tag:blogger.com,1999:blog-8231784566931768362.post3104673726572355820..comments2023-09-09T08:21:55.454-04:00Comments on MathNotations: Another View of sin(A-B) for a Special Case: An InvestigationDave Marainhttp://www.blogger.com/profile/13321770881353644307noreply@blogger.comBlogger12125tag:blogger.com,1999:blog-8231784566931768362.post-66286546746429884022007-11-12T15:06:00.000-05:002007-11-12T15:06:00.000-05:00Looks good, tc!I got to the 3θ form but I th...Looks good, tc!<BR/>I got to the 3θ form but I thought you wanted all expressions in terms of the single angle. But you're right -- this is much more elegant.<BR/>Also, I'm hoping you get a chance to see the new posting as of 11-12-07 which develops the sum/difference identities using matrices. This activity/investigation is a reach but some motivated high school students might enjoy it. Please check it for accuracy!Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-45497644070961653012007-11-12T13:27:00.000-05:002007-11-12T13:27:00.000-05:00An update: On further investigation, I get a much ...An update: On further investigation, I get a much nicer form:<BR/><BR/>x=-2cos(t)cos(3t)<BR/>y=+-2cos(t)sin(3t)<BR/><BR/>Need to double check, but this is surprisingly elegant (so most likely correct?)<BR/><BR/>TCUnknownhttps://www.blogger.com/profile/06449079338919787252noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-13874336074534272802007-11-11T10:00:00.000-05:002007-11-11T10:00:00.000-05:00For the wrong problem, I obtain the same rectangul...For the wrong problem, I obtain the same rectangular form as you do. <BR/><BR/>The parametric form I get, however, looks different, though it might be the same:<BR/>x = 2 cos^2(t) (4 sin^2(t) -1)<BR/>y=sqrt(2) cos(t)sqrt(4 cos^2(t)-1)<BR/><BR/>I will have to spend some more time checking, but that will have to wait until later. <BR/><BR/>TCAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-77434923533860751182007-11-10T14:23:00.000-05:002007-11-10T14:23:00.000-05:00Oh, No!I meant to say angle ABC = 2*angle BAC. I'l...Oh, No!<BR/><BR/>I meant to say angle ABC = 2*angle BAC. <BR/><BR/>I'll have to look through your solution carefully, though. (Actually, I will just try to work it out myself first).<BR/><BR/><BR/>TC.Unknownhttps://www.blogger.com/profile/06449079338919787252noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-59568213404014725712007-11-10T10:31:00.000-05:002007-11-10T10:31:00.000-05:00tc--sorry for not responding sooner...This is a wo...tc--<BR/>sorry for not responding sooner...<BR/>This is a wonderful problem and it lead me down several paths.<BR/>I may have made it much harder than it is but I'll get started:<BR/><BR/>We'll follow your convention:<BR/>A(0,0), B(1,0) and C(x,y)<BR/><BR/>(1) From the Law of Sines, AC = 2cost, where t = theta or angle ACB.<BR/><BR/>Drop a perpendicular from C to line AB to show that x^2 + y^2 = 4(cos^2)(t) < 4 since t>0. From this we obtain that x<2 and y<2. This answers part (1), i.e., C is to the left of the line x = 2.<BR/><BR/>I then proved that BC = 4cos^2(t)-1 . This took some work because I had to consider both cases 0<t≤45 and 45<t<60. I'll provide details for this if you want.<BR/><BR/>I was able to eliminate the parameter before expressing x and y in terms of t:<BR/>(x-1)^2+y^2 = (x^2+y^2-1)^2.<BR/><BR/>I'll derive this for the case 0<t≤45 (having a diagram here would be much easier to follow):<BR/>In this case, the altitude from C falls within segment AB. This leads to 2 equations:<BR/>(i) x^2+y^2=4cos^2(t)<BR/>(ii) (1-x)^2 + y^2 = (4cos^2(t)-1)^2<BR/>Replacing 4cos^2(t) by x^2+y^2 in (ii), we obtain the rectangular equation of the curve, although I have to believe there's a much easier form for this (and an easier derivation than mine!).<BR/>We can express x and y in terms of 3t but I assumed you did not want that.<BR/>Here's the parametric form:<BR/>x = 1-(cos2t)(4cos^2(t)-1)<BR/>y = (sin2t)(4cos^2(t)-1).<BR/>Double angle identities can be used to express x and y in terms of trig functions of t.<BR/><BR/>If we fix the locations of C and B and let A vary, we get a much nicer result! In this case the locus of A is the polar curve r = 2cos(t) or the circle x^2+y^2 = 2x, whose center is (1,0) and whose radius is 1. I was hoping this is what you meant to avoid all of the work above, but I guess not. <BR/>I may have messed up, but I really enjoyed this. I think the time has come, tc, for you and I to co-author this blog. You have a wealth of knowledge and great problems to share. We could complement each other. Let me know...Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-19497932982088041332007-11-07T09:31:00.000-05:002007-11-07T09:31:00.000-05:00While we are on the subject of double angles, a su...While we are on the subject of double angles, a suggested activity:<BR/><BR/>(1) Given a segment AB (with end-points say (0,0) and (1,0)). Let C be a point above the X-axis, so that ABC is a triangle, and angle ABC = 2* angle ACB. The point C has to lie to the left of a certain line. Find the equation of the line. <BR/><BR/>(2) Find a parametric equation (in terms of t = angle (ACB)) for the co-ordinates of the point C.<BR/><BR/>(3) From the above, find a non-parametric equation for the locus of C.<BR/><BR/>TCUnknownhttps://www.blogger.com/profile/06449079338919787252noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-83606876365850219102007-11-05T20:10:00.000-05:002007-11-05T20:10:00.000-05:00jonathan--I'm not ignoring you! I missed that 'con...jonathan--<BR/>I'm not ignoring you! I missed that 'constructive' comment you made! Now that I see it, I love it. I know that I learned so much from doing constructions, even more, when my geometry teacher asked us <B>why</B> a particular construction worked. Every pair of those intersecting arcs leads to a rich discussion of properties of circles and congruence proofs (or similar triangles). <BR/><BR/>By the way, do you use Geometer's SketchPad or traditional tools or both? It might be interesting to start a thread about the relative merits of virtual tools vs. traditional. Having experienced the software, I see both sides of that coin, but I haven't read too much about this on others' blogs.Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-89914947214954436482007-11-05T19:59:00.000-05:002007-11-05T19:59:00.000-05:00tc--Yes! Just another innocent observation that op...tc--<BR/>Yes! Just another innocent observation that opens a door...<BR/><BR/>This reminds me of a conversation I was having with another math nerd like me. My wife overheard me say, "What a great problem!" In her inimitable deadpan style, she retorted, "That's funny. I've never had any problems I would consider <I>great</I>."Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-4484776083800374952007-11-05T18:20:00.000-05:002007-11-05T18:20:00.000-05:00Cool!From that, you can derive the formula for cos...Cool!<BR/><BR/>From that, you can derive the formula for cos(2B) in terms of cos(B), tan (B) in terms of sin(2B) and cos(2B), and sin(2B) from the other two.<BR/><BR/>TC.Unknownhttps://www.blogger.com/profile/06449079338919787252noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-39691210672776733402007-11-05T17:11:00.000-05:002007-11-05T17:11:00.000-05:00tc--I played around with a generalization and I ob...tc--<BR/>I played around with a generalization and I obtained some interesting trig relationships.<BR/>Referring to the figure:<BR/>For the purposes of this comment, we will let AD = DB = 1. We can then show that AC =sin(2B) and CD = cos(2B). AB can then be obtained two ways - either from sin(B) = AC/AB or from the Pythagorean Theorem in triangle ACB. The first leads to AB = 2cos(B).The second gives (AB)^2 = (sin^2)(2B) + (1+cos(2B))^2 which simplifies to 4cos^2(B). These are just a few rambling thoughts.Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-45092389464061738402007-11-05T08:18:00.000-05:002007-11-05T08:18:00.000-05:00Nice, Dave.It appears that we can generalize this ...Nice, Dave.<BR/><BR/>It appears that we can generalize this procedure. Given an angle y for which we know the sin, we can find the sin of x such that y+2x=90. The procedure is similar to what you indicate: have the upper angle be y+x, and the lower angle be x, divide the upper angle into y and x, and follow the increasingly cumbersome numerical manipulations of the irrationals.<BR/><BR/>For example, using the value of sin 75 derived, we can find the sin of 7.5.<BR/><BR/>TCUnknownhttps://www.blogger.com/profile/06449079338919787252noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-43863386865035021382007-11-03T23:50:00.000-04:002007-11-03T23:50:00.000-04:00This year we are constructing everything. We might...This year we are constructing everything. We might construct AD...Anonymousnoreply@blogger.com