, the y values are all positive) and the second semi-circle is below the x-axis (the y values are all negative because the equation involves the negative root). The graphs of these semi-circles are thus as follows: From the above graphs we can see that, for
the semi-circle y = p9 the domain is [ x2, 3;3] and the range is [0;3] and, for the semi-circle y = ¡ p9 ¡ ¡ ¡ x2, the domain is [ ¡ 3;3] and the range is [-3,0]. 179 3 3 ¡ 3 3 ¡ 0 3 3 ¡ Figure 10.13: Semi-circles of radius 3, centered at the origin. The equation for the left semi-circle is y = p9 x2 whereas the right semi-circle is governed by y = p9 x2 ¡ ¡ ¡ The equation x2 + y2 = r2 can also be used to solve for x which gives x = r2 y2 (10.102) §p Again the positive and negative roots each describe a semi-circle, but in this case, on either side of the y-axis. Therefore the equations for two other types of semi-circles are ¡ x = r2 p y2 ¡ and x = r2 ¡p y2 ¡ (10.103) The domains of these semi-circles are [0;r] (in the ﬂrst case) and [ (in the second case). In both cases, the range is [ r;r]. r;0] ¡ ¡ Note: Again y = ¡ Thus these semi-circles are not functions. r and y = r corresponds to the same x = 0 value. 180 Worked Example: Q: Plot the graphs of the following relations 1 ¡p p State the domains and ranges of these relations. x = x = and ¡ 1 y2 y2 ¡ (10.104) A: The above equations describe semi-circles of radius 1 on either side of the y axis (in the ﬂrst case x is always positive and in the second x is always negative). The graphs of these relations are as follows ¡ Figure 10.14: Semi-circles of radius 1 on either side of the y axis. In the ﬂrst y2 and in the second case x is always case x is always positive as x = y2. (NOTE: This entire chapter needs a rethink on negative as x = 1 how it references graphs... instead of non-descriptive sentences like \in the ﬂrst
p case", we need to use 1 p ¡ ¡ ¡ ref a lot more) n For the ﬂrst semi-circle, the domain is [0;1] and the range is [ the case of the second semi-circle, the domain is [ [ ¡ 1;1]. ¡ 1;1] and in ¡ 1;0] and the range is Worked Example: Q: Plot a graph of the function f (x) = x + 2 j j ¡ 5. A: Let us ﬂrst work out a table of x and f (x) values as follows: x: f (x) -8 1 -7 0 -6 -1 -5 -2 -4 -3 -3 -4 -2 -5 -1 -4 0 -3 1 -2 2 - The graph of the function f (x) is thus shown below. 181 Figure 10.15: Graph of the function f (x) = x + 2 j j ¡ 5 Worked Examples: Example 1: Q: Consider the function y = 2 4. Plot a graph of this absolute x ¡ j value function, showing all the intercepts, the turning point and the axis of symmetry. 1 j ¡ 4) A: We can see that, since b = 1 and c = and the axis of symmetry is x = 1. Also, because a = 2 is positive, the absolute value function will be V-shaped. 4, the turning point is (1; ¡ ¡ Now let us ﬂnd the y-intercept. At 0(1) 4 = ¡ 2 (10.105) ¡ Since the turning point is below the x-axis and the graph points upwards, we suspect that this function does have x-intercepts. Let us try to ﬂnd these intercepts, which are the x-intercepts of the two straight lines y = 2(x 4 and y = 2( x + 1) 1) 4. ¡ ¡ ¡ ¡ y = 2(x 2x ¡ 6 = 0 1) ¡ 2x = 6 x = 3 4 = 0 ¡ (10.106) (10.107) (10.108) (10.109) (10.110) ) ) ) and 182,,,,,,,,,,,,,,,,, y =
2( 2x x + 1) ¡ 2 = 0 ) ¡ ¡ 2x = 10.111) (10.112) (10.113) (10.114) (10.115) Therefore the graph of the absolute values function is as follows Figure 10.16: Graph of the absolute values function y = 2 j 4 1 j ¡ ¡ Example 2: Q: Plot a graph of the function f (x) = turning point and axis of symmetry. x+3 1 showing the intercepts, ¡j j¡ A: The turning point is of this function is ( symmetry is given by x = like an upside down V. 3. As a = ¡ ¡ 1) and the axis of 1 is negative the graph is shaped ¡ ¡ 3; At x = 0, the y-intercept is ¡j j ¡ 1 = 1 = f (0) = (0) + 3 3 j ¡ ¡j Now this function points downwards and obtains its maximum at the turning point ( 1 is negative). Since the function is never greater than -1, it cannot become positive and therefore cannot cross the x-axis. Thus this absolute value function has no x-intercepts. 1). This point is below the x-axis (since y = (10.116; The graph of the function is shown below. 183 ¡ Figure 10.17: Graph of the function f (x) = x + 3 1 ¡j j ¡ Worked Example: Q: What the does the relation x2 graph of this relation. ¡ 2x + y2 + 4y = 4 describe? Plot a ¡ A: We can see that it is not going to be easy to solve to x and y, but let us try another trick, which is to write the relation as the sum of perfect squares. We know that x2 ¡ 2x = (x2 ¡ 2x + 1) and also that y2 + 4y = (y2 + 4y + 4) 1 = (x 1)2 1 ¡ ¡ 4 = (y + 2)2 4 ¡ ¡ ¡ Therefore the relation can be written as follows: and thus 1)2 (x ¡ ¡ 1 + (y + 2)2 4 = 4 ¡ ¡ 1)2 + (y + 2)2 = 1 (x ¡ (10.117
) (10.118) (10.119) (10.120) But this is just the equation for a circle centered at the origin with radius 1 (x2 + y2 = 1), where x and y have been replaced by x 1 and y + 2. So we can see that this is a circle which has been moved 1 to the right and 2 downwards. Therefore this relation describes a circle centered at the point (1,-2) with radius 1 (as shown in the diagram below). ¡ Note: point (a;b) is In general, the equation for a circle of radius r centered at the a)2 + (y (x ¡ ¡ b)2 = r2 (10.121) 184.. 1 2 (1; 2) ¡ 1 ¡ 2 ¡ 3 ¡ Figure 10.18: Graph of the circle (x 1)2 + (y + 2)2 = 1 ¡ Worked Example: Q: Show that the equation for an ellipse centered at the origin can be derived from that of a circle centered at the origin by performing suitable transformations. Draw a graph of the resulting ellipse. A: We shall start with the equation of a circle centered at the origin, which is given by x2 + y2 = r2, and replace x by r a x and y by r b y. We can see that the equation for the circle becomes y)2 = r2 r b b2 = r2 x)2 + ( r ( a r2 x2 a2 + r2 y2 y2 x2 b2 = 1 a2 + ) ) (10.122) (10.123) (10.124) which is the equation for an ellipse centred at the origin with y-intercepts (0; b) and (0;b) and x-intercepts ( a;0) and (a;0). ¡ ¡ The graph of this ellipse is as follows: a b b a Figure 10.19: Graph of an ellipse centered at the origin with y-intercepts (0; a;0) and (a;0) and (0;b) and x-intercepts ( ¡ b) ¡ 185 / 0 0 0 0 10.3.1 Worked Examples: Example 1: Q: Consider the parabolic function f (x) = x2. What is
the equation for the function which can be obtained by stretching f (x) vertically by a factor of a, then shifting this function to the right by p and upwards by q? A: We start with the function y = x2 and must be very careful to apply these changes in the right order. First we stretch f (x) vertically by a factor of a. This means that we must change y to y a, which gives = x2 y a y = ax2 ) (10.125) (10.126) Now we shift y = ax2 to the right by p. In other words, we change x to x p, which gives ¡ y = a(x p)2 ¡ (10.127) p)2 upwards by q. Therefore we must replace y Finally we shift y = a(x by y ¡ q. This means that ¡ q = a(x y y = a(x ¡ p)2 ¡ p)2 + q ¡ ) (10.128) (10.129) which is one way of writing the general formula for a quadratic function (see section 1.4). Example 2: Let f (x) = x + 6. What is the function which is the result of Q: reecting f (x) about the y-axis and then shrinking this function by a factor of 2 horizontally and then vertically? Plot graphs of the initial and ﬂnal functions. There are other ways of getting this ﬂnal function from f (x) by performing only two changes. Give an example of such a method. A: We start with the initial function y = x + 6. If we reect this about the x-axis, we must change x to x and this gives ¡ ¡ Now we must shrink this function by a factor of 2 horizontally. Thus changing x to 2x result in the equation y = x + 6 (10.130) ¡ If we now shrink this by a factor of 2 vertically (changing y to 2y) we get y = 2x + 6 (10.131) 186 2y = y = 2x + 6 ¡ x + 3 ¡ ) (10.132) (10.133) The graphs of the initial function y = x + 6 and the ﬂnal function y = x + 3 are shown below Figure 10.20: Graph of the initial function y = x + 6 (
left) and the ﬂnal function y = x + 3 (right) ¡ Finally we must look for two transformations which give this same result. We suspect that there must be a reection involved, because the ﬂnal funcx and the initial function contains x (also the above graph tion involves shows that the straight lines are perpendicular). Therefore, as before, we start by reecting the function f (x) about the y-axis. This gives ¡ y = x + 6 ¡ (10.134) Now we see that to get the ﬂnal equation y = x + 3 we must also add -3 to this equation. This is the same as replacing x by x + 3, so let us try shifting the equation by 3 to the left. This gives the equation ¡ ¡ ¡ Therefore, two changes which give the same ﬂnal function are a reection about the y-axis followed by a shift by 3 to left. ) y = y = (x + 3) + 6 x + 3 (10.135) (10.136) 187 1 2 2 Appendix A GNU Free Documentation License 2000,2001,2002 Free Software Foundation, Inc. Version 1.2, November 2002 Copyright c 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA Everyone is permitted to copy and distribute verbatim copies of this license document, but changing it is not allowed. PREAMBLE The purpose of this License is to make a manual, textbook, or other functional and useful document \free" in the sense of freedom: to assure everyone the eﬁective freedom to copy and redistribute it, with or without modifying it, either commercially or non-commercially. Secondarily, this License preserves for the author and publisher a way to get credit for their work, while not being considered responsible for modiﬂcations made by others. This License is a kind of \copyleft", which means that derivative works of the document must themselves be free in the same sense. It complements the GNU General Public License, which is a copyleft license designed for free software. We have designed this License in order to use it for manuals for free software, because free software needs free documentation: a free program should come with manuals providing the same freedoms that the software does. But this License is not limited to software manuals; it can be used for any textual work, regardless of
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of ABC, mABD 15, and mDBC 90. a. Find mABC. b. Name an acute angle. A D c. Name a right angle. d. Name an obtuse angle. B C 14365C01.pgs 7/9/07 4:37 PM Page 18 18 Essentials of Geometry Solution a. mABC mABD mDBC 15 90 105 b. ABD is an acute angle. c. DBC is a right angle. d. ABC is an obtuse angle. A D B C Answers a. mABC 105 b. ABD c. DBC d. ABC Exercises Writing About Mathematics 1. Explain the difference between a half-line and a ray. h PS 2. Point R is between points P and S. Are h PR and the same ray, opposite rays, or neither the same nor opposite rays? Explain your answer. Developing Skills 3. For the figure shown: a. Name two rays. b. Name the vertex of the angle. c. Name the sides of the angle. d. Name the angle in four ways. 4. a. Name the vertex of BAD in the figure. F y E D b. Name the sides of BAD. c. Name all the angles with A as vertex. h AB d. Name the angle whose sides are e. Name the ray that is a side of both BAD and BAC. f. Name two angles in whose interior regions point R lies. and h. AC g. Name the angle in whose exterior region point S lies. A h AB h AC h. Are i. Is BAC a straight angle? Explain your answer. opposite rays? Explain your answer. and C S D R B 14365C01.pgs 7/9/07 4:37 PM Page 19 5. For the figure shown: a. Name angle 1 in four other ways. b. Name angle 2 in two other ways. c. Name angle 3 in two other ways. d. Name the point of intersection of AC and. BD e. Name two pairs of opposite rays. f. Name two straight angles each of which has its vertex at E. g. Name two angles whose sum is ABC. More Angle Definitions 19 D 1 2 E C 3 B A 1-6 MORE ANGLE DEFINITIONS Congruent Angles DEFINITION Congruent angles are angles that have the same measure. C F B A E D In the figures
, ABC and DEF have the same measure. This is written as mABC mDEF. Since congruent angles are angles that have the same measure, we may also say that ABC and DEF are congruent angles, symbolized as ABC DEF. We may use either notation. ABC DEF mABC mDEF The angles are congruent. The measures of the angles are equal. Note: We do not write ABC DEF. Two angles are equal only if they name the union of the same rays. For example, ABC CBA. When two angles are known to be congruent, we may mark each one with the same number of arcs, as shown in the figures above, where each angle is marked with a single arc. 14365C01.pgs 7/9/07 4:37 PM Page 20 20 Essentials of Geometry Bisector of an Angle DEFINITION A bisector of an angle is a ray whose endpoint is the vertex of the angle, and that divides that angle into two congruent angles OC For example, if mAOC mCOB. If 1 m/COB 5 1 2m/AOB, 2mCOB. is the bisector of AOB, then AOC COB and, mAOB 2mAOC, and mAOB bisects AOB, we may say that h OC 2m/AOB m/AOC g AB, ACB is a straight angle, and mACB = 180. If Let C be any point on h is the bisector of ACB, each of the two congruent angles, ACD and CD DCB, has degree measure 90 and is a right angle. Any line, ray, or line segment at C. that is a subset of, and contains C, is said to be perpendicular to g AB h CD For example, CD is perpendicular to g. AB DEFINITION Perpendicular lines are two lines that intersect to form right angles. Since rays and line segments are contained in lines, any rays or line seg- ments that intersect to form right angles are also perpendicular. DEFINITION The distance from a point to a line is the length of the perpendicular from the point to the line. P A P is a point not on g AB, and PR ⊥ g AB. The segment PR is called the per- B pendicular from P to. The point R at which the perpendicular meets the line g AB g AB is PR (the is
called the foot of the perpendicular. The distance from P to R length of PR ). Adding and Subtracting Angles T P S R DEFINITION If point P is a point in the interior of RST and RST is not a straight angle, or if P is any point not on straight angle RST, then RST is the sum of two angles, RSP and PST. 14365C01.pgs 7/9/07 4:37 PM Page 21 More Angle Definitions 21 h SP In the figure, separates RST into two angles, RSP and PST, such that the following relations are true for the measures of the angles: mRST mRSP mPST mRSP mRST mPST mPST mRST mRSP EXAMPLE 1 h CD In the figure, bisects ACB. a. Name 1 in two other ways. b. Write a conclusion that states that two angles are con- gruent. c. Write a conclusion that states that two angles have C equal measures. Solution a. ACD and DCA. Answer b. ACD BCD or 2 1. Answer c. mACD mBCD or m2 m1. Answer P R S T B 2 1 D A Exercises Writing About Mathematics 1. Tanya said that if g RST separates PSQ into two congruent angles, then g RST is the bisector of PSQ. Do you agree with Tanya? Explain why or why not. 2. If an obtuse angle is bisected, what kind of an angle is each of the two angles formed? Developing Skills In 3–10, if an angle contains the given number of degrees, a. state whether the angle is an acute angle, a right angle, an obtuse angle, or a straight angle, b. find the measure of each of the angles formed by the bisector of the given angle. 3. 24 7. 82 4. 98 8. 180 5. 126 9. 57 6. 90 10. 3 14365C01.pgs 7/9/07 4:37 PM Page 22 22 Essentials of Geometry In 11–13, find the measure of each of the following: 11. of a right angle 1 2 12. of a right angle 1 3 13. of a straight angle 4 5 In 14 and 15, in each case, use the information to: a. Draw a diagram. b.
Write a conclusion that states the congruence of two angles. c. Write a conclusion that states the equality of the measures of two angles. h CD h AC 14. 15. bisects ACB. is the bisector of DAB. 16. If a straight angle is bisected, what types of angles are formed? 17. If a right angle is bisected, what types of angles are formed? In 18–20, complete each statement, which refers to the figure shown. 18. mLMN mLMP m______ 19. mLMP mLMN m______ 20. mLMN m______ mNMP In 21–24, use the figure to answer each question. 21. mABE mEBC m______ 22. mBEC mCED m______ 23. mADC mCDE m______ 24. mAEC mAEB m______ N P M L D C E A B 25. Every horizontal line is perpendicular to every vertical line. Draw, using pencil and paper or geometry software, a horizontal line ACB and a vertical line DCF. a. Name four congruent angles in your diagram. b. What is the measure of each of the angles that you named in part a? c. Name two rays that bisect ACB. d. Name two rays that bisect DCF. 26. and Applying Skills h h PQ PR what is the value of x? h BD sented by 5a 6, what is mABC? are opposite rays and 27. h PS bisects QPR. If mQPS is represented by 4x 30, bisects ABC. If mABD can be represented by 3a 10 and mDBC can be repre- 14365C01.pgs 7/9/07 4:37 PM Page 23 Triangles 23 h RT h BD 28. 29. bisects QRS. If mQRS 10x and mSRT 3x 30, what is mQRS? bisects ABC and h MP bisects LMN. If mCBD = mPMN, is ABC LMN? Justify your answer. Hands-On Activity For this activity, use a sheet of wax paper, patty paper, or other paper thin enough to see through. You may work alone or with a partner. 1. Draw a straight angle, ABC. 2. Fold the paper through the vertex of the angle so that the two opposite rays correspond,
and crease the paper. Label two points, D and E, on the crease with B between D and E. 3. Measure each angle formed by the crease and one of the opposite rays that form the straight angle. What is true about these angles? 4. What is true about DE and AC? 1-7 TRIANGLES DEFINITION A polygon is a closed figure in a plane that is the union of line segments such that the segments intersect only at their endpoints and no segments sharing a common endpoint are collinear. A polygon consists of three or more line segments, each of which is a side of the polygon. For example, a triangle has three sides, a quadrilateral has four sides, a pentagon has five sides, and so on. In the definition of a polygon, we used the word closed. We understand by the word closed that, if we start at any point on the figure and trace along the sides, we will arrive back at the starting point. In the diagram, we see a figure that is not closed and is, therefore, not a polygon. DEFINITION A triangle is a polygon that has exactly three sides. The polygon shown is triangle ABC, written as ABC. In ABC, each of the points A, B, and C is a vertex of the triangle. are the AB sides of the triangle., and CA BC, C A B 14365C01.pgs 7/9/07 4:37 PM Page 24 24 Essentials of Geometry Included Sides and Included Angles of a Triangle If a line segment is the side of a triangle, the endpoints of that segment are the vertices of two angles. For example, in ABC, the endpoints of are the vertices of A and B. We say that the side,, is included between the angles, A and B. In ABC: AB AB C 1. AB 2. BC 3. CA is included between A and B. is included between B and C. is included between C and A. A B In a similar way, two sides of a triangle are subsets of the rays of an angle, and we say that the angle is included between those sides. In ABC: 1. A is included between 2. B is included between 3. C is included between AC and AB. BA CA and BC. and CB. Opposite Sides and Opposite Angles in a Triangle For each side of a triangle,
there is one vertex of the triangle that is not an endpoint of that side. For example, in ABC, C is not an endpoint of. We say that. AB AB is opposite A Similarly, and A is opposite is the side opposite C and that C is the angle opposite side AC is opposite B and B is opposite ; also AB AC BC BC. The length of a side of a triangle may be represented by the lowercase form of the letter naming the opposite vertex. For example, in ABC, BC a, CA b, and AB c. Classifying Triangles According to Sides Scalene Triangle Z Isosceles Triangle T Equilateral Triangle N X Y R S L M 14365C01.pgs 7/9/07 4:37 PM Page 25 Triangles 25 DEFINITION A scalene triangle is a triangle that has no congruent sides. An isosceles triangle is a triangle that has two congruent sides. An equilateral triangle is a triangle that has three congruent sides. Parts of an Isosceles Triangle In isosceles triangle RST, the two congru, are called the legs ent sides, and TS of the triangle. The third side,, is called the base. TR RS The angle formed by the two congruent sides of the isosceles triangle, T, is called the vertex angle of the isosceles triangle. The vertex angle is the angle opposite the base. T vertex angle leg l e g base angle base angle R base S The angles whose vertices are the endpoints of the base of the triangle, R and S, are called the base angles of the isosceles triangle. The base angles are opposite the legs. Classifying Triangles According to Angles Acute Triangle B A C DEFINITION G H Right Triangle Obtuse Triangle M L K L An acute triangle is a triangle that has three acute angles. A right triangle is a triangle that has a right angle. An obtuse triangle is a triangle that has an obtuse angle. In each of these triangles, two of the angles may be congruent. In an acute triangle, all of the angles may be congruent. DEFINITION An equiangular triangle is a triangle that has three congruent angles. 14365C01.pgs 7/9/07 4:37 PM Page 26 26 Essentials of Geometry Right Triangles In right triangle GHL, the
two sides of the triangle, are called the that form the right angle, GH legs of the right triangle. The third side of the triangle,, the side opposite the right angle, is called the hypotenuse. and HL GL G g e l hyp oten use H leg L EXAMPLE 1 E In DEF, mE 90 and EF ED. D F a. Classify the triangle according The triangle is an isosceles triangle. to sides. Answers b. Classify the triangle according The triangle is a right triangle. to angles. c. What sides are the legs of the triangle? The legs of the triangle are and. ED EF d. What side is opposite the right FD is opposite the right angle. angle? e. What angle is opposite? EF Angle D is opposite EF. f. What angle is included between Angle F is between EF and. FD EF and? FD Using Diagrams in Geometry In geometry, diagrams are used to display and relate information. Diagrams are composed of line segments that are parallel or that intersect to form angles. From a diagram, we may assume that: • A line segment is a part of a straight line. • The point at which two segments intersect is a point on each segment. • Points on a line segment are between the endpoints of that segment. • Points on a line are collinear. • A ray in the interior of an angle with its endpoint at the vertex of the angle separates the angle into two adjacent angles. From a diagram, we may NOT assume that: • The measure of one segment is greater than, or is equal to, or is less than that of another segment. • A point is a midpoint of a line segment. 14365C01.pgs 7/9/07 4:37 PM Page 27 Triangles 27 • The measure of one angle is greater than, is equal to, or is less than that of another angle. • Lines are perpendicular or that angles are right angles (unless indicated with an angle bracket). • A given triangle is isosceles or equilateral (unless indicated with tick marks). • A given quadrilateral is a parallelogram, rectangle, square, rhombus, or trapezoid. EXAMPLE 2 Tell whether or not each statement may be assumed from the given diagram. g PR (1) is a straight line. Yes (2) S, T, and R are collinear. (3) PSR is a right
angle. (4) QTR is isosceles. No (5) QRW is adjacent to WRU. Yes Yes No P V Q W S RT U (6) Q is between V and T. (7) PQR and VQT intersect. (8) R is the midpoint of. TU Yes Yes No (9) QT PS No (10) mQRW mVQP No (11) PQTS is a trapezoid. No Exercises Writing About Mathematics 1. Is the statement “A triangle consists of three line segments” a good definition? Justify your answer. 2. Explain the difference between the legs of an isosceles triangle and the legs of a right triangle. Developing Skills In 3 and 4, name the legs and the hypotenuse in each right triangle shown. 3. C 4. L K A B J 14365C01.pgs 7/9/07 4:37 PM Page 28 28 Essentials of Geometry In 5 and 6, name the legs, the base, the vertex angle, and the base angles in each isosceles triangle shown. 5. N 6. T L M R S In 7–10, sketch, using pencil and paper or geometry software, each of the following. Mark congruent sides and right angles with the appropriate symbols. 7. An acute triangle that is isosceles 8. An obtuse triangle that is isosceles 9. A right triangle that is isosceles 11. The vertex angle of an isosceles triangle is ABC. Name the base angles of this triangle. 12. In DEF, 13. In RST, S is included between which two sides? is included between which two angles? 10. An obtuse triangle that is scalene DE 14. a. Name three things that can be assumed from the diagram to the right. b. Name three things that can not be assumed from the diagram to the right. Applying Skills D G A F C E B 15. The degree measures of the acute angles of an obtuse triangle can be represented by 5a 12 and 3a 2. If the sum of the measures of the acute angles is 82, find the measure of each of the acute angles. 16. The measures of the sides of an equilateral triangle are represented by x + 12, 3x 8, and 2x 2. What is the measure of each side
of the triangle? 17. The lengths of the sides of an isosceles triangle are represented by 2x 5, 2x 6, and 3x 3. What are the lengths of the sides of the triangle? (Hint: the length of a line segment is a positive quantity.) 18. The measures of the sides of an isosceles triangle are represented by x 5, 3x 13, and 4x 11. What are the measures of each side of the triangle? Two answers are possible. 14365C01.pgs 7/9/07 4:37 PM Page 29 Chapter Summary 29 CHAPTER SUMMARY Undefined Terms • set, point, line, plane Definitions to Know • A collinear set of points is a set of points all of which lie on the same straight line. • A noncollinear set of points is a set of three or more points that do not all lie on the same straight line. • The distance between two points on the real number line is the absolute value of the difference of the coordinates of the two points. • B is between A and C if and only if A, B, and C are distinct collinear points and AB BC AC. • A line segment, or a segment, is a set of points consisting of two points on a line, called endpoints, and all of the points on the line between the endpoints. • The length or measure of a line segment is the distance between its end- points. • Congruent segments are segments that have the same measure. • The midpoint of a line segment is a point of that line segment that divides the segment into two congruent segments. • The bisector of a line segment is any line, or subset of a line, that intersects the segment at its midpoint. • A line segment, RS, is the sum of two line segments, RP and PS, if P is between R and S. • Two points, A and B, are on one side of a point P if A, B, and P are collinear and P is not between A and B. • A half-line is a set of points on one side of a point. • A ray is a part of a line that consists of a point on the line, called an end- point, and all the points on one side of the endpoint. • Opposite rays are two rays of the same line with a common endpoint and no other point in common.
• An angle is a set of points that is the union of two rays having the same endpoint. • A straight angle is an angle that is the union of opposite rays and whose degree measure is 180. • An acute angle is an angle whose degree measure is greater than 0 and less than 90. • A right angle is an angle whose degree measure is 90. • An obtuse angle is an angle whose degree measure is greater than 90 and less than 180. 14365C01.pgs 7/9/07 4:37 PM Page 30 30 Essentials of Geometry P R S T Properties of the Real Number System • Congruent angles are angles that have the same measure. • A bisector of an angle is a ray whose endpoint is the vertex of the angle, and that divides that angle into two congruent angles. • Perpendicular lines are two lines that intersect to form right angles. • The distance from a point to a line is the length of the perpendicular from the point to the line. • If point P is a point in the interior of RST and RST is not a straight angle, or if P is any point not on straight angle RST, then RST is the sum of two angles, RSP and PST. • A polygon is a closed figure in a plane that is the union of line segments such that the segments intersect only at their endpoints and no segments sharing a common endpoint are collinear. • A triangle is a polygon that has exactly three sides. • A scalene triangle is a triangle that has no congruent sides. • An isosceles triangle is a triangle that has two congruent sides. • An equilateral triangle is a triangle that has three congruent sides. • An acute triangle is a triangle that has three acute angles. • A right triangle is a triangle that has a right angle. • An obtuse triangle is a triangle that has an obtuse angle. • An equiangular triangle is a triangle that has three congruent angles. Addition Multiplication Closure Commutative Property a b b a Associative Property a b is a real number. a b is a real number. a b b a (a b) c a (b c) a (b c) (a b) c a 0 a and 0 a a a 1 a and 1 a a a 0, a (a) 0 a? and ab ac a (b c) 1
a 5 1 Distributive Property a(b c) ab ac Identity Property Inverse Property Multiplication Property ab 0 if and only if a 0 or b 0 of Zero VOCABULARY 1-1 Undefined term • Set • Point • Line • Straight line • Plane 1-2 Number line • Coordinate • Graph • Numerical operation • Closure property of addition • Closure property of multiplication • Commutative 14365C01.pgs 7/9/07 4:37 PM Page 31 Review Exercises 31 property of addition • Commutative property of multiplication • Associative property of addition • Associative property of multiplication • Additive identity • Multiplicative identity • Additive inverses • Multiplicative inverses • Distributive property • Multiplication property of zero • 1-3 Definition • Collinear set of points • Noncollinear set of points • Distance between two points on the real number line • Betweenness • Line segment • Segment • Length (Measure) of a line segment • Congruent segments 1-4 Midpoint of a line segment • Bisector of a line segment • Sum of two line segments 1-5 Half-line • Ray • Endpoint • Opposite rays • Angle • Sides of an angle • Vertex • Straight angle • Interior of an angle • Exterior of an angle • Degree • Acute angle • Right angle • Obtuse angle • Straight angle 1-6 Congruent angles • Bisector of an angle • Perpendicular lines • Distance from a point to a line • Foot • Sum of two angles 1-7 Polygon • Side • Triangle • Included side • Included angle • Opposite side • Opposite angle • Scalene triangle • Isosceles triangle • Equilateral triangle • Legs • Base • Vertex angle • Base angles • Acute triangle • Right triangle • Obtuse triangle • Equiangular triangle • Hypotenuse REVIEW EXERCISES 1. Name four undefined terms. 2. Explain why “A line is like the edge of a ruler” is not a good definition. In 3–10, write the term that is being defined. 3. The set of points all of which lie on a line. 4. The absolute value of the difference of the coordinates of two points. 5. A polygon that has exactly three sides. 6. Any line or subset of a line that intersects a line segment at its midpoint.
7. Two rays of the same line with a common endpoint and no other points in common. 8. Angles that have the same measure. 9. A triangle that has two congruent sides. 10. A set of points consisting of two points on a line and all points on the line between these two points. 11. In right triangle LMN, mM 90. Which side of the triangle is the hypotenuse? 14365C01.pgs 7/9/07 4:37 PM Page 32 32 Essentials of Geometry 12. In isosceles triangle RST, RS ST. a. Which side of the triangle is the base? b. Which angle is the vertex angle? 13. Points D, E, and F lie on a line. The coordinate of D is 3, the coordinate of E is 1, and the coordinate of F is 9. a. Find DE, EF, and DF. b. Find the coordinate of M, the midpoint of DF. g AB c. intersects EF at C. If g AB is a bisector of EF, what is the coordinate of C? 14. Explain why a line segment has only one midpoint but can have many bisectors. In 15–18, use the figure shown. In polygon ABCD, the midpoint of BD. AC and BD intersect at E, 15. Name two straight angles. 16. What angle is the sum of ADE and EDC? 17. Name two congruent segments. 18. What line segment is the sum of AE and EC? B A E C In 19–22, use the figure shown. AB'BC and h BD bisects ABC. 19. Name two congruent angles. 20. What is the measure of ABD? 21. Name a pair of angles the sum of whose measures is 180. 22. Does AB BC AC? Justify your answer. D A B D C Exploration Euclidean geometry, which we have been studying in this chapter, focuses on the plane. Non-Euclidian geometry focuses on surfaces other than the plane. For instance, spherical geometry focuses on the sphere. In this Exploration, you will draw on a spherical object, such as a grapefruit or a Styrofoam ball, to relate terms from Euclidean geometry to spherical geometry. 14365C01.pgs 7/9/07 4:37 PM Page 33 Review Exercises 33 1. Draw a point on your
sphere. How does this point compare to a point in Euclidean geometry? 2. The shortest distance between two points is called a geodesic. In Euclidean geometry, a line is a geodesic. Draw a second point on your sphere. Connect the points with a geodesic and extend it as far as possible. How does this geodesic compare to a line in Euclidean geometry? 3. Draw a second pair of points and a second geodesic joining them on your sphere. The intersection of the geodesics forms angles. How do these angles compare to an angle in Euclidean geometry? 4. Draw a third pair of points and a third geodesic joining them on your sphere. This will form triangles, which by definition are polygons with exactly three sides. How does a triangle on a sphere compare to a triangle in Euclidean geometry? Consider both sides and angles. 14365C02.pgs 7/9/07 4:40 PM Page 34 CHAPTER 2 CHAPTER TABLE OF CONTENTS 2-1 Sentences, Statements, and Truth Values 2-2 Conjunctions 2-3 Disjunctions 2-4 Conditionals 2-5 Inverses, Converses, and Contrapositives 2-6 Biconditionals 2-7 The Laws of Logic 2-8 Drawing Conclusions Chapter Summary Vocabulary Review Exercises Cumulative Review 34 LOGIC Mathematicians throughout the centuries have used logic as the foundation of their understanding of the relationships among mathematical ideas. In the late 17th century, Gottfried Leibniz (1646–1716) organized logical discussion into a systematic form, but he was ahead of the mathematical thinking of his time and the value of his work on logic was not recognized. It was not until the 19th century that George Boole (1815–1864), the son of an English shopkeeper, developed logic in a mathematical context, representing sentences with symbols and carefully organizing the possible relationships among those sentences. Boole corresponded with Augustus DeMorgan (1806–1871) with whom he shared his work on logic. Two important relationships of logic are known today as DeMorgan’s Laws. not (p and q) (not p) or (not q) not (p or q) (not p) and (not q) Boolean algebra is key in the development of com- puter science and circuit design. 14365C02.pgs 7/9/
07 4:40 PM Page 35 2-1 SENTENCES, STATEMENTS, AND TRUTH VALUES Sentences, Statements, and Truth Values 35 Logic is the science of reasoning. The principles of logic allow us to determine if a statement is true, false, or uncertain on the basis of the truth of related statements. We solve problems and draw conclusions by reasoning from what we know to be true. All reasoning, whether in mathematics or in everyday living, is based on the ways in which we put sentences together. Sentences and Their Truth Values When we can determine that a statement is true or that it is false, that statement is said to have a truth value. In this chapter we will study the ways in which statements with known truth values can be combined by the laws of logic to determine the truth value of other statements. In the study of logic, we use simple declarative statements that state a fact. That fact may be either true or false. We call these statements mathematical sentences. For example: 1. Congruent angles are angles that True mathematical sentence have the same measure. 2. 17 5 12 3. The Brooklyn Bridge is in New York. 4. 17 3 42 True mathematical sentence True mathematical sentence False mathematical sentence 5. The Brooklyn Bridge is in California. False mathematical sentence Nonmathematical Sentences and Phrases Sentences that do not state a fact, such as questions, commands, or exclamations, are not sentences that we use in the study of logic. For example: 1. Did you have soccer practice today? This is not a mathematical sentence because it asks a question. 2. Go to your room. This is not a mathematical sentence because it gives a command. An incomplete sentence or part of a sentence, called a phrase, is not a math- ematical sentence and usually has no truth value. For example: 1. Every parallelogram This is not a mathematical sentence. 2. 19 2 This is not a mathematical sentence. 14365C02.pgs 7/9/07 4:40 PM Page 36 36 Logic Some sentences are true for some persons and false for others. For example: 1. I enjoy reading historical novels. 2. Summer is the most pleasant season. 3. Basketball is my favorite sport. Conclusions based on sentences such as these do not have the same truth value for all persons. We will not use sentences such as these in the study of logic. Open Sentences In the study of algebra, we worked with
open sentences, that is, sentences that contain a variable. The truth value of the open sentence depended on the value of the variable. For example, the open sentence x 2 5 is true when x 3 and false for all other values of x. In some sentences, a pronoun, such as he, she, or it, acts like a variable and the name that replaces the pronoun determines the truth value of the sentence. 1. x 2 8 Open sentence: the variable is x. 2. He broke my piggybank. Open sentence: the variable is he. 3. Jenny found it behind the sofa. Open sentence: the variable is it. In previous courses, we learned that the domain or replacement set is the set of all elements that are possible replacements for the variable. The element or elements from the domain that make the open sentence true is the solution set or truth set. For instance: Open sentence: 14 x 9 Variable: x Domain: {1, 2, 3, 4, 5} Solution set: {5} When x 5, then 14 5 9 is a true sentence. The method we use for sentences in mathematics is the same method we apply to sentences in ordinary conversation. Of course, we would not use a domain like {1, 2, 3, 4} for the open sentence “It is the third month of the year.” Common sense tells us to use a domain consisting of the names of months. The following example compares this open sentence with the algebraic sentence used above. Open sentences, variables, domains, and solution sets behave in exactly the same way. Open sentence: It is the third month of the year. Variable: It Domain: {Names of months} Solution set: {March} When “It” is replaced by “March,” then “March is the third month of the year” is a true sentence. 14365C02.pgs 7/9/07 4:40 PM Page 37 Sentences, Statements, and Truth Values 37 Sometimes a solution set contains more than one element. If Elaine has two brothers, Ken and Kurt, then for her, the sentence “He is my brother” has the solution set {Ken, Kurt}. Here the domain is the set of boys’ names. Some people have no brothers. For them, the solution set for the open sentence “He is my brother” is the empty set, ∅ or { }. Identify each of the following as
a true sentence, a false sentence, an open sentence, or not a mathematical sentence at all. Answers a. Football is a water sport. False sentence b. Football is a team sport. True sentence c. He is a football player. Open sentence: the variable is he. d. Do you like football? Not a mathematical sentence: this is a question. e. Read this book. f. 3x 7 11 g. 3x 7 Not a mathematical sentence: this is a command. Open sentence: the variable is x. Not a mathematical sentence: this is a phrase or a binomial. EXAMPLE 1 EXAMPLE 2 Use the replacement set 2, tence “It is an irrational number.”, 2.5, U 2p 3 2 2 " V to find the truth set of the open sen- Solution Both 2 and 2.5 are rational numbers. 2 Both p and rational number and an irrational number is an irrational number, are irrational. are irrational numbers. Since the product or quotient of a and 2p 3 " 2 2 " Answer 2p 3, 2 U 2 V " Statements and Symbols A sentence that can be judged to be true or false is called a statement or a closed sentence. In a statement, there are no variables. A closed sentence is said to have a truth value. The truth values, true and false, are indicated by the symbols T and F. 14365C02.pgs 7/9/07 4:40 PM Page 38 38 Logic Negations In the study of logic, you will learn how to make new statements based upon statements that you already know. One of the simplest forms of this type of reasoning is negating a statement. The negation of a statement always has the opposite truth value of the given or original statement and is usually formed by adding the word not to the given statement. For example: 1. Statement: Neil Armstrong walked on the moon. Negation: Neil Armstrong did not walk on the moon. 2. Statement: A duck is a mammal. Negation: A duck is not a mammal. (True) (False) (False) (True) There are other ways to insert the word not into a statement to form its negation. One method starts the negation with the phrase “It is not true that...” For example: 3. Statement: A carpenter works with wood. Negation: Negation: A carpenter does not work with wood
. It is not true that a carpenter works with wood. (True) (False) (False) Both negations express the same false statement. Logic Symbols The basic element of logic is a simple declarative sentence. We represent this basic element by a single, lowercase letter. Although any letter can be used to represent a sentence, the use of p, q, r, and s are the most common. For example, p might represent “Neil Armstrong walked on the moon.” The symbol that is used to represent the negation of a statement is the symbol placed before the letter that represents the given statement. Thus, if p represents “Neil Armstrong walked on the moon,” then p represents “Neil Armstrong did not walk on the moon.” The symbol p is read “not p.” Symbol Statement in Words Truth Value p p q q There are 7 days in a week. There are not 7 days in a week. 8 9 16 8 9 16 True False False True When p is true, then its negation p is false. When q is false, then its nega- tion q is true. A statement and its negation have opposite truth values. 14365C02.pgs 7/9/07 4:40 PM Page 39 Sentences, Statements, and Truth Values 39 It is possible to use more than one negation in a sentence. Each time another negation is included, the truth value of the statement will change. For example: Symbol r r (r) Statement in Words Truth Value A dime is worth 10 cents. A dime is not worth 10 cents. It is not true that a dime is not worth 10 cents. True False True We do not usually use sentences like the third. Note that just as in the set of real numbers, (a) a, (r) always has the same truth value as r. We can use r in place of (r). Therefore, we can negate a sentence that contains the word not by omitting that word. r: A dime is not worth 10 cents. (r): A dime is worth 10 cents. The relationship between a statement p and its negation p can be summarized in the table at the right. When p is true, p is false. When p is false, p is true. p T F p F T In this example, symbols are used to represent statements. The truth value of each statement is given. k: Oatmeal is a cereal
. m: Massachusetts is a city. (True) (False) For each sentence given in symbolic form: a. Write a complete sentence in words to show what the symbols represent. b. Tell whether the statement is true or false. (1) k (2) m Answers a. Oatmeal is not a cereal. a. Massachusetts is not a city. b. False b. True EXAMPLE 3 Exercises Writing About Mathematics 1. Explain the difference between the use of the term “sentence” in the study of grammar and in the study of logic. 14365C02.pgs 7/9/07 4:40 PM Page 40 40 Logic 2. a. Give an example of a statement that is true on some days and false on others. b. Give an example of a statement that is true for some people and false for others. c. Give an example of a statement that is true in some parts of the world and false in others. Developing Skills In 3–10, tell whether or not each of the following is a mathematical sentence. 3. Thanksgiving is on the fourth Thursday 4. Albuquerque is a city in New Mexico. in November. 5. Where did you go? 7. Be quiet. 9. y 7 3y 4 6. Twenty sit-ups, 4 times a week 8. If Patrick leaps 10. Tie your shoe. In 11–18, all of the sentences are open sentences. Find the variable in each sentence. 11. She is tall. 13. 2y 17 12. We can vote at age 18. 14. 14x 8 9 15. This country has the third largest population. 16. He hit a home run in the World Series. 17. It is my favorite food. 18. It is a fraction. In 19–26: a. Tell whether each sentence is true, false, or open. b. If the sentence is an open sentence, identify the variable. 19. The Statue of Liberty was given to the United States by France. 20. They gained custody of the Panama Canal on December 31, 1999. 21. Tallahassee is a city in Montana. 23. 6x 4 16 25. 6(2) 4 16 22. A pentagon is a five-sided polygon. 24. 6(10) 4 16 26. 23 32 In 27–31, find the truth set for each open sentence using the replacement set {Nevada, Illinois, Massachusetts, Alaska, New York}.
27. Its capital is Albany. 28. It does not border on or touch an ocean. 29. It is on the east coast of the United States. 30. It is one of the states of the United States of America. 31. It is one of the last two states admitted to the United States of America. 14365C02.pgs 7/9/07 4:40 PM Page 41 Sentences, Statements, and Truth Values 41 In 32–39, use the domain {square, triangle, rectangle, parallelogram, rhombus, trapezoid} to find the truth set for each open sentence. 32. It has three and only three sides. 33. It has exactly six sides. 34. It has fewer than four sides. 35. It contains only right angles. 36. It has four sides that are all equal in measure. 37. It has two pairs of opposite sides that are parallel. 38. It has exactly one pair of opposite sides 39. It has interior angles with measures that are parallel. whose sum is 360 degrees. In 40–47, write the negation of each sentence. 40. The school has an auditorium. 41. A stop sign is painted red. 42. The measure of an obtuse angle is greater than 90°. 44. Michigan is not a city. 46. 3 4 5 6 43. There are 1,760 yards in a mile. 45. 14 2 16 12 47. Today is not Wednesday. In 48–56, for each given sentence: a. Write the sentence in symbolic form using the symbols shown below. b. Tell whether the sentence is true, false, or open. Let p represent “A snake is a reptile.” Let q represent “A frog is a snake.” Let r represent “Her snake is green.” 48. A snake is a reptile. 49. A snake is not a reptile. 50. A frog is a snake. 51. A frog is not a snake. 52. Her snake is green. 53. Her snake is not green. 54. It is not true that a frog is a snake. 55. It is not the case that a snake is not a reptile. 56. It is not the case that a frog is not a snake. In 57–64, the symbols represent sentences. p: Summer follows spring. q: August is a summer month. r: A year has 12 months. s: She
likes spring. For each sentence given in symbolic form: a. Write a complete sentence in words to show what the symbols represent. b. Tell whether the sentence is true, false, or open. 57. p 61. (p) 58. q 62. (q) 59. r 63. (r) 60. s 64. (s) 14365C02.pgs 7/9/07 4:40 PM Page 42 42 Logic 2-2 CONJUNCTIONS We have identified simple sentences that have a truth value. Often we wish to use a connecting word to form a compound sentence. In mathematics, sentences formed by connectives are also called compound sentences or compound statements. One of the simplest compound statements that can be formed uses the connective and. In logic, a conjunction is a compound statement formed by combining two simple statements using the word and. Each of the simple statements is called a conjunct. When p and q represent simple statements, the conjunction p and q is written in symbols as p ∧ q. For example: p: A dog is an animal. q: A sparrow is a bird. p ∧ q: A dog is an animal and a sparrow is a bird. This compound sentence is true because both parts are true. “A dog is an animal is true and “A sparrow is a bird” is true. When one or both parts of a conjunction are false, the conjunction is false. For example, • “A dog is an animal and a sparrow is not a bird” is false because “A spar- row is not a bird” is the negation of a true statement and is false. • “A dog is not an animal and a sparrow is a bird” is false because “A dog is not an animal” is the negation of a true statement and is false. • “A dog is not an animal and a sparrow is not a bird” is false because both “A dog is not an animal” and “A sparrow is not a bird” are false. We can draw a diagram, called a tree diagram, to show all possible combinations of the truth values of p and q that are combined to make the compound statement p ∧ q. p is true p is false q is true p is true and q is true. q is false p is true and q is false. q is true p is false and
q is true. q is false p is false and q is false. These four possible combinations of the truth values of p and q can be displayed in a chart called a truth table. The truth table can be used to show the possible truth values of a compound statement that is made up of two simple statements. For instance, write a truth table for p ∧ q. 14365C02.pgs 7/9/07 4:40 PM Page 43 STEP 1. In the first column, we list the truth values of p. For each possible truth value of p, there are two possible truth values for q. Therefore, we list T twice and F twice. STEP 2. In the second column, we list the truth values of q. In the two rows in which p is true, list q as true in one and false in the other. In the two rows in which p is false, list q as true in one and false in the other. STEP 3. In the last column, list the truth values of the conjunction, p ∧ q. The conjunction is true only when both p and q are true. The conjunction is false when one or both conjuncts are false. Conjunctions 43 The conjunction, p and q, is true only when both parts are true: p must be true and q must be true. For example, let p represent “It is spring,” and let q represent “It is March.” CASE 1 Both p and q are true. On March 30, “It is spring” is true and “It is March” is true. Therefore, “It is spring and it is March” is true. CASE 2 p is true and q is false. On April 30, “It is spring” is true and “It is March” is false. Therefore, “It is spring and it is March” is false. CASE 3 p is false and q is true. On March 10, “It is spring” is false and “It is March” is true. Therefore, “It is spring and it is March” is false. CASE 4 Both p and q are false. On February 28, “It is spring” is false and “It is March” is false. Therefore, “It is spring and it is March” is false. 14365C02.pgs 7/9

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