A 5-7-8 Triangle for Starters -- An In-Depth Exploration in Algebra, Geometry and Trigonometry

As promised, here is an investigation/activity/challenge/.../ (Anyone remember a certain star college and pro football player, with initials K.S., who was nicknamed 'Slash' because he could play several positions?).

The following extensive activity is designed as a long-term assignment, however, modify it as you wish. Don't forget to forgive proper attribution as indicated in the sidebar.


A Possible Instructional Scenario...
"Girls and boys, what do you think of when you're told that a triangle has a 60° angle? What if you're given that all the sides have integer lengths? Why can't it be 30-60-90 in that case? Of course, it could still be equilateral, but for this exploration, we are looking for scalene triangles with integer sides and a 60° angle. By the way, if it's not equilateral, how do we know it must be scalene? Why couldn't it be isosceles?"
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The Exploration

Let's agree on the following labeling for our triangle:
ΔABC with ∠C = 60° with opposite side c. Other angles are labeled A and B with opposite sides a and b, respectively. For this investigation, b > a.

(a) Show that the angle opposite the '7' side in the 5-7-8 triangle in the title of this post is 60°. Using the labeling above: If a=5, b=8, c=7, show that ∠C = 60°.

Note: This can be done with or without the Law of Cosines. As Joshua pointed out in a previous post, the student can work with the altitude on the '5' side and use the Pythagorean Theorem and algebra to show that a 30-60-90 triangle is formed. This method is instructive (and constructive too!).

For parts (b) - (e), we will no longer be focusing only on the 5-7-8 triangle. Consider any integer-sided ΔABC with ∠C = 60° and with opposite side of length 7.
(b) Show that a² + b² - ab = 49. [*]
See note after part (a) for two possible methods.

(c) We are looking for integer solutions to [*]. There are several methods including a Pell equation approach (we will not go in that direction). Students can certainly try a guess-test strategy, however we can initially restrict the possible values of b, can't we?
As an initial boundary, show that 13 >b ≥ 8 using basic geometry.

(d) Show that a better restriction for b is 14/√3 ≥ b ≥ 8 by:

(i) [Trig] Using the Law of Sines
(ii) [Advanced Algebra] Use the quadratic formula in [*] to solve for a in terms of b. Using the discriminant, show that b ≤ 14/√3.

From this result, explain why it follows that b must equal 8.

(e) Solve for a.
Suggestions: From [*] OR use your result from the quadratic formula in (d)(ii) to show that there are exactly 2 scalene triangles satisfying the above conditions.
The answers for this are:
a=3, b=8, c=7;
a=5, b=8, c=7
BTW, is it a coincidence that the two values of a happen to add up to b?

(f) REPEAT PARTS (b) - (e) for an integer-sided scalene triangle with a 60° angle and an opposite side of length 13. There will be some slight modifications needed such as formula [*]. State and use a similar inequality from (d) to show that there are two possible values for b, namely 14 and 15. Go further and show b=14 is not possible (e.g., using the discriminant).
Then show that b = 15 leads to two solutions (triangles) - sorry, I'm not giving these away yet!

(g) REPEAT PARTS (b) - (e) for an integer-sided scalene triangle with a 60° angle and an opposite side of length 19. Again, you should find that the inequality from (d) leads to two possible values for b, only one of which works. Give the two solutions (triangles).

(h) Some of you will no doubt wonder why we did 3 separate analyses, when a slightly more general approach could have been used, specifically a more general inequality in part (d). Ok, so do that (keep all conditions, except side 'c' will now be a parameter).
Show that, in general, 2c/√3 ≥ b ≥ c.

(j) Surely, we can't go further other than searching for a general solution for the 60° problem. Of course, we can: Come up with similar questions and solutions to all parts above if ∠C = 120°. I'll start you off: a=3, b=5, c=7. Show that ∠C = 120°, etc...

I'm sure our astute and talented readers will pick up on my usual errors or omissions or make suggestions to improve the flow of the activity. I'm counting on you! Enjoy...

The Matrix Reloaded: Deriving sin(A+B), cos(A+B) by Rotation Matrices

In our previous post we asked students to verify the sin(A+B) identity for an angle of 75°. Although one might try to generalize the result, there are many other derivations for the sum and difference identities that teachers have seen or used. Those who teach this topic whether it be in an Advanced Algebra/Trig, Advanced Math, Precalculus, or some similarly named course have the choice of deriving the formula, outlining a proof and having students attempt to provide the details or simply motivating the formula. My guess is that, unless you have a highly motivated and strong group of students, the full derivation is not done in class. By the way, an excellent applet for helping students visualize these formulas is located here. One could use this for classroom demonstration or in tutorial mode.

The forms of these trig identities often engage students and some may wonder if there's a pattern one can recognize that occurs elsewhere in mathematics. Using nonsense names for functions, these rules seem to have the form: the squiggle of the first times the squeegee of the second ± the squiggle of the second times the squeegee of the first. Students will hopefully recognize this pattern again when they see the product rule in calculus. Is there an overarching concept that encompasses forms like this? Well, I'd like to suggest one in this post. You can decide for yourself if it's worth developing the required machinery for secondary students (or it can be an enrichment topic or project).

The trig identities mentioned above have the form of a sum (or difference) of products. Where might one encounter this in mathematics? For myself, it's when I studied the products of matrices or the dot product of 2 vectors. v₁⋅v₂ = a₁b₁+a₂b₂. Matrix products and vectors are required in some states' curricula (CA for example), so there is a basis for this approach (pun intended!!).

To simplify , we will focus on position vectors with a length of 1 so that their terminal points are all on the unit circle. Students learn early on in trig that each point on the unit circle can be represented using the parameter t or θ as in (cosθ, sinθ) and that θ represents an angle or rotation, counterclockwise for θ>0, etc. Another way of viewing this is that the unit vectors (1,0) and (0,1), (which are labeled i and j by most physics teachers), are transformed by this rotation as follows:
(1,0) ---> (cos θ, sin θ) (*)
(0,1) ---> (-sin θ, cos θ).
These results are straightforward and students should be able to demonstrate them graphically for values of θ in Quadrant I. For those who recall the terminology from linear algebra, (1,0) and (0,1) are referred to as an orthornormal basis.

[We will use the notation R_θ to denote the rotation transformation by an angle θ about the origin]:

STUDENT (or reader) ACTIVITY
1. Have students verify or find the following (either graphically or using (*) above) :
R_90° (1,0) = (0,1)
R_90° (0,1) = (-1,0)
R_60° (1,0) = (1/2,√3/2)
R_60° (0,1) = ______
R_45° (1,0) = ______
R_45° (0,1) = ______

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We will now show how the following 2x2 rotation matrix can accomplish this transformation:
(**)

Note that the first column (vector) is the rotation image of the vector (1,0) and similarly for the 2nd column. This idea of using (orthonormal) basis vectors to construct a matrix representing a transformation is crucial in linear algebra. This requires more development and that is not the purpose of this post. We are asking students to accept this for now, although we will have them verify that this matrix approach produces correct results in specific instances. We will now represent the rotation image of any point on the unit circle by multiplying this matrix by the column vector which represents that point. In rectangular (x,y) notation:

(***)

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STUDENT ACTIVITY (cont.)

2. Write the matrix form (**) of R_θ for θ =
(a) 90°
(b) 60°
(c) 45°
(d) -90°

3. Re-do problem #1 above using the matrix multiplication formula (***). Make sure your results match!
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So how does all of this relate to the sum/difference trig identities? We're almost there! Instead of using (x,y) to represent an arbitrary point on the unit circle, we will use the trigonometric form (cosα,sinα) in column vector form. For now we will assume α is between 0 and π/2:

STUDENT ACTIVITY (cont.)

4. Perform the following matrix multiplication:


Does the resulting column vector look familiar? Answer the next two questions and maybe you'll figure out why!

5. The result of #4 is equivalent to a single rotation of what angle?

6. What can we conclude?
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There's much more to say here but this is enough for now. As usual, I take full responsibility for errors or ambiguities. I await your thoughts. If this is a derivation you've seen before, let me know. One could also derive this using complex numbers or polar coordinates or ...

Another View of sin(A-B) for a Special Case: An Investigation

[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!

A Trig Mnemonic Revisited with Texify!

Some time ago, I posted a piece about math mnemonics. Buried near the bottom was my feeble attempt to make a table showing a well-known (?) fascinating pattern for sin and cos values for the common angles in Quadrant I. Over the years, some students have found this to be as useful as memorizing ordered pairs on the unit circle or deriving everything from 30-60-90 and 45-45-90 (which I still prefer personally). I've seen students make this table at the top of their trig unit exam - they figured it was worth the effort! I'm reprinting this today using an image created in LaTeX and the absolutely wonderful and easy to use Texify website. This has been a real boon for those using Blogger since LaTeX has not yet been supported. Many math bloggers have been using it for a while now and I'm sure they appreciate its power and simplicity as much as I do. Its author is Andrey Burkov and Ars Mathematica gives him proper credit here. Certainly if an old dog like me can learn new tricks like this, anyone can! By the way at the Texify site, there is an extremely well-written tutorial with many examples to follow. I suspect I will be using this a lot for my new posts and perhaps cleaning up my old. Let me know if the table below is as readable as it appears to me. and, of course, if you like the pattern, you can tell me that too!

The original post used the klutziest of notations and was barely readable. This should be a lot better! I omitted the row for the tan function which is just the quotient of rows 2 and 3:

A Challenge Problem: Ellipses and Tangents and Normals

MAA members will likely recognize the following challenge that appeared on the outside of the envelope in the mailing to members or prospective members. I plan on giving this to my AP Calculus students as review for the exam or afterwards. As usual I will modify it for the student, place it in the context of an activity, broken into several parts with some hints. The original problem comes with a helpful diagram, however, unless I scan it, it would be difficult to reproduce. The problem involves a property of a point on an ellipse and requires basic understanding of the parametric form of this curve and some basic calculus and trig. The last part of the activity suggests a possible significance of this property but I'll leave the details to our astute readers.

Consider a standard ellipse, center at (0,0), with major axis of length 2a on the x-axis and minor axis of length 2b.
Let P(x,y) be a generic point on this ellipse with the restriction that P is not one of the endpoints of the major or minor axes. Consider the tangent and normal lines at P. Let P denote the point of intersection of the normal line with the x-axis and Q, the point of intersection of the tangent line with the x-axis.
Prove that (OP)(OQ) = a²-b², where OP represents the distance between the origin and P and similarly for OQ.
Here is an outline with several parts for the student:
(a) Show that x = acos(t), y = bsin(t), 0<=t<2pi,>2-b²)/a)cos(t).
(f) Use (d) and (e) to derive the desired result: (OP)(OQ) = a²-b²
(g) Explain why we did not allow P to be an endpoint of the major or minor axes.
(h) What does the expression a²-b² have to do with the foci of the ellipse? For EXTRA CREDIT, investigate this 'focal' property further.