An algebra and trigonometry–based proof of Kepler’s First Law

The proofs of Kepler’s Three Laws are usually included in textbooks for multivariable calculus. So I was very intrigued when I saw, in the Media Reviews of College Mathematics Journal, that somebody had published a proof of Kepler’s First Law that only uses algebra and trigonometry. Let me quote from the review:

Kepler’s first law states that bounded planetary orbits are elliptical. This law is presented in introductory textbooks, but the proof typically requires intricate integrals or vector analysis involving an accidental degeneracy. Simha offers an elementary proof of Kepler’s first law using algebra and trigonometry at the high school level.
https://doi.org/10.1080/07468342.2022.2026089

Once upon a time, I taught Precalculus for precocious high school students. I wish I had known of this result back then, as it would have been a wonderful capstone to their studies of trigonometry and the conic sections.

The preprint of this result can be found on arXiv. (The proof only addresses Kepler’s First Law and not the Second and Third Laws.) The actual article, for those with institutional access, was published in American Journal of Physics Vol. 89 No. 11 (2021): 1009-1011.

Engaging students: Law of Sines

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 Tiger Hersh. His topic, from Precalculus: the Law of Sines.

How does this topic extend what your students should have learned in previous courses?

This topic can be extended to geometry where students must be able to use trigonometric identities (1) to identify the degree or length in order to use the Law of Sines. The issue about trigonometric identities is that you can only use them on right triangles (2). However, with the Law of Sines, students are able to use the trigonometric identities they have learned in Geometry and are able to draw a perpendicular line across a non-right triangle (3) and then apply the Law of Sines to solve either the height of the triangle, the length of the side of the triangle, or the degree of an angle of the triangle. So, the Law of Sines use the idea of trigonometric identities from Geometry in order to be applicable.

How can this topic be used in your students’ future courses in mathematics or science? Unit circle calculus / solving for height of triangles

Students are able to the Law of Sines in order to find the height or degree of a triangle on the unit circle in precalculus or to calculator vector quantities in physics. The Law of Sines is prominent in the unit circle which is noticeable in the linked website which will provide students a connection from the Law of Sines to the unit circle. The Law of Sines also connects to physics where vectors used to show motion and direction in two dimensional space. The Law of sines may also be applied in physics where in (2); The vectors form a non-right triangle. The vectors ‘length’ can be determined by identifying the magnitude of each vector and then using the method as described before to use the Law of Sines in-order to find vector r.

https://image.businessinsider.com/51910f38ecad040506000002?width=800&format=jpeg&auto=webp

How has this topic appeared in pop culture (movies, TV, current music, video games, etc.)?

The law of sines has appeared in almost every 3 dimensional video games known to exist that has characters that are rendered with polygons. To note: it’s not just any polygon that can be used to create the characters you see in video games but specifically, they usually use triangles to render the characters. Even some movies that use animation software use these triangular polygons to render the figures in the movie; like for example Woody from the movie Toy Story (as seen below with polygons). We can use the Law of Sines in order to find the length or degree of each triangle on the figure if we were willing so.

Dabbing and the Pythagorean Theorem

I enjoyed this article from Fox Sports. Apparently, a French Precalculus textbook created a homework problem asking if football (soccer) superstar Paul Pogba is doing the perfect dab by creating two right triangles.

Langley’s Adventitious Angles: 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 short series on a couple of easily stated but remarkably difficult geometry problems.

Part 1: The world’s second hardest easy geometry problem.

Part 2: The world’s hardest easy geometry problem.

Langley’s Adventitious Angles (Part 1)

Math With Bad Drawings had an interesting post about solving for $x$ in the following picture (this picture is taken from http://thinkzone.wlonk.com/MathFun/Triangle.htm):

I had never heard of this problem before, but it’s apparently well known and is called Langley’s Adventitious Angles. See Math With Bad Drawings, Wikipedia, and Math Pages for more information about the solution of this problem. Math Pages has a nice discussion about mathematical aspects of this problem, including connections to the Laws of Sines and Cosines and to various trig identities.

I’d encourage you to try to solve for $x$ without clicking on any of these links… a certain trick out of the patented Bag of Tricks is required to solve this problem using only geometry (as opposed to the Law of Cosines and the Law of Sines). I have a story that I tell my students about the patented Bag of Tricks: Socrates gave the Bag of Tricks to Plato, Plato gave it to Aristotle, it passed down the generations, my teacher taught the Bag of Tricks to me, and I teach it to my students. In the same post, Math With Bad Drawings has a nice discussion about pedagogical aspects of this problem concerning when a “trick” becomes a “technique”.

I recommend this problem for advanced geometry students who need to be challenged; even bright students will be stumped concerning coming up with the requisite trick on their own. Indeed, the problem still remains quite challenging even after the trick is shown.

Proving theorems and special cases (Part 11): The Law of Cosines

In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:

Is it ever possible to prove a statement or theorem by proving a special case of the statement or theorem?

Usually, the answer is no. In this series of posts, we’ve seen that a conjecture could be true for the first 40 cases or even the first $10^{316}$ cases yet not always be true. We’ve also explored the computational evidence for various unsolved problems in mathematics, noting that even this very strong computational evidence, by itself, does not provide a proof for all possible cases.

However, there are plenty of examples in mathematics where it is possible to prove a theorem by first proving a special case of the theorem. For the remainder of this series, I’d like to list, in no particular order, some common theorems used in secondary mathematics which are typically proved by first proving a special case.

2. Theorem. In $\triangle ABC$ where $a = BC$ , $b = AC$ , and $c = AB$ , we have $c^2 = a^2 + b^2 - 2 a b \cos (m \angle C)$ .

This is typically proven using the Pythagorean theorem:

Lemma. In right triangle $\triangle ABC$ , where $\angle C$ is a right angle, we have $c^2 = a^2 + b^2$ .

Though it usually isn’t thought of this way, the Pythagorean theorem is a special case of the Law of Cosines since $\cos 90^\circ = 0$ .

There are well over 100 different proofs of the Pythagorean theorem that do not presuppose the Law of Cosines. The standard proof of the Law of Cosines then uses the Pythagorean theorem. In other words, a special case of the Law of Cosines is used to prove the Law of Cosines.

Another proof of the Law of Cosines

Source: https://www.facebook.com/AmerMathMonthly/photos/a.250425975006394.53155.241224542593204/755597884489198/?type=1&theater

Inverse Functions: Arccosine and SSS (Part 21)

Arccosine has an important advantage over arcsine when solving for the parts of a triangle: there is no possibility ambiguity about the angle.

Solve $\triangle ABC$ if $a = 16$ , $b = 20$ , and $c = 25$ .

When solving for the three angles, it’s best to start with the biggest angle (that is, the angle opposite the biggest side). To see why, let’s see what happens if we first use the Law of Cosines to solve for one of the two smaller angles, say $\alpha$ :

$a^2 = b^2 + c^2 - 2 b c \cos \alpha$

$256 = 400 + 625 - 1000 \cos \alpha$

$-769 = -1000 \cos \alpha$

$0.769 = \cos \alpha$

$\alpha \approx 39.746^\circ$

So far, so good. Now let’s try using the Law of Sines to solve for $\gamma$ :

$\displaystyle \frac{\sin \alpha}{a} = \displaystyle \frac{\sin \gamma}{c}$

$\displaystyle \frac{\sin 39.746^\circ}{16} \approx \displaystyle \frac{\sin \gamma}{25}$

$0.99883 \approx \sin \gamma$

Uh oh… there are two possible solutions for $\gamma$ since, hypothetically, $\gamma$ could be in either the first or second quadrant! So we have no way of knowing, using only the Law of Sines, whether $\gamma \approx 87.223^\circ$ or if $\gamma \approx 180^\circ - 87.223^\circ = 92.777^\circ$ .

For this reason, it would have been far better to solve for the biggest angle first. For the present example, the biggest answer is $\gamma$ since that’s the angle opposite the longest side.

$c^2 = a^2 + b^2 - 2 a b \cos \gamma$

$625 = 256 + 400 - 640 \cos \gamma$

$-31 =-640 \cos \gamma$

$0.0484375 = \cos \gamma$

Using a calculator, we find that $\gamma \approx 87.223^\circ$ .

We now use the Law of Sines to solve for either $\alpha$ or $\beta$ (pretending that we didn’t do the work above). Let’s solve for $\alpha$ :

$\displaystyle \frac{\sin \alpha}{a} = \displaystyle \frac{\sin \gamma}{c}$

$\displaystyle \frac{\sin \alpha}{16} \approx \displaystyle \frac{\sin 87.223}{25}$

$\sin \alpha \approx 0.63949$

This equation also has two solutions in the interval $[0^\circ, 180^\circ]$ , namely, $\alpha \approx 39.736^\circ$ and $\alpha \approx 180^\circ - 39.736^\circ = 140.264^\circ$ . However, we know full well that the answer can’t be larger than $\gamma$ since that’s already known to be the largest angle. So there’s no need to overthink the matter — the answer from blindly using arcsine on a calculator is going to be the answer for $\alpha$ .

Naturally, the easiest way of finding $\beta$ is by computing $180^\circ - \alpha - \gamma$ .

Inverse Functions: Arccosine and SSS (Part 20)

The Law of Cosines also recognizes when the purported sides of a triangle are impossible.

Solve $\triangle ABC if$ latex a = 16$, $b = 20$ , and $c = 40$ .

Hopefully students would recognize that $c > a + b$ , thus quickly demonstrating that the triangle is impossible. However, this also falls out of the Law of Cosines:

$c^2 = a^2 + b^2 - 2 a b \cos \gamma$

$1600 = 256 + 400 - 640 \cos \gamma$

$944 =-640 \cos \gamma$

$-1.475 = \cos \gamma$

Since the cosine of an angle can’t be less than -1, we can conclude that this is impossible.

Stated another way, we have the implications (since $a$ , $b$ , and $c$ are all positive)

$c > a + b \Longleftrightarrow c^2 > (a+b)^2$

$\Longleftrightarrow a^2 + b^2 - 2 a b \cos \gamma > a^2 + 2 a b + b^2$

$\Longleftrightarrow -2 a b \cos \gamma > 2 a b$

$\Longleftrightarrow \cos \gamma < -1$

Since the last statement is impossible, so is the first one.

Inverse Functions: Arccosine and SSS (Part 19)

Arccosine has an important advantage over arcsine when solving for the parts of a triangle: there is no possibility ambiguity about the angle.

Solve $\triangle ABC if$ latex a = 16$, $b = 20$ , and $c = 25$ .

To solve for, say, the angle $\gamma$ , we employ the Law of Cosines:

$c^2 = a^2 + b^2 - 2 a b \cos \gamma$

$625 = 256 + 400 - 640 \cos \gamma$

$-31 =-640 \cos \gamma$

$0.0484375 = \cos \gamma$

Using a calculator, we find that $\gamma \approx 87.2^\circ$ . And the good news is that there is no need to overthink this… this is guaranteed to be the angle since the range of $y = \cos^{-1} x$ is $[0,\pi]$ , or $[0^\circ, 180^\circ]$ in degrees. So the equation

$\cos x = \hbox{something}$

is guaranteed to have a unique solution between $0^\circ$ and $180^\circ$ . (But there are infinitely many solutions on $\mathbb{R}$ . And since an angle in a triangle must lie between $0^\circ$ and $180^\circ$ , the practical upshot is that just plugging into a calculator blindly is perfectly OK for this problem. This is in stark contrast to the Law of Sines, for which some attention must be paid for solutions in the interval $[0^\circ,90^\circ]$ and also the interval $[90^\circ, 180^\circ]$ .

From this point forward, the Law of Cosines could be employed again to find either $\alpha$ or $\beta$ . Indeed, this would be my preference since the sides $a$ , $b$ , and $c$ are exactly. However, my experience is that students prefer the simplicity of the Law of Sines to solve for one of these angles, using the now known pair of $c$ (exactly known) and $\gamma$ (approximately known with a calculator).