My Favorite One-Liners: Part 106

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Years ago, when I first taught Precalculus at the college level, I was starting a section on trigonometry by reminding my students of the acronym SOHCAHTOA for keeping the trig functions straight:

\sin \theta = \displaystyle \frac{\hbox{Opposite}}{\hbox{Hypotenuse}},

\cos \theta = \displaystyle \frac{\hbox{Adjacent}}{\hbox{Hypotenuse}},

\tan \theta = \displaystyle \frac{\hbox{Opposite}}{\hbox{Adjacent}}.

At this point, one of my students volunteered that a previous math teacher had taught her an acrostic to keep these straight: Some Old Hippie Caught Another Hippie Tripping On Acid.

Needless to say, I’ve been passing this pearl of wisdom on to my students ever since.

My Favorite One-Liners: Part 69

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

This story, that I’ll share with my Precalculus students, comes from Fall 1996, my first semester as a college professor. I was teaching a Precalculus class, and the topic was vectors. I forget the exact problem (believe me, I wish I could remember it), but I was going over the solution of a problem that required finding \tan^{-1}(7). I told the class that I had worked this out ahead of time, and that the approximate answer was 82^o. Then I used that angle for whatever I needed it for and continued until obtaining the eventual solution.

(By the way, I now realize that I was hardly following best practices by computing that angle ahead of time. Knowing what I know now, I should have brought a calculator to class and computed it on the spot. But, as a young professor, I was primarily concerned with getting the answer right, and I was petrified of making a mistake that my students could repeat.)

After solving the problem, I paused to ask for questions. One student asked a good question, and then another.

Then a third student asked, “How did you know that \tan^{-1}(7) was 82^o?

Suppressing a smile, I answered, “Easy; I had that one memorized.”

The class immediately erupted… some with laughter, some with disbelief. (I had a terrific rapport with those students that semester; part of the daily atmosphere was the give-and-take with any number of exuberant students.) One guy in the front row immediately challenged me: “Oh yeah? Then what’s \tan^{-1}(9)?

I started to stammer, “Uh, um…”

“Aha!” they said. “He’s faking it.” They start pulling out their calculators.

Then I thought as fast as I could. Then I realized that I knew that \tan 82^o \approx 7, thanks to my calculation prior to class. I also knew that \displaystyle \lim_{x \to 90^-} \tan x = \infty since the graph of y = \tan x has a vertical asymptote at x = \pi/2 = 90^o. So the solution to \tan x = 9 had to be somewhere between 82^o and 90^o.

So I took a total guess. “84^o,” I said, faking complete and utter confidence.

Wouldn’t you know it, I was right. (The answer is about 83.66^o.)

In stunned disbelief, the guy who asked the question asked, “How did you do that?”

I was reeling in shock that I guessed correctly. But I put on my best poker face and answered, “I told you, I had it memorized.” And then I continued with the next example. For the rest of the semester, my students really thought I had it memorized.

To this day, this is my favorite stunt that I ever pulled off in front of my students.

Engaging students: Defining sine, cosine and tangent in a right triangle

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Loc Nguyen. His topic, from Geometry: defining sine, cosine and tangent in a right triangle.

green line

What interesting (i.e., uncontrived) word problems using this topic can your students do now?

There are many real world applications that involve in this topic and I will incorporate some problems in real life to engage the students.  Suppose I have a classroom that has the shape of rectangular prism.  I will begin my lesson by challenging the students to find the height of the classroom and of course I will award them with something cool.  I believe this will ignite students’ curiosity and excitement to participate into the problem.  In the process of finding the height, I will gradually introduce the concept of right triangle trigonometry.  The students will learn the relationship of ratios of the sides in the triangle.  Eventually, the students will realize that they need this concept for finding the height of the classroom.  I will pose some guiding questions to drive them toward the solution.  Such questions could be: what can I measure? Can we measure the angle from our eyes to the opposite corner of the ceiling point?   What formula will help me to find the height?
trig1

 

After this problem I will provide them many different real world problems to practice such as:

trig2 trig3

 

 

green line

How can this topic be used in your students’ future courses in mathematics or science?

Knowing how to compute sine, cosine or tangent in the right triangle will help students a lot when they get to higher level math or other science class, especially Physics.  In higher level math, students will always have the chance to encounter this concept.  For example, in Pre-Calculus, the students will likely learn about polar system.  This requires students to have the strong fundamental understandings of sine, cosine and tangent in a right triangle.  Students will be asked to convert from the Cartesian system to polar system, or vice versa.  If they do not grasp the ideas of this topic, they will eventually encounter huge obstacles in future.  In science, especially physics, the students will learn a lot about the motions of an objects.  This will involve concepts of force, velocity, speed, momentum.  The students will need to understand the how to compute sine, cosine and tangent in the right triangle so that they can easily know how to approach the problems in physics.

green line

How can technology (YouTube, Khan Academy [khanacademy.org], Vi Hart, Geometers Sketchpad, graphing calculators, etc.) be used to effectively engage students with this topic?

This website, https://www.geogebra.org/material/simple/id/48148 , can be a great tool for the students to understand the relationships of the sides in the right triangle.  The website creates an activity for students to explore the ratios of the sides such as AC/BC, AC/AB, and BC/AB.  The students will observe the changes of the ratios based on the changes of theta and side BC which is the hypotenuse.  At this point, the students will be introduced the name of each side of the right triangle which corresponds to theta such as opposite, adjacent and hypotenuse.  This activity allows the students to visualize what happens to the triangle when we change the angle or its side lengths.  The students will then explore the activity to find interesting facts about the side ratios.  I will pose some questions to help the students understand the relationships of side ratios. Such questions could be:  What type of triangle is it?  Tell me how the triangle changes as we change the hypotenuse or angle.  If we know one side length and the angle, how can we find the other side lengths?  Those questions allow me to introduce the terms sine, cosine, and tangent in the right triangle.

trig4

References

https://en.wikibooks.org/wiki/High_School_Trigonometry/Applications_of_Right_Triangle_Trigonometry

https://www.geogebra.org/material/simple/id/48148

 

 

SOHCAHTOA

Years ago, when I first taught Precalculus at the college level, I was starting a section on trigonometry by reminding my students of the acronym SOHCAHTOA for keeping the trig functions straight:

\sin \theta = \displaystyle \frac{\hbox{Opposite}}{\hbox{Hypotenuse}},

\cos \theta = \displaystyle \frac{\hbox{Adjacent}}{\hbox{Hypotenuse}},

\tan \theta = \displaystyle \frac{\hbox{Opposite}}{\hbox{Adjacent}}.

At this point, one of my students volunteered that a previous math teacher had taught her an acrostic to keep these straight: Some Old Hippie Caught Another Hippie Tripping On Acid.

Needless to say, I’ve been passing this pearl of wisdom on to my students ever since.

10 Secret Trig Functions Your Math Teachers Never Taught You

Students in trigonometry are usually taught about six functions:

\sin \theta, \cos \theta, \tan \theta, \cot \theta, \sec \theta, \csc \theta

I really enjoyed this article about trigonometric functions that were used in previous generations but are no longer taught today, like \hbox{versin} \theta and \hbox{havercosin} \theta:

http://blogs.scientificamerican.com/roots-of-unity/10-secret-trig-functions-your-math-teachers-never-taught-you/

Naturally, Math With Bad Drawings had a unique take on this by adding a few more suggested functions to the list. My favorites:

Hippopotenuse

Source: https://www.facebook.com/sandraboynton/photos/a.206184542749790.56269.114554295246149/1075341509167418/?type=3&theater

See also http://www.zazzle.com/hippopotenuse_t_shirt_by_sandra_boynton-235290474700412664

The antiderivative of 1/(x^4+1): Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The following links comprised my series on the computation of

\displaystyle \int \frac{dx}{x^4+1}

Part 1: Introduction.

Part 2: Factoring the denominator using De Moivre’s Theorem.

Part 3: Factoring the denominator using the difference of two squares.

Part 4: The partial fractions decomposition of the integrand.

Part 5: Partial evaluation of the resulting integrals.

Part 6: Evaluation of the remaining integrals.

Part 7: An apparent simplification using a trigonometric identity.

Part 8: Discussion of the angles for which the identity holds.

Part 9: Proof of the angles for which the identity holds.

Part 10: Implications for using this identity when computing definite integrals.

 

 

Different Ways of Solving a Contest Problem: Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The following links comprised my series on different ways of solving the contest problem “If 3 \sin \theta = \cos \theta, what is \sin \theta \cos \theta?”

Part 1: Drawing the angle \theta

Part 2: A first attempt using a Pythagorean identity.

Part 3: A second attempt using a Pythagorean identity and the original hypothesis for \theta.

 

 

The antiderivative of 1/(x^4+1): Part 9

In the course of evaluating the antiderivative

\displaystyle \int \frac{1}{x^4 + 1} dx,

I have stumbled across a very curious trigonometric identity:

\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) -\pi if x < x_2,

\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) if x_2 < x < x_1,

\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) + \pi if x> x_1,

where x_1 and x_2 are the unique values so that

\tan^{-1} ( x_1\sqrt{2} - 1 ) + \tan^{-1}( x_1 \sqrt{2} + 1) = \displaystyle \frac{\pi}{2},

\tan^{-1} ( x_2\sqrt{2} - 1 ) + \tan^{-1}( x_2 \sqrt{2} + 1) = -\displaystyle \frac{\pi}{2}.

I will now show that x_1 = 1 and x_2 = -1. Indeed, it’s apparent that these have to be the two transition points because these are the points where \displaystyle \frac{x \sqrt{2}}{1 - x^2} is undefined. However, it would be more convincing to show this directly.

To show that x_1 = 1, I need to show that

\tan^{-1} (\sqrt{2} - 1 ) + \tan^{-1}( \sqrt{2} + 1) = \displaystyle \frac{\pi}{2}.

I could do this with a calculator…

arctangent…but that would be cheating.

Instead, let \alpha = \tan^{-1} (\sqrt{2} - 1 ) and \beta = \tan^{-1} (\sqrt{2} + 1 ), so that

\tan \alpha = \sqrt{2} - 1,

\tan \beta = \sqrt{2} + 1.

Indeed, by SOHCAHTOA, the angles \alpha and \beta can be represented in the figure below:

arctangenttriangle2The two small right triangles make one large triangle, and I will show that the large triangle is also a right triangle. To do this, let’s find the lengths of the three sides of the large triangle. The length of the longest side is clearly \sqrt{2} - 1 + \sqrt{2} + 1 = 2\sqrt{2}. I will use the Pythagorean theorem to find the lengths of the other two sides. For the small right triangle containing \alpha, the missing side is

\sqrt{ \left(\sqrt{2} - 1 \right)^2 + 1^2} = \sqrt{2 - 2\sqrt{2} + 1 + 1} = \sqrt{4-2\sqrt{2}}

Next, for the small right triangle containing \beta, the missing side is

\sqrt{ \left(\sqrt{2} + 1 \right)^2 + 1^2} = \sqrt{2 + 2\sqrt{2} + 1 + 1} = \sqrt{4+2\sqrt{2}}

So let me redraw the figure, eliminating the altitude from the previous figure:

arctangenttriangle3

Notice that the condition of the Pythagorean theorem is satisfied, since

\left( \sqrt{4-2\sqrt{2}} \right)^2 + \left( \sqrt{4+2\sqrt{2}} \right)^2 = 4 - 2\sqrt{2} + 4 + 2 \sqrt{2} = 8,

or

\left( \sqrt{4-2\sqrt{2}} \right)^2 + \left( \sqrt{4+2\sqrt{2}} \right)^2 = \left( 2\sqrt{2} \right)^2.

Therefore, by the converse of the Pythagorean theorem, the above figure must be a right triangle (albeit a right triangle with sides of unusual length), and so \alpha + \beta = \pi/2. In other words, x_1 = 1, as required.

To show that x_2 = -1, I will show that the function f(x) = \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) is an odd function using the fact that \tan^{-1} x is also an odd function:

f(-x) = \tan^{-1} ( -x\sqrt{2} - 1 ) + \tan^{-1}( -x \sqrt{2} + 1)

= \tan^{-1} ( -[x\sqrt{2} + 1] ) + \tan^{-1}( -[x \sqrt{2} - 1])

= -\tan^{-1} ( x\sqrt{2} + 1 ) - \tan^{-1}( x \sqrt{2} - 1)

= - \left[ \tan^{-1} ( x\sqrt{2} + 1 ) + \tan^{-1}( x \sqrt{2} - 1) \right]

= -f(x).

Therefore, f(-1) = -f(1) = -\displaystyle \frac{\pi}{2}, and so x_2 = -1.

The antiderivative of 1/(x^4+1): Part 8

In the course of evaluating the antiderivative

\displaystyle \int \frac{1}{x^4 + 1} dx,

I’ve accidentally stumbled on a very curious looking trigonometric identity:

\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) -\pi if x < -1,

\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) if -1 < x < 1,

\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) + \pi if x> 1.

The extra -\pi and \pi are important. Without them, the graphs of the left-hand side and right-hand sides are clearly different if x < -1 or x > 1:

TwoArctangents1

However, they match when those constants are included:

TwoArctangents2

Let’s see if I can explain why this trigonometric identity occurs without resorting to the graphs.

Since \tan^{-1} x assumes values between -\pi/2 and \pi/2, I know that

-\displaystyle \frac{\pi}{2} < \tan^{-1} ( x\sqrt{2} - 1 ) < \frac{\pi}{2},

-\displaystyle \frac{\pi}{2} < \tan^{-1} ( x\sqrt{2} + 1 ) < \frac{\pi}{2},

and so

-\pi< \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1} ( x\sqrt{2} + 1 ) < \pi.

However,

-\displaystyle \frac{\pi}{2} < \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) < \frac{\pi}{2},

and so \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1} ( x\sqrt{2} + 1 ) and \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) must differ if \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1} ( x\sqrt{2} + 1) is in the interval [-\pi,-\pi/2] or in the interval [\pi/2,\pi].

I also notice that

-\pi< \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1} ( x\sqrt{2} + 1 ) < \pi,

-\displaystyle \frac{\pi}{2} < -\tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) < \frac{\pi}{2},

and so

-\displaystyle \frac{3\pi}{2} < \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1} ( x\sqrt{2} + 1 )-\tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) < \frac{3\pi}{2}.

However, this difference can only be equal to a multiple of \pi, and there are only three multiples of \pi in the interval \displaystyle \left( -\frac{3\pi}{2}, \frac{3\pi}{2} \right), namely -\pi, 0, and \pi.

To determine the values of x where this happens, I also note that f_1(x) = x \sqrt{2} - 1, f_2(x) = x \sqrt{2} + 1, and f_3(x) = \tan^{-1} x are increasing functions, and so f(x) = \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) must also be an increasing function. Therefore, to determine where f(x) lies in the interval [\pi/2,\pi],it suffices to determine the unique value x_1 so that f(x_1) = \pi/2. Likewise, to determine where f(x) lies in the interval [-\pi,-\pi/2],it suffices to determine the unique value x_2 so that f(x_2) = -\pi/2.

In summary, I have shown so far that

\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) -\pi if x < x_2,

\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) if x_2 < x < x_1,

\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) + \pi if x> x_1,

where x_1 and x_2 are the unique values so that

\tan^{-1} ( x_1\sqrt{2} - 1 ) + \tan^{-1}( x_1 \sqrt{2} + 1) = \displaystyle \frac{\pi}{2},

\tan^{-1} ( x_2\sqrt{2} - 1 ) + \tan^{-1}( x_2 \sqrt{2} + 1) = -\displaystyle \frac{\pi}{2}.

So, to complete the proof of the trigonometric identity, I need to show that x_1 = 1 and x_2 = -1. I will do this in tomorrow’s post.