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.

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?

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

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.

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.

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:

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…

…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:

The 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:

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 $x > 1$:

However, they match when those constants are included:

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.

The antiderivative of 1/(x^4+1): Part 7

This antiderivative has arguable the highest ratio of “really hard to compute” to “really easy to write”:

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

As we’ve seen in this series, the answer is

$\displaystyle \frac{\sqrt{2}}{8} \ln \left(\frac{x^2 + x\sqrt{2} + 1}{x^2 - x\sqrt{2} + 1} \right) + \frac{\sqrt{2}}{4} \tan^{-1} ( x\sqrt{2} - 1 ) + \frac{\sqrt{2}}{4} \tan^{-1}( x \sqrt{2} + 1) + C$

It turns out that this can be simplified somewhat as long as $x \ne 1$ and $x \ne -1$. I’ll use the trig identity

$\tan(\alpha + \beta) = \displaystyle \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$

When I apply this trig identity for $\alpha = \tan^{-1} ( x\sqrt{2} - 1 )$ and $\beta = \tan^{-1} ( x\sqrt{2} + 1 )$, I obtain

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

$= \displaystyle \frac{2x \sqrt{2}}{1 - (2x^2 - 1)}$

$= \displaystyle \frac{2x \sqrt{2}}{2 - 2x^2}$

$= \displaystyle \frac{x \sqrt{2}}{1 - x^2}$.

So we can conclude that

$\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) + n\pi$

for some integer $n$ that depends on $x$. The $+n\pi$ is important, as a cursory look reveals that $y = \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1)$ and $y = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right)$ have different graphs. (The vertical lines in the orange graph indicate where the right-hand side is undefined when $x = 1$ or $x = -1$.

The two graphs coincide when $-1 < x < 1$ but differ otherwise. However, it appears that the two graphs differ by a constant. Indeed, if I subtract $\pi$ from the orange graph if $x < -1$ and add $\pi$ to the orange graph if $x > 1$, then they match:

So, evidently

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

So as long as $x \ne 1$ and $x \ne -1$, this constant $-\pi$, $0$, or $\pi$ can be absorbed into the constant $C$:

$\displaystyle \int \frac{1}{x^4 + 1} dx = \displaystyle \frac{\sqrt{2}}{8} \ln \left(\frac{x^2 + x\sqrt{2} + 1}{x^2 - x\sqrt{2} + 1} \right) + \frac{\sqrt{2}}{4} \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) + C$.

However, a picture may be persuasive but is not a proof, and there are some subtle issues with this simplification. I’ll discuss these further details in tomorrow’s post.

Different ways of solving a contest problem (Part 1)

The following problem appeared on the American High School Mathematics Examination (now called the AMC 12) in 1988:

If $3 \sin \theta = \cos \theta$, what is $\sin \theta \cos \theta$?

When I presented this problem to a group of students, I was pleasantly surprised by the amount of creativity shown when solving this problem.

Here’s the first solution that I received: draw the appropriate triangles for the angle $\theta$:

$3 \sin \theta = \cos \theta$

$\tan \theta = \displaystyle \frac{1}{3}$

Therefore, the angle $\theta$ must lie in either the first or third quadrant, as shown. (Of course, $\theta$ could be coterminal with either displayed angle, but that wouldn’t affect the values of $\sin \theta$ or $\cos \theta$.)

In Quadrant I, $\sin \theta = \displaystyle \frac{1}{\sqrt{10}}$ and $\cos \theta = \displaystyle \frac{3}{\sqrt{10}}$. Therefore,

$\sin \theta \cos \theta = \displaystyle \frac{1}{\sqrt{10}} \times \frac{3}{\sqrt{10}} = \displaystyle \frac{3}{10}$.

In Quadrant III, $\sin \theta = \displaystyle -\frac{1}{\sqrt{10}}$ and $\cos \theta = -\displaystyle \frac{3}{\sqrt{10}}$. Therefore,

$\sin \theta \cos \theta = \displaystyle \left( - \frac{1}{\sqrt{10}} \right) \times \left( -\frac{3}{\sqrt{10}} \right) = \displaystyle \frac{3}{10}$.

Either way, we can be certain that $\sin \theta \cos \theta = \displaystyle \frac{3}{10}$.