[As always, don't forget to give proper attribution when using this in the classroom or elsewhere as indicated in the sidebar]
In a standard trig unit, students learn those wonderful formulas for the sin and cos of the sum and difference of angles. Many creative methods have been developed to derive these formulas and, depending on the ability of the group and teacher preference, these are demonstrated or not. Students are typically shown various mnemonics for recalling them on the big test, but, in this investigation, students will derive sin 15° using only 30-60-90 triangle ratios and the Pythagorean Theorem. We will then compare the result to that obtained by the traditional formulas for sin(45°-30°) or sin(60°-45°) and show equivalence by algebraic methods using radicals. This is not an attempt to develop a general approach to deriving sum/difference formulas, although readers are invited to try a generalization. You may recall other posts on this blog of a similar nature.
THE PROBLEM/INVESTIGATION
Refer to the triangle above. If the print is too small, click on the image to magnify.
∠A = 75° and ∠B = 15°
(a) In the triangle above, locate point D on side BC such that ∠CAD = 60° . Express the lengths of the sides of triangle CAD in terms of a.
[Note: We could avoid the variable a altogether and assign a value of 1 since this is a ratio problem.]
(b) Show that CB = a √3 +2a.
(c) Use the Pythagorean Theorem to show that AB = a √(8+4 √ 3)
(d) Verify the identity (√ 6 + √ 2)2 = 8+4 √ 3. Use this to rewrite AB.
(e) Use above results to obtain an expression for sin 15°.
(f) Use the standard trig formula for sin(45°-30°) to obtain an expression for sin 15°.
(g) Show your results in (e) and (f) are equivalent.
An Instructional Aside
When introducing the formula for sin(A+B), for example, teachers sometimes motivate it effectively using numerical values or considering the special case A=B. Here's an alternative:
Consider the special case A+B = 90°
Ask students to verify the formula for sin(A+B) in this special case. Simple, but at least it's something slightly different to pique their curiosity.
Standard Disclaimer:
This investigation is not copied from some other source. As it is original and has not been edited by others, there's always the possibility of error. Please feel free to suggest corrections/edits/extensions...
You know I welcome your comments!
Saturday, November 3, 2007
Another View of sin(A-B) for a Special Case: An Investigation
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12 comments:
This year we are constructing everything. We might construct AD...
Nice, Dave.
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.
For example, using the value of sin 75 derived, we can find the sin of 7.5.
TC
tc--
I played around with a generalization and I obtained some interesting trig relationships.
Referring to the figure:
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.
Cool!
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.
TC.
tc--
Yes! Just another innocent observation that opens a door...
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 great."
jonathan--
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 why 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).
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.
While we are on the subject of double angles, a suggested activity:
(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.
(2) Find a parametric equation (in terms of t = angle (ACB)) for the co-ordinates of the point C.
(3) From the above, find a non-parametric equation for the locus of C.
TC
tc--
sorry for not responding sooner...
This is a wonderful problem and it lead me down several paths.
I may have made it much harder than it is but I'll get started:
We'll follow your convention:
A(0,0), B(1,0) and C(x,y)
(1) From the Law of Sines, AC = 2cost, where t = theta or angle ACB.
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.
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.
I was able to eliminate the parameter before expressing x and y in terms of t:
(x-1)^2+y^2 = (x^2+y^2-1)^2.
I'll derive this for the case 0<t≤45 (having a diagram here would be much easier to follow):
In this case, the altitude from C falls within segment AB. This leads to 2 equations:
(i) x^2+y^2=4cos^2(t)
(ii) (1-x)^2 + y^2 = (4cos^2(t)-1)^2
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!).
We can express x and y in terms of 3t but I assumed you did not want that.
Here's the parametric form:
x = 1-(cos2t)(4cos^2(t)-1)
y = (sin2t)(4cos^2(t)-1).
Double angle identities can be used to express x and y in terms of trig functions of t.
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.
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...
Oh, No!
I meant to say angle ABC = 2*angle BAC.
I'll have to look through your solution carefully, though. (Actually, I will just try to work it out myself first).
TC.
For the wrong problem, I obtain the same rectangular form as you do.
The parametric form I get, however, looks different, though it might be the same:
x = 2 cos^2(t) (4 sin^2(t) -1)
y=sqrt(2) cos(t)sqrt(4 cos^2(t)-1)
I will have to spend some more time checking, but that will have to wait until later.
TC
An update: On further investigation, I get a much nicer form:
x=-2cos(t)cos(3t)
y=+-2cos(t)sin(3t)
Need to double check, but this is surprisingly elegant (so most likely correct?)
TC
Looks good, tc!
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.
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!
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