A Bisecting Angle

This is a bit of fun geometry that doesn’t have much to do with what’s going on in class, but does reflect on mathematical thinking.

An article promoting the use of technology in the classroom began: “Draw a perfect circle. Now bisect that with a 45-degree angle, the perfect slice of geometric pizza. Now, using your drawing, find the area of the rest of the circle. ”

What they meant was that the 45-degree angle divides the circle (or, more properly speaking, the region contained by the circle*). “Bisect” has a rigorous definition in mathematics: It means to divide an object into two objects of the same exact size. So if we bisect a circular region with anything, then the area of the two pieces will have half the area of the circular region, by definition.

Rather than accepting this and moving on, though, I asked myself: Can you bisect a circular region with a 45-degree angle? I shifted this to: Given a circle and an angle whose vertex is on or in the circle, what is the smallest angle that will cut out half of its region?

I’m going to start with a conjecture: Our target angle is an inscribed angle bisected by a diameter. In other words, here’s a picture of circle A, which has a radius of one unit:

My conjecture is that the shaded region, where angles BEA and CEA are congruent,  represents the largest portion of the circular region that can be covered by angle BEC. Let $$\alpha = m\angle BEC$$.

So what is the area of the shaded region? First, we have sector BAC. The area of a sector is $$\beta r^2/2$$, where $$\beta$$ is the measure of the central angle (in radians). Since it’s a unit circle, $$r^2 = 1$$. Since angle BEC is the inscribed angle that corresponds to the central angle BAC, it has half the measure, and the area of the sector is $$\beta/2 = \alpha$$.

Since AE bisects angle BEC, the two triangles are congruent. Since AC, AE, and AB are all radii, the two triangles are isosceles. We can determine the area of triangle BAE by first dropping a line perpendicular to BE. This divides the triangle into two congruent right triangles, AFE and AFB.

Call $$\gamma = m\angle AEF$$, so $$\gamma = \alpha/2$$. To find the area of $$\Delta AFE$$, we need a height and a width. Since AE is a radius, the height AF is $$\sin\gamma$$ and the width FE is $$\cos\gamma$$. This gives an area of $$\sin\gamma\cos\gamma/2$$.

I’ll write about the double- and half-angle trigonometric formulas in a separate post, but one of these is: $\sin 2\theta = 2\sin\theta\cos\theta$

Applying this gives us an area of triangle AEF of $$(\sin 2\gamma)/4 = (\sin\alpha)/4$$. Since there are four such congruent triangles, the total area of the two larger shaded triangles is $$\sin\alpha$$, and the area of the entire shaded region is $$\sin\alpha + \alpha$$.

If the shaded region is half the area bounded by the circle, and since the area bounded by a unit circle is $$\pi$$, $$\sin\alpha + \alpha = \pi/2$$. I’m not sure how or if that can be solved analytically, but we can use a calculator to graph $$\sin\alpha + \alpha – \pi/2$$ and find its solution.

The solution function of the calculator gives a zero at 0.8317112.

Throughout this post, I have worked in radians. For the Algebra 2 students, we haven’t gotten there yet. Radians represent a different way to measure angles, but it’s a straightforward conversion: $$2\pi = 360^o$$, and $$1 = 180^o/\pi$$. So we have $$0.8317112 \times 180^o/\pi \approx 47.65^o$$.

So: An inscribed angle slightly larger than 47.65 degrees bisected by a diameter will create a rounded wedge that bounds half as many points as a circle.

Going back to the original inspiration for this item, this means that the 45 degree angle with a vertex that is inside a circle cannot bisect the circle’s region.

I started with a conjecture. This conjecture can be proven, but the formal proof requires more trigonometry, so I’ll leave it for the reader, or for another time.

* Since this article is about rigorous definitions, I’m using “circle” to refer to the set of points equidistant from a point (i.e., $$(x-h)^2 + (y-k)^2 = r^2$$) and “circular region” to refer to all points satisfying $$(x-h)^2 + (y-k)^2 \le r^2$$. In high school, we often conflate these two with the word “circle”.

The Derivative of the Sine and Cosine Functions

In AP Calculus, we saw the following derivatives: $\sin’ x = \cos x \\ \cos’ x = -\sin x$

That is, the derivative of the sine is the cosine, and the derivative of the cosine is the opposite of the sine.

I showed one way to see why this is true using Desmos. The purple line is the tangent, while the black dot is the slope of that tangent. As you move the slider a from side to side, the black dot traces the cosine function.

The book has an algebraic proof of the first rule (p. 112), using the limit process. In this post, I’ll discuss a more visual, geometry-based explanation.

We’ll start with a few triangles and a portion of the unit circle.

The angle $$\theta$$ represents our original angle, while $$\delta = \Delta\theta$$. Note that we are NOT looking at the change in the $$x$$ value; we’re looking at the change in the angle measurement. This is key!

So what we want to know is: How does $$\sin\theta$$ change as $$\theta$$ changes?

$$\overline{AC}$$ is the radius of the unit circle, so its length is 1; $$\sin\theta = m\overline{CF}/m\overline{AC} = f$$. Likewise, $$\cos\theta = g$$.

$$\overline{EC}$$ is on the tangent to the circle at point $$C$$. $$\overline{CD}$$ is collinear with $$\overline{CF}$$; $$\overline{ED}\perp\overline{CD}$$.

So let’s consider the blue and green triangles. Using geometry, we know that since the blue triangle is a right triangle, $$\theta$$ and $$\gamma$$ are complementary. Since $$\overline{DF}$$ is on a line, and since $$\angle ECA$$ is right, $$\alpha$$ and $$\gamma$$ are complementary… meaning that $$\alpha\cong\theta$$.

More about the green triangle: $$\cos\alpha = \cos\theta = e/c \Rightarrow e = c \cos\theta$$. But $$c$$ is also opposite $$\delta$$ in $$\triangle ECA$$, meaning that $$\sin\delta = c/a \Rightarrow c = a\sin\delta$$ and $$e = a\sin\delta\cos\theta$$.

We’re ready to write our limit now. We want to know what happens to $$e$$ when $$\delta$$ gets smaller and smaller. Specifically: $\lim_{\Delta\theta\rightarrow 0}\frac{\Delta y}{\Delta\theta} = \lim_{\delta\rightarrow 0}\frac{e}{\delta} \\ = \lim_{\delta\rightarrow 0}\frac{a\sin\delta\cos\theta}{\delta} \\ = \lim_{\delta\rightarrow 0} a \cdot \lim_{\delta\rightarrow 0}\frac{\sin\delta}{\delta} \cdot \lim_{\delta\rightarrow 0} \cos\theta$

Consider each in turn. As $$\delta$$ gets smaller, the difference between $$a$$ and the radius of the unit circle gets smaller. That is, $\lim_{\delta\rightarrow 0} a = 1$

From our examination of the Squeeze Theorem, we know the important identity $\lim_{x\rightarrow 0}\frac{\sin x}{x} = 1$ which we can apply here (where $$x = \delta$$).

Finally, since $$\cos\theta$$ is not affected by $$\delta$$ at all, $\lim_{\delta\rightarrow 0} \cos\theta = \cos\theta$

Putting these together, we get  $\lim_{\delta\rightarrow 0} a \cdot \lim_{\delta\rightarrow 0}\frac{\sin\delta}{\delta} \cdot \lim_{\delta\rightarrow 0} \cos\theta = 1 \cdot 1 \cdot \cos\theta = \cos\theta$ QED.

The process for demonstrating the cosine is very similar (using $$d = a\sin\delta\sin\theta$$), but the question that comes up is: Why is the cosine’s derivative the opposite of the sine? Notice what happens in the green triangle: While the height goes upward, the width (corresponding to $$\sin\alpha = \sin\theta$$) goes to the left. This is why the cosine’s derivative is the opposite of the sine.

Welcome to 2018!

This website has been updated for the new school year.

If you want to know what you missed or review the notes, select the appropriate class.

This tab will contain announcements and some additional thoughts.

All posts that are older than this one are for previous years. You’re welcome to go through them if you’re curious.

Equations, expressions, and functions

Three things that students of algebra often confuse are the notions of equations, expressions, and functions.

Expressions

An expression is any set of constant numeric values, variables, and operations. The idea is that, if we know the values of all the variables at a given point, we can determine the numeric value of the entire expression.

Some expressions don’t have any variables at all. This usually means that they have a specific, unchanging mathematical value. The only exception to this is when some portion of the expression breaks math. Dividing by zero, for instance, always breaks math, so $$5/0$$ has no mathematical value. Taking square roots of negative numbers leads to mathematical values, but not real ones, so depending on our goal, $$\sqrt(-5)$$ may not have a valid value.

Since some values can create problems, this means that some expressions with variables have specific values or sets of values that create problems. For instance, if we’re limiting ourselves to real numbers, $$\sqrt(x)$$ prohibits $$x$$ from being negative. And $$1/x$$ prohibits $$x$$ from being zero.

Equations

An equation is a statement of fact about two expressions. An equation is true for any values of variables that make the two expressions have the same mathematical value, and otherwise it is false.

For instance, consider $$x^2 = 4$$. This is true for any numbers which, when squared, have a value of 4. These are 2 and -2. Those are the only values that make that statement true.

We can have variables on both sides of the equation. Consider $$x^2 = x + 2$$. This is also true for two values, 2 and -1.

If an equation involves two or more variables, it can be true for an infinite number of values. For instance, $$y = x$$ is true for all pairs $$(x, y)$$ where $$x$$ and $$y$$ have the same value.

Functions

A function is a relationship between sets of data. It is often described as being a machine: If you put a specific value into the machine, you can predict exactly what output you’ll get.

It can very often by described by an expression, but it is rarely described by an equation. This is confusing because we normally state a function by giving an equation. But let’s take a careful look at a function. Here’s an example: $f(x) = x^2 + 7$

This is an equation, but it’s connecting two expressions. One expression consists of the name of a function ($$f$$) and its input ($$x$$). The other expression describes what operations are going to be applied to the input. The expression on the right is the actual function; the expression on the left gives its name.

Consider this: “My car is a silver Honda.” What is my car? It’s not the entire sentence “My car is a silver Honda.” That would be silly. It’s the object that both “my car” and “a silver Honda” refer to. In a similar way, $$f(x)$$ and $$x^2 + 7$$ are two ways to refer to a machine that takes any value, squares it, and adds seven to the result.

We use $$f(x)$$ when we want to make general comments, or when we want to make other statements about the function. We use the specific form of the function ($$x^2 + 7$$) when we want to see what it actually does.

Today in class I presented a technique for factoring quadratics. Here are the steps. Remember that we’re starting from the right. This only works when the coefficient on the $$x^2$$ term is 1.

1. List the pairs of numbers that multiply to the constant (ignoring the sign!).
2. The sign on the constant tells us whether we’re looking for a sum or a difference.
3. The coefficient on the $$x$$ term (ignoring the sign!) tells us what sum or difference we’re looking for.
4. Now, using the sign on the $$x$$ term, decide the signs for the factors.
1. The higher factor will use the same sign.
2. The lower factor will use the same sign if it’s a sum, and the opposite sign if it’s a difference.

Here are some examples.

$x^2 + 7x + 12$

1. $$12 = 1 \times 12 = 2 \times 6 = 3 \times 4$$
2. We’re looking for a sum. Our options are:
1. $$1 + 12 = 13$$
2. $$2 + 6 = 8$$
3. $$3 + 4 = 7$$
3. Our sum is 7. That means we want to use 3 and 4.
4. We’re using +. So our factors are:
1. $$x + 4$$
2. $$x + 3$$

That means that $$x^2 + 7x + 12 = (x + 3)(x + 4)$$.

$x^2 – 2x – 63$

1. $$63 = 1 \times 63 = 3 \times 21 = 7 \times 9$$
2. We’re looking for a difference. Our options are:
1. $$63 – 1 = 62$$
2. $$21 – 3 = 18$$
3. $$9 – 7 = 2$$
3. Our difference is 2. That means we want to use 7 and 9.
4. We’re using -. So our factors are:
1. $$x – 9$$
2. $$x + 7$$

That means that $$x^2 – 2x – 63 = (x – 9)(x + 7)$$.

Remember that the solutions will have the opposite signs to the factors, because solutions are values that make the factors equal to zero. So, in our first example, our solutions are $$x = {-3, -4}$$, while in the second example, they’re $$x = {9, -7}$$.

If this method doesn’t work, it means that at least one of the solutions of the quadratic expression isn’t an integer.

Answers to Operations on Rational Functions worksheet

These are the answers to Monday’s worksheet (Honors version).

Answers to Operations on Rational Functions worksheet

These are the answers to Monday’s worksheet (regular version).

Week of Apr 23, 2018

Powerpoints: Apr 25 Apr 24 Apr 23

Monday
Topic: Graphing Rational Functions
Homework: Rational Functions Graphing 2

Tuesday
Topic: Simplifying and Multiplying Rational Functions
Homework: Simplifying and Multiplying Rational Functions

Wednesday
Topic: Simplifying, Multiplying, and Dividing Rational Functions

Week of Apr 23, 2018

Powerpoints: Apr 25 Apr 24 Apr 23

Monday
Topic: Graphing Rational Functions
Homework: Rational Functions Graphing 2

Tuesday
Topic: Simplifying and Multiplying Rational Functions
Homework: Simplifying and Multiplying Rational Functions

Wednesday
Topic: Simplifying, Multiplying, and Dividing Rational Functions

Week of Apr 16, 2018

Powerpoints: Apr 20 Apr 19 Apr 16

Monday
Topic: Direct and Inverse Variation
Classwork: Direct and Inverse Variation

Tuesday
Graphing (Mr. Hartzer absent)
Classwork: Graphing Polynomials and Rational Functions (Kuta)

Wednesday
Polynomials (Mr. Hartzer absent)
Classwork: Polynomial Review (Kuta)

Thursday
Topic: Rational Functions

Friday
Topic: Graphing Rational Functions
Classwork: Rational Functions Graphing 1