# 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 100

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.

Today’s quip is one that I’ll use surprisingly often:

If you ever meet a mathematician at a bar, ask him or her, “What is your favorite application of the Cauchy-Schwartz inequality?”

The point is that the Cauchy-Schwartz inequality arises surprisingly often in the undergraduate mathematics curriculum, and so I make a point to highlight it when I use it. For example, off the top of my head:

1. In trigonometry, the Cauchy-Schwartz inequality states that

$|{\bf u} \cdot {\bf v}| \le \; \parallel \!\! {\bf u} \!\! \parallel \cdot \parallel \!\! {\bf v} \!\! \parallel$

for all vectors ${\bf u}$ and ${\bf v}$. Consequently,

$-1 \le \displaystyle \frac{ {\bf u} \cdot {\bf v} } {\parallel \!\! {\bf u} \!\! \parallel \cdot \parallel \!\! {\bf v} \!\! \parallel} \le 1$,

which means that the angle

$\theta = \cos^{-1} \left( \displaystyle \frac{ {\bf u} \cdot {\bf v} } {\parallel \!\! {\bf u} \!\! \parallel \cdot \parallel \!\! {\bf v} \!\! \parallel} \right)$

is defined. This is the measure of the angle between the two vectors ${\bf u}$ and ${\bf v}$.

2. In probability and statistics, the standard deviation of a random variable $X$ is defined as

$\hbox{SD}(X) = \sqrt{E(X^2) - [E(X)]^2}$.

The Cauchy-Schwartz inequality assures that the quantity under the square root is nonnegative, so that the standard deviation is actually defined. Also, the Cauchy-Schwartz inequality can be used to show that $\hbox{SD}(X) = 0$ implies that $X$ is a constant almost surely.

3. Also in probability and statistics, the correlation between two random variables $X$ and $Y$ must satisfy

$-1 \le \hbox{Corr}(X,Y) \le 1$.

Furthermore, if $\hbox{Corr}(X,Y)=1$, then $Y= aX +b$ for some constants $a$ and $b$, where $a > 0$. On the other hand, if $\hbox{Corr}(X,Y)=-1$, if $\hbox{Corr}(X,Y)=1$, then $Y= aX +b$ for some constants $a$ and $b$, where $a < 0$.

Since I’m a mathematician, I guess my favorite application of the Cauchy-Schwartz inequality appears in my first professional article, where the inequality was used to confirm some new bounds that I derived with my graduate adviser.

# My Favorite One-Liners: Part 76

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.

Here’s a problem that might arise in trigonometry:

Compute $\cos \displaystyle \frac{2017\pi}{6}$.

To begin, we observe that $\displaystyle \frac{2017}{6} = 336 + \displaystyle \frac{1}{6}$, so that

$\cos \displaystyle \frac{2017\pi}{6} = \cos \left( \displaystyle 336\pi + \frac{\pi}{6} \right)$.

We then remember that $\cos \theta$ is a periodic function with period $2\pi$. This means that we can add or subtract any multiple of $2\pi$ to the angle, and the result of the function doesn’t change. In particular, $-336\pi$ is a multiple of $2 \pi$, so that

$\cos \displaystyle \frac{2017\pi}{6} = \cos \left( \displaystyle 336\pi + \frac{\pi}{6} \right)$

$= \cos \left( \displaystyle 336\pi + \frac{\pi}{6} - 336\pi \right)$

$= \cos \displaystyle \frac{\pi}{6}$

$= \displaystyle \frac{\sqrt{3}}{2}$.

Said another way, $336\pi$ corresponds to $336/2 = 168$ complete rotations, and the value of cosine doesn’t change with a complete rotation. So it’s OK to just throw away any even multiple of $\pi$ when computing the sine or cosine of a very large angle. I then tell my class:

In mathematics, there’s a technical term for this idea; it’s called $\pi$ throwing.

# My Favorite One-Liners: Part 9

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.

Today, I’d like to discuss a common mistake students make in trigonometry… as well as the one-liner that I use to (hopefully) help students not make this mistake in the future.

Question. Find all solutions (rounded to the nearest tenth of a degree) of $\sin x = 0.8$.

Erroneous Solution. Plugging into a calculator, we find that $x \approx 53.1^o$.

The student correctly found the unique angle $x$ between $-90^o$ and $90^o$ so that $\sin x = 0.8$. That’s the definition of the arcsine function. However, there are plenty of other angles whose sine is equal to $0.7$. This can happen in two ways.

First, if $\sin x > 0$, then the angle $x$ could be in either the first quadrant or the second quadrant (thanks to the mnemonic All Scholars Take Calculus). So $x$ could be (accurate to one decimal place) equal to either $53.1^o$ or else $180^o - 53.1^o = 126.9^o$. Students can visualize this by drawing a picture, talking through each step of its construction (first black, then red, then brown, then green, then blue).

However, most students don’t really believe that there’s a second angle that works until they see the results of a calculator.

Second, any angle that’s coterminal with either of these two angles also works. This can be drawn into the above picture and, as before, confirmed with a calculator.

So the complete answer (again, approximate to one decimal place) should be $53.1^{\circ} + 360n^o$ and $126.9 + 360n^{\circ}$, where $n$ is an integer. Since integers can be negative, there’s no need to write $\pm$ in the solution.

Therefore, the student who simply answers $53.1^o$ has missed infinitely many solutions. The student has missed every nontrivial angle that’s coterminal with $53.1^o$ and also every angle in the second quadrant that also works.

Here’s my one-liner — which never fails to get an embarrassed laugh — that hopefully helps students remember that merely using the arcsine function is not enough for solving problems such as this one.

You’ve forgotten infinitely many solutions. So how many points should I take off?

For further reading, here’s my series on inverse functions.

# A Visual Proof of a Remarkable Trig Identity

Strange but true (try it on a calculator):

$\displaystyle \cos \left( \frac{\pi}{9} \right) \cos \left( \frac{2\pi}{9} \right) \cos \left( \frac{4\pi}{9} \right) = \displaystyle \frac{1}{8}$.

Richard Feynman learned this from a friend when he was young, and it stuck with him his whole life.

Recently, the American Mathematical Monthly published a visual proof of this identity using a regular 9-gon:

This same argument would work for any $2^n+1$-gon. For example, a regular pentagon can be used to show that

$\displaystyle \cos \left( \frac{\pi}{5} \right) \cos \left( \frac{2\pi}{5} \right) = \displaystyle \frac{1}{4}$,

and a regular 17-gon can be used to show that

$\displaystyle \cos \left( \frac{\pi}{17} \right) \cos \left( \frac{2\pi}{17} \right) \cos \left( \frac{4\pi}{17} \right) \cos \left( \frac{8\pi}{17} \right) = \displaystyle \frac{1}{16}$.

# A natural function with discontinuities (Part 3)

This post concludes this series about a curious function:

In the previous post, I derived three of the four parts of this function. Today, I’ll consider the last part ($90^\circ \le \theta \le 180^\circ$).

The circle that encloses the grey region must have the points $(R,0)$ and $(R\cos \theta, R \sin \theta)$ on its circumference; the distance between these points will be $2r$, where $r$ is the radius of the enclosing circle. Unlike the case of $\theta < 90^\circ$, we no longer have to worry about the origin, which will be safely inside the enclosing circle.

Furthermore, this line segment will be perpendicular to the angle bisector (the dashed line above), and the center of the enclosing circle must be on the angle bisector. Using trigonometry,

$\sin \displaystyle \frac{\theta}{2} = \frac{r}{R}$,

or

$r = R \sin \displaystyle \frac{\theta}{2}$.

We see from this derivation the unfortunate typo in the above Monthly article.

# A natural function with discontinuities (Part 2)

Yesterday, I began a short series motivated by the following article from the American Mathematical Monthly.

Today, I’d like to talk about the how this function was obtained.

If $180^\circ \le latex \theta \le 360^\circ$, then clearly $r = R$. The original circle of radius $R$ clearly works. Furthermore, any circle that inscribes the grey circular region (centered at the origin) must include the points $(-R,0)$ and $(R,0)$, and the distance between these two points is $2R$. Therefore, the diameter of any circle that works must be at least $2R$, so a smaller circle can’t work.

The other extreme is also easy: if $\theta =0^\circ$, then the “circular region” is really just a single point.

Let’s now take a look at the case $0 < \theta \le 90^\circ$. The smallest circle that encloses the grey region must have the points $(0,0)$, $(R,0)$, and $(R \cos \theta, R \sin \theta)$ on its circumference, and so the center of the circle will be equidistant from these three points.

The center must be on the angle bisector (the dashed line depicted in the figure) since the bisector is the locus of points equidistant from $(R,0)$ and $(R \cos \theta, R \sin \theta)$. Therefore, we must find the point on the bisector that is equidistant from $(0,0)$ and $(R,0)$. This point forms an isosceles triangle, and so the distance $r$ can be found using trigonometry:

$\cos \displaystyle \frac{\theta}{2} = \displaystyle \frac{R/2}{r}$,

or

$r = \displaystyle \frac{R}{2} \sec \frac{\theta}{2}$.

This logic works up until $\theta = 90^\circ$, when the isosceles triangle will be a 45-45-90 triangle. However, when $\theta > 90^\circ$, a different picture will be needed. I’ll consider this in tomorrow’s post.

# 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.