# 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.

# 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): 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.

# Different ways of solving a contest problem (Part 3)

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.

Yesterday, I presented a solution using a Pythagorean identity, but I was unable to be certain if the final answer was a positive or negative without drawing a picture. Here’s a third solution that also use a Pythagorean trig identity but avoids this difficulty. Again, I begin by squaring both sides.

$9 \sin^2 \theta = \cos^2 \theta$

$9 (1 - \cos^2 \theta) = \cos^2 \theta$

$9 - 9 \cos^2 \theta = \cos^2 \theta$

$9 = 10 \cos^2 \theta$

$\displaystyle \frac{9}{10} = \cos^2 \theta$

$\displaystyle \pm \frac{3}{\sqrt{10}} = \cos \theta$

Yesterday, I used the Pythagorean identity again to find $\sin \theta$. Today, I’ll instead plug back into the original equation $3 \sin \theta = \cos \theta$:

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

$3 \sin \theta = \displaystyle \frac{3}{\sqrt{10}}$

$\sin \theta = \displaystyle \pm \frac{1}{\sqrt{10}}$

Unlike the example yesterday, the signs of $\sin \theta$ and $\cos \theta$ must agree. That is, if $\cos \theta = \displaystyle \frac{3}{\sqrt{10}}$, then $\sin \theta = \displaystyle \frac{1}{\sqrt{10}}$ must also be positive. On the other hand, if $\cos \theta = \displaystyle -\frac{3}{\sqrt{10}}$, then $\sin \theta = \displaystyle -\frac{1}{\sqrt{10}}$ must also be negative.

If they’re both positive, then

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

and if they’re both negative, then

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

Either way, the answer must be $\displaystyle \frac{3}{10}$.

This is definitely superior to the solution provided in yesterday’s post, as there’s absolutely no doubt that the product $\sin \theta \cos \theta$ must be positive.

# Different ways of solving a contest problem (Part 2)

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.

Yesterday, I presented a solution using triangles. Here’s a second solution that I received: begin by squaring both sides and using a Pythagorean trig identity.

$9 \sin^2 \theta = \cos^2 \theta$

$9 (1 - \cos^2 \theta) = \cos^2 \theta$

$9 - 9 \cos^2 \theta = \cos^2 \theta$

$9 = 10 \cos^2 \theta$

$\displaystyle \frac{9}{10} = \cos^2 \theta$

$\displaystyle \pm \frac{3}{\sqrt{10}} = \cos \theta$

We use the Pythagorean identity again to find $\sin \theta$:

$\displaystyle \frac{9}{10} = \cos^2 \theta$

$\displaystyle \frac{9}{10} = 1 - \sin^2 \theta$

$\sin^2 \theta = \displaystyle \frac{1}{10}$

$\sin \theta = \displaystyle \pm \frac{1}{\sqrt{10}}$

Therefore, we know that

$\sin \theta \cos \theta = \displaystyle \left( \pm \frac{1}{\sqrt{10}} \right) \left( \pm \frac{3}{\sqrt{10}} \right) = \displaystyle \pm \displaystyle \frac{3}{10}$,

so the answer is either $\displaystyle \frac{3}{10}$ or $\displaystyle -\frac{3}{10}$. However, this was a multiple-choice contest problem and $\displaystyle -\frac{3}{10}$ was not listed as a possible answer, and so the answer must be $\displaystyle \frac{3}{10}$.

For a contest problem, the above logic makes perfect sense. However, the last step definitely plays to the fact that this was a multiple-choice problem, and the concluding step would not have been possible had $\displaystyle -\frac{3}{10}$ been given as an option.