Tamarind Posted July 6, 2012 Share Posted July 6, 2012 I have never studied trigonometry. I would like to understand trigonometry well enough to help my daughters with their ACT prep. Can you recommend a crash course? I'd like a text I can hold in my hand, not a youtube video. Also, I need something that takes it step by step--the ACT prep books don't cut it for me. Thanks! Tamarind Quote Link to comment Share on other sites More sharing options...

Colleen in NS Posted July 6, 2012 Share Posted July 6, 2012 (edited) I have never studied trigonometry. I would like to understand trigonometry well enough to help my daughters with their ACT prep. Can you recommend a crash course? I'd like a text I can hold in my hand, not a youtube video. Also, I need something that takes it step by step--the ACT prep books don't cut it for me. Thanks!Tamarind Maybe this old Dolciani text: http://www.amazon.com/Modern-Trigonometry-Mathematics-Series/dp/B000J4N96Y/ref=wl_it_dp_o_pC_nS_nC?ie=UTF8&coliid=I3A3UJY9CX3SQ3&colid=2MDDG6R2HCGMC Look for Jane in NC's posts on Dolciani trigonometry. You can pm her, too, for advice - she loves to help people with this. edit: I found some posts of hers: http://forums.welltrainedmind.com/showpost.php?p=2584852&postcount=9 http://forums.welltrainedmind.com/showpost.php?p=88740&postcount=1 Edited July 6, 2012 by Colleen in NS Quote Link to comment Share on other sites More sharing options...

SwallowTail Posted July 6, 2012 Share Posted July 6, 2012 I would get a textbook and just work through it. Although, I also think Khan academy online is *awesome*. Quote Link to comment Share on other sites More sharing options...

mumto2 Posted July 6, 2012 Share Posted July 6, 2012 Ck 12 has a trig course. You could print it as you go. The khan academy really is good. The videos are really well done. Another easy option for you would be a dummies book. We bought a huge stack cheap at a going out of business sale and they get lots of use. We don't have trig but my dd uses the calculus one frequently for quick explanations of unfamiliar terms etc. Plus your library might have it so you could try it for free. Quote Link to comment Share on other sites More sharing options...

Caroline Posted July 6, 2012 Share Posted July 6, 2012 Cliff Notes has a Trigonometry book. It's pretty good and fast and to the point. I also like Forster's Precalculus text. Quote Link to comment Share on other sites More sharing options...

AngieW in Texas Posted July 6, 2012 Share Posted July 6, 2012 Go through the trig videos at brightstorm. http://www.brightstorm.com/math/trigonometry/ Quote Link to comment Share on other sites More sharing options...

mathwonk Posted July 6, 2012 Share Posted July 6, 2012 (edited) Maybe I will try to write a short trig book. Basically I hated trig in high school and never learned it until decades later. It seemed like a bunch of meaningless complicated formulas. Then I eventually found out how simple it is. Basically it is a way to measure angles. If you put the eraser of your pencil on the table in front of you and raise the pointed end up off the table, the pencil makes a certain angle with the table. How to measure that angle? One way is to measure how high off the table is the point. Of course that depends both on the angle and the length of the pencil. So use a pencil of length one. Then the height of the point off the table is called the "sine" of the angle. (If you don't have a "length one" pencil, then the height divided by the length, is the sine.) The sine is a vertical measure. The other way to measure the angle is a horizontal measure. Shine a light straight down on the pencil and measure how long the shadow is on the table. If the pencil has length one, that shadow length is the "cosine" of the angle. (If the pencil has any other length than one, the cosine is the length of the shadow on the table divided by the pencil length, as before.) There is one basic fact about sine and cosine that rules all others, the "pythagorean theorem". I.e. because the vertical distance from the point to the table forms a right angle with the shadow on the table, and the shaft of the pencil is opposite that right angle, we get the relation sine^2 + cosine^2 = 1^2 = 1, true for every angle. See if you agree that when the pencil is vertical, sine = 1, and cosine = 0. Thus sine(90degrees) = 1 and cosine(90degrees) = 0. Also sine(0degrees) = 0, while cosine(0degrees) = 1, (when the pencil is lying flat on the table.) That's about all there is to trig, except for a couple of tricky formulas that tell you how sine and cosine change when you double the angle, and that are hard to remember. (But there is a cute trick for remembering those too.) There are also some other angle measures that are not really new, but they are taught in school, like "tangent" = sine/cosine. There is also cotangent = cosine/sine, secant = 1/cosine, and cosecant = 1/sine, but you don't need them as much. If you are familiar with the unit circle in the (x,y) plane, then an angle is represented just by a point on the circle. I.e. the angle is the one made by the positive x axis and the radius from the center of the circle to the point on the circumference. The angle vertex is at the center. Then the x and y coordinates of that point on the circle are exactly the cosine and sine of the angle. This is the right way to do trig, using the unit circle. The size of the angle is measured, in "radians" rather than degrees, by the length of the circular arc cut off by the angle. Thus in radian measure a full 360 degree angle is 2pi radians, since 2pi = the length of a circle of radius one. Thus 90 degrees = pi/2 radians. Hence in radians, which are always used in calculus, sine(pi/2) = 1, and cosine(pi/2) = 0, while cosine(0) = 1, and sine(0) = 0. Homework: The other useful facts are that sine(pi/4) = sqrt(2)/2, and sine(pi/6) = 1/2, and sine(pi/3) = sqrt(3)/2. I did these in my head so see if they are correct, using pythagoras on a "30-60-90" triangle, and a "45-45-90" triangle. And see if you can use the pythagorean formula above to deduce that tan^2 + 1 = sec^2. Suggestion: if this is too opaque now, learn from one of those other longer sources and then come back and read this again. This is really about all there is to it, honest. Edited July 6, 2012 by mathwonk Quote Link to comment Share on other sites More sharing options...

mathwonk Posted July 6, 2012 Share Posted July 6, 2012 Question: if an angle can be precisely measured just by its radian measure, why do we use two numbers, sine and cosine? Answer: it's easier to compute sine and cosine, because they are lengths of straight line segments, whereas radian measure is the length of a circular arc. In fact that is the whole reason for the existence of the subject of trig, to relate "linear" measure, to circular measure. Quote Link to comment Share on other sites More sharing options...

Twigs Posted July 6, 2012 Share Posted July 6, 2012 Thanks for the trig helps -very much apprecieated. â€¦ That's about all there is to trig, except for a couple of tricky formulas that tell you how sine and cosine change when you double the angle, and that are hard to remember. (But there is a cute trick for remembering those too.) â€¦ Please don't leave us hanging - what is the cute trick?? :bigear: Thanks Quote Link to comment Share on other sites More sharing options...

Colleen in NS Posted July 6, 2012 Share Posted July 6, 2012 mathwonk, I love your explanations!! Stick around! Quote Link to comment Share on other sites More sharing options...

Jann in TX Posted July 6, 2012 Share Posted July 6, 2012 (edited) All you need to know about Trig for the ACT test is found in most Geometry texts. You need to be able to solve right triangles using Sine, Cosine and Tangent ratios: Soh Cah Toa There is usually one question on the ACT involving the "Law of Sines" and one involving the ''Law of Cosines"-- these are 2 formulas that have to be memorized--but are easy to use once the student knows Soh Cah Toa (the right triangle ratios). I teach it in 2-3 class periods. You can search the Khan Academy videos for the right triangle trig ratios and the Law of Sines and Law of Cosines. For extra practice look at Kutasoftware.com free worksheets in their Algebra 2 section for those topics. Edited July 7, 2012 by Jann in TX duh Quote Link to comment Share on other sites More sharing options...

mathwonk Posted July 6, 2012 Share Posted July 6, 2012 (edited) Wow! You guys are a great class! The cute trick is the fact that if you use complex numbers, then sine and cosine are a special case of the exponential function! And the laws of exponents are simpler than those for sine and cosine. Lets just treat this as a magic trick that works. If you have an angle of t radians, i.e. that an angle with vertex at the center of a unit radius circle, that cuts off an arc of length t on the circle, then cos(t) + i.sin(t) = e^(it), where e is a certain beautiful positive real number between 2 and 3, and i is a new "complex" number with i^2 = -1. Think of the numbers on the x axis as the usual "real" numbers, whose squares are all positive, and then i is a number on the y axis, at unit distance from the origin. Real numbers make angle zero (with the x axis) and i makes angle pi/2 (I started to say 90 degrees, as old habits die hard). Every point in the x,y plane represents a complex number, and to multiply them you multiply their lengths and add their angles. So i^2 has length 1^2 = 1, and angle pi/2 + pi/2 = pi = 180 degrees. So i^2 is at -1 on the x axis! Just as you can take powers and exponents of real numbers you can also take complex exponents of real numbers like e, and then for any real number t, the complex number e^(it) is the same as the complex number cos(t) + i.sin(t). I.e. it has x coordinate cos(t) and y coordinate (or "i coordinate") sin(t). Now here is the cool part, just as e^(a+b) = e^a.e^b, when a,b are real, also e^i(s+t) = e^is.e^it. that gives us relations for cos(s+t) and sin(s+t) and the same functions of s and t. E.g. suppose we want to know what cos(2t) is, Well, e^i(2t) = e^i(t+t) = e^it.e^it. Now substitute the cos and sin formula in and multiply out. I.e. e^i(2t) = cos(2t) + i.sin(2t), and e^it = cos(t)+i.sin(t), so from e^i(2t) = e^it.e^it, we get (purely mechanically, as in Saxon!), that cos(2t) + i.sin(2t) = (cos(t)+i.sin(t)).(cos(t)+isin(t)) = cos^2(t) + i^2 sin^2(t) + 2i.cos(t)sin(t), I hope. Since i^2 = -1, that gives cos(2t) + i.sin(2t) = [cos^2(t)-sin^2(t)] + i.[2cos(t)sin(t)]. equating real and imaginary (i.e. x and y) parts, gives both formulas at once! cos(2t) = cos^2(t) - sin^2(t), and sin(2t) = 2cos(t)sin(t), the so - called "double angle formulas for cos and sin. Homework: check similarly the "addition laws" for cos and sin: cos(s+t) = cos(s)cos(t) - sin(s)sin(t), and sin(s+t) = cos(s)sin(t) + sin(s)cos(t). OK this was tedious, but it all follows from just remembering two formulas: e^it = cos(t) + i.sin(t), and e^(a+b) = e^a.e^b, (the addition law for exponentials) and most people find those easier to remember than the addition formulas for sin and cos. Now you know why I didn't give this before! I wanted it to look easy! Edited July 6, 2012 by mathwonk Quote Link to comment Share on other sites More sharing options...

mathwonk Posted July 6, 2012 Share Posted July 6, 2012 (edited) Ah yes, Jann in TX reminds us there is also the law of cosines, which actually occurs in Book II of Euclid as Prop. II.12 and Prop.II.13 I think, (and law of sines). So one already knows basic trig if one has had a classical course in Euclidean geometry, I will discuss this after dinner. The law of cosines is the pythagorean theorem for non right triangles, and follows from the right triangle case. Edited July 7, 2012 by mathwonk Quote Link to comment Share on other sites More sharing options...

Twigs Posted July 7, 2012 Share Posted July 7, 2012 Wow! You guys are a great class! The cute trick is the fact that if you use complex numbers, then sine and cosine are a special case of the exponential function! And the laws of exponents are simpler than those for sine and cosine. â€¦ ! Thanks! Quote Link to comment Share on other sites More sharing options...

letsplaymath Posted July 7, 2012 Share Posted July 7, 2012 Here's a series of lessons on trig with sticks and shadows that goes through it a little more slowly than mathwonk's quick summary -- and with interactive pictures to help! Quote Link to comment Share on other sites More sharing options...

mathwonk Posted July 7, 2012 Share Posted July 7, 2012 (edited) Ok here's the law of cosines, from euclid. Prop. II.12. Hmmm... the data limit here is exceeded even by a one page pdf file. Here is the word file but without the pictures. the original pdf file is week 2 day three in this link (epsilon camp notes): http://www.math.uga.edu/~roy/camp2011/10.pdf teaser: there is a point of view in which the law of cosines becomes just the familiar rule (a-b)^2 = a^2 - 2ab + b^2. (hooo hahahhaaaa). Law of Cosines Pythagoras says the square on the side opposite a right angle “equals” the two squares on the sides containing the angle. If the angle is acute, the square on the side opposite it is smaller than the two squares on the sides containing it, and if obtuse the square it is greater. The law of cosines tells exactly how much less or how much greater; in particular it says the discrepancy is twice the area of a certain rectangle. These are propositions 12-13, Book II of Euclid. Prop. II.12 (Law of cosines, obtuse case): Let ABC be a triangle on the base BC, with obtuse angle at C, and vertex at A. Drop a perpendicular from A to the line extending base BC, meeting it at X, outside segment BC. Then (AB)^2 = (AC)^2 + (BC)^2 + 2 (BC)(CX). Proof: By Pythagoras applied to right triangle AXB, we have (AB)^2 = (AX)^2 + (BX)^2. From IV.4, this equals (AX)^2 + (CX)^2 + (BC)^2 + 2 (BC)(CX). By Pythagoras applied to triangle AXC, this equals (AC)^2 + (BC)^2 + 2 (BC)(CX). QED. Exercise: Prove: Prop. II.13: (Law of cosines, acute case): Let triangle ABC on base BC have an acute angle at C, and vertex A. Drop a perpendicular from A to base BC, and assume it meets the base at X, between B and C. Then (AB)^2 = (AC)^2 + (BC)^2 - 2 (BC)(CX). Remarks: What does this theorem have to do with cosines? If you recall the definition of the cosine of angle <C in the picture for the acute case above, cos(<C) = |XC|/|AC|, the ratio of the numerical lengths of the two sides. Hence cos(<C).(AC) = XC, an equality of segments. Substituting this into Euclid’s formula above gives us (AB)^2 = (AC)^2 + (BC)^2 - 2 (AC)(BC).cos(<C), and this is the usual law of cosines in trigonometry. It also works for the obtuse case, since the cosine of an obtuse angle is negative, so the minus signs cancel and give us the formula in II.12 above. Edited July 7, 2012 by mathwonk Quote Link to comment Share on other sites More sharing options...

Kathy in Richmond Posted July 7, 2012 Share Posted July 7, 2012 (edited) Hi mathwonk, Did you watch the Project Mathematics trig videos with us at camp last summer? It was one of the optional evening activities, so you might have been busy with other stuff. Thanks to quark, I now have a link to public domain copies of those videos on the internet.:) These videos were produced by Tom Apostol out of Caltech, and they were a big hit with the camp kids. If you like to think visually, you might enjoy these, too. They brilliantly illustrate what mathwonk was explaining above, with the bonus of funky humor, animations, cool classical music, and even lessons in how to use that new-fangled device called a VCR. And beautiful math! :D There are even some pamplets with accompanying problems if you want that sort of practice. The first video explains similarity, radian measure and the unit circle, sin and cos on the unit circle, some basic trig identies, symmetries, and periodic curves. At the very end of the first video, right triangle trig (SOH-CAH-TOA) is explained as a special case. He also gives some applications of trig functions to sound waves & harmonics. The second video covers the Law of Cosines and the Law of Sines. It also explains the Taylor Series approximations to the sine and cosine functions (i.e., how your calculator finds trig values), and goes into lots of applications (oscillations: spring motion, water waves, alternating currents, light as a wave). Anyway, I'd highly recommend these as a mathematically sound but accessible intro to trig to anyone out there who likes the visual approach! Edited July 7, 2012 by Kathy in Richmond Quote Link to comment Share on other sites More sharing options...

Caroline Posted July 7, 2012 Share Posted July 7, 2012 Except it is Soh Cah Toa All you need to know about Trig for the ACT test is found in most Geometry texts. You need to be able to solve right triangles using Sine, Cosine and Tangent ratios: Soa Cah Toa There is usually one question on the ACT involving the "Law of Sines" and one involving the ''Law of Cosines"-- these are 2 formulas that have to be memorized--but are easy to use once the student knows Soa Cah Toa (the right triangle ratios). I teach it in 2-3 class periods. You can search the Khan Academy videos for the right triangle trig ratios and the Law of Sines and Law of Cosines. For extra practice look at Kutasoftware.com free worksheets in their Algebra 2 section for those topics. Quote Link to comment Share on other sites More sharing options...

mathwonk Posted July 7, 2012 Share Posted July 7, 2012 (edited) Thank you Kathy! Before i look at those I will explain my take on the law of sines, which seems rather easy. (mathematicians have a habit of calling "easy" anything they can even come close to understanding, probably to put subtle pressure on anyone who may think it is harder. This is a sort of childish one upsmanship. So to a mathematician there are two categories, easy, and I don't get it. Technically, in this case, easy means the explanation is short and uses no tricky business.) Take a triangle with base AB and vertex C, and side a opposite angle A, side b opposite angle B, and so on. Assume angles A and B are both acute to make it easy. Now drop a perpendicular from vertex C to side AB hitting it at point Q between A and B. Let the length of that perpendicular CQ be x. Now by the very definition of sin and cos, we have that sin(A) = x/b, and sin(B) = x/a. Hence x = bsin(A) = asin(B), so sin(A)/a = sin(B)/b. voila! I don't use this much. It is also true that sin(A)/a = sin©/c. Edited July 7, 2012 by mathwonk Quote Link to comment Share on other sites More sharing options...

Jann in TX Posted July 7, 2012 Share Posted July 7, 2012 that is what I get for typing with a 90 pound German Shepherd in my face (I probably should have fed him first!)... What a silly mistake as I've been teaching this for YEARS (correctly!).:tongue_smilie: Quote Link to comment Share on other sites More sharing options...

mathwonk Posted July 7, 2012 Share Posted July 7, 2012 (edited) Coordinate geometry and trig There is a nice way to do geometry in the (x,y) plane, using coordinates that allows the algebraic tools of addition and multiplication to enhance the geometry. For each point (a1,a2) in the (x,y) plane, we imagine an arrow going from the origin (0,0) to the point (a1,a2). We call this arrow A. Let B be another arrow fro the origin (0,0) to the point (b1,b2). If we add the coordinates getting another point (a1+b1, a2+b2), we can ask how the arrow A+B from the origin (0,0) to (a1+b1, a2+b2) is related to the first two arrows. It turns out that the arrow A+B forms the diagonal of a parallelogram with sides A and B. The vertices of this parallelogram are the points (0,0), (a1,a2), (a1+b1, a2+b2), and (b1,b2). E.g. if A = (1,0), and B = (0,1), then A+B = (1,1) is the diagonal of the square with vertices (0,0), (1,0), (1,1), and (0,1). Consider the triangle with vertices (0,0), (a1,a2), and (a1+b1, a2+b2). The arrows A and A+B form two sides of this triangle with common vertex (0,0). The third side, which goes from the point (a1,a2) to the point (a1+b1, a2+b2), is parallel to the arrow B, and has the same length. Thus if the arrows A and B are perpendicular, the arrow A+B is the hypotenuse of a right triangle whose sides have the same lengths as the arrows A and B. If we denote length of an arrow by | |, then by Pythagoras we get |A+B|^2 = |A|^2 + |B|^2. By the same reasoning, if we consider the triangle with two sides A and B, its third side is parallel to the arrow A-B, and has the same length. Hence by Pythagoras, if A and B are perpendicular, then |A-B|^2 = |A|^2 + |B|^2. We will define a multiplication of arrows that captures this theorem and also the more general law of cosines. To multiply two arrows A = (a1,a2) and B = (b1,b2), we define their “dot product”as A.B = a1b1 + a2b2, which is a number, rather than an arrow. It is easy to check that this multiplication has some of the properties of usual multiplication, like commutativity and distributivity for addition, and so on, but the product of two non - zero arrows can be zero. E.g. (1,0).(0,1) = 1.0 + 0.1 = 0+0 = 0. Moreover, the product of an arrow with itself is exactly the square of its length, i.e. A.A = (a1)^2 + (a2)^2 = |A|^2, by Pythagoras. In fact the dot product of two arrows is zero exactly when the arrows are perpendicular. I.e. consider A and B as two sides of a triangle. Then the third side is parallel to the arrow A-B, and has the same length. Hence |A-B|^2 = (A-B).(A-B) = A.A – 2A.B + B.B =|A|^2 +|B|^2 – 2A.B. But if A and B are perpendicular, then by Pythagoras we must have |A-B|^2 = |A|^2 + |B|^2, so A.B must be zero. If A and B are sides of any triangle with third side parallel to and of same length as A-B, then again |A-B|^2 = A|^2 +|B|^2 – 2A.B. This looks exactly like the law of cosines except that we have 2A.B in place of 2|A||B|cos© where c is the angle between A and B. Thus in fact A.B must equal |A||B|cos©. If on the other hand we knew that A.B = |A||B|cos©, then we get the law of cosines by expanding the dot product |A-B|^2 = (A-B).(A-B) = A.A – 2A.B + B.B = |A|^2 +|B|^2 – 2A.B =|A|^2 +|B|^2 – 2|A||B|cos©. This allows us to remember the easier formula A.B = |A||B|cos©, and then to recover the more complicated law of cosines. It also gives a way to calculate cosines without a calculator. E.g. the angle between the arrows A and B has cosine equal to (A.B)/|A||B|. E.g. the angle between A = (1,0) and B = (1,sqrt(3)) has cosine equal to 1/2. Remember what angle that is? Edited July 7, 2012 by mathwonk Quote Link to comment Share on other sites More sharing options...

mathwonk Posted July 7, 2012 Share Posted July 7, 2012 (edited) The thread killer strikes again! To summarize that mess, define a product of two arrows A,B in the plane, beginning at the origin, as: A.B = |A||B|cos©, where | | denotes length, and c is the angle between the two arrows. Also define an addition of arrows, by placing the tail of one arrow at the head of the other and taking the sum to go from the tail of the first to the head of the second arrow. Then the third side of the triangle with arrows A and B as two sides, is parallel to the arrow A-B, and of same length. Then the law of cosines becomes the usual rule for expanding a square: (A-B).(A-B) = A.A - 2A.B + B.B and since the angle between any vector and itself is zero, we get: |A-B||A-B|cos(0) = |A||A|cos(0) - 2 |A||B| cos© + |B||B|cos(0). Then since cos(0)m = 1, we get |A-B|^2 = |A|^2 - 2|A||B| cos© + |B|^2, which is exactly the usual law of cosines. Caveat: I have not proved here that this multiplication is distributive, For that I probably need the law of cosines! (sometime later)... Well, duh, I guess that's the whole point: the law of cosines is equivalent to saying this multiplication is distributive, i.e. is a multiplication! So this does not reprove the law of cosines, it just restates it in a more natural form. Then the three term principle says that since the explicit multiplication rule A.B = a1b1 + a2b2, is also distributive, and agrees with the first one on vectors that are equal, it must agree as well on all vectors. My point here is to try tio show how elementary math is illuminated when viewed from a higher point of view. No advanced math is being done here, but we are seeing elementary math more clearly I hope, by exposing its structure. This is what is not obtained from plug and chug treatments. If anyone has a book on trig, Saxon or otherwise, it might be interesting to compare its treatment with the one given here. Edited July 7, 2012 by mathwonk Quote Link to comment Share on other sites More sharing options...

mathwonk Posted July 7, 2012 Share Posted July 7, 2012 (edited) By the way, if anyone has a child who looks at the law of cosines |C|^2 = |A|^2 +|B|^2 – 2|A||B|cos©, and remarks that the right hand side looks kind of like what you get when you expand (A-B)^2, or asks whether C = A-B, then you have a potential mathematician - certainly a child who is thinking like one. Mathematics is about looking for patterns, and analogies. (By the way, I believe the college board recently removed analogies from the verbal SAT, which renders it much less useful as a gauge of reasoning ability. We noticed long ago at UGA that the verbal SAT measured math aptitude, or at least success in our entry level math courses, better than the quantitative part. In fact there were two measures that tracked entry level math success, the verbal SAT and the social security number, which latter seems to reflect what state the child went to school in. The quantitative SAT was unrelated to entry level college math course success. Hence efforts to raise that number are misguided, unless of course it is related to college entrance. Oddly there is apparently a big difference between what admissions officials look for and what professors look for. I may have damaged my child's entrance chances by recommending he skip taking advanced college level courses while in high school and master basic high school level courses thoroughly instead.) Edited July 7, 2012 by mathwonk Quote Link to comment Share on other sites More sharing options...

Tamarind Posted July 9, 2012 Author Share Posted July 9, 2012 Thanks for all the great trig advice! Quote Link to comment Share on other sites More sharing options...

mathwonk Posted July 9, 2012 Share Posted July 9, 2012 (edited) You are so welcome! I enjoy crafting explanations for individuals. Whenever I try to offer my old notes to anyone I always feel they aren't quite suitable for that particular person and prefer to write a fresh version. This sequence of explanations here started at least as a response to signals picked up from questions here. When there is a little back and forth, one gets a better sense of what to say. Of course I usually go off the deep end after a while and say whatever comes to mind, but even that is inspired by the conversation, and may become useful later. This last stuff e.g. is essentially an intro to vector algebra, usually taught in 3rd semester calculus, but without using those words. I think I was in the second or third grade (??) when some ambitious student teacher showed us the law of cosines, but I probably had my PhD before I realized it is just the distributivity of the dot product. And it was only recently that I found out it had an area formulation in Euclid! I think young kids should be shown these connections as soon as feasible, independent of the artificial separation of topics in usual curricula. Edited July 9, 2012 by mathwonk Quote Link to comment Share on other sites More sharing options...

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