More trig notes

 

For the earlier notes, see Wednesday 20 November

The notes below are still incomplete – and I want to add illustrations as well. But the ancient computer I am using these days connects to the internet slowly if at all, so I am already late posting this. I will add to it and update when I can.

 

Important versions of the two special triangles: when the hypotenuse is 1.

You can see these two right triangles on this blog post from squarerootofnegativeoneteachmath (which also shows how they are used in the unit circle definitions of the trig functions, which we will discuss next time).

Embedding a right triangle in the coordinate plane:

We pick out one of the acute angles in the right triangle: let’s call it angle $\theta$. We say that the right triangle is in standard position in the coordinate plane if the vertex of angle $\theta$ is at the origin, the side adjacent to angle $\theta$ is on the positive x-axis, and the hypotenuse extends into the first quadrant. This desmos graph shows a right triangle in standard position. Notice that the coordinates of the top right vertex are $(a, b)$ where $a$ and $b$ are the lengths of the two legs of the right triangle.

We are going to use this embedding to extend the definitions of the trig functions to angles which are not acute angles, so they are not angles of right triangles. (That is the coordinate plane definition of the trig functions.)

Important:

Now that we are temporarily leaving the world of right triangles, we will be thinking of an angle as a rotation. The rotation rotates one side of the angle, called the initial side, over to the other side of the angle, called the terminal side.

The angle is positive if the rotation is counter-clockwise, and is negative if the rotation is clockwise. I will indicate this direction of rotation by putting a curved arrow near the vertex.

An angle (rotation) in the coordinate plane is in standard form if its vertex is at the origin $(0,0)$, and its initial side is along the positive x-axis (horizontal axis).

For the sake of what we will be doing in the rest of this course, it is useful (and will be necessary) to refer to the horizontal axis (rather than x-axis) and the vertical axis (rather than y-axis). One big reason for this is that we will often use the variable $x$ to refer to the angle (in radians) – see further on.

Radian measure of angles (rotations): For simplicity I will define radian measure in the setting of angles in standard position in the coordinate plane.

Starting with some angle $\theta$ which is in standard position, draw (or imagine) a circle of radius $r$ centered at the origin. As we travel around this circle from the initial side to the terminal side of our angle, that marks out an arc on the circle, which is called the subtended arc for the angle $\theta$. Then the radian measure of $\theta$ is defined as the ratio $\frac{\text{length of the subtended arc}}{r}$.

For example, consider the case $\theta$ = a right angle. This angle encloses an arc whose length is $\frac{1}{4}$ of the circumference, which is $\frac{1}{4}(2\pi r) = \frac{\pi r}{2}$. So the radian measure of the right angle is this length divided by $r$:

$\frac{\left(\frac{\pi r}{2}\right)}{r} = \left(\frac{\pi r}{2}\right)\left(\frac{1}{ r}\right) = \frac{\pi}{2}$

So a right angle has radian measure $\frac{\pi}{2}$.

If you think about this example, it shows that the radian measure does not depend on the choice of the radius $r$, so we might as well use a circle of radius 1. This is called the unit circle.

The circumference of the unit circle is $2\pi$. From this we can easily see that a rotation one time around in the counter-clockwise direction has radian measure $2\pi$ (because $r=1$).

A “straight angle” is a rotation halfway around the circle, so it has radian measure $\pi$.

We already know that a right angle has radian measure $\frac{\pi}{2}$, and that makes sense also because it is half of a straight angle.

Half a right  angle, which is the acute angle that appears in the isosceles right triangle, has radian measure $\frac{1}{2}\cdot \frac{\pi}{2} = \frac{\pi}{4}$.

Here are two more important angles:

The angle which is in the equilateral triangle is $\frac{1}{3}$ of a straight angle (because all 3 angles must add up to a straight angle, and they are all equal in an equilateral triangle). So the angle of an equilateral triangle is $\frac{\pi}{3}$.

This is the larger of the two acute angles in our “half-equilateral” right triangle. The smaller one is half of that, so is $\frac{1}{2}\cdot \frac{\pi}{3} = \frac{\pi}{6}$.

Become very familiar with those angles and their radian measure, preferably without trying to translate or even think of degrees! You are learning a second language, and it will harm your progress if you keep translating back into your first language. You should learn to think in radians, which are a much more important way to measure in terms of the appli9cations of trig functions. You want to immediately associate a visual image of an angle with its radian measure. Discipline!

(To be continued…)

 

 

 

Old notes on the three definitions of the trig functions

Notes are coming; grades are posted

This is just a quick post to tell you what’s been posted recently.

I have posted the Test 3 scores to the gradebook here in OpenLab, so you can check them. Make sure also to read  this post.

I am working on my notes on the last couple of classes in-between other rather urgent matters today, and hope to have them finished by noon. In the meantime, I recommend these resources:

Please make sure that you work on the WeBWorK even if it’s not due yet! There will be a quiz tomorrow which will be on rationalizing denominators (including complex numbers) and on radian measure of angles.

Test 3 answers and some comments

The solutions to the Test 3 problems are posted on the Test solutions page (warning: an earlier version was posted there which had not been fully edited. Please make sure you look at this version.) There were, as usual, two versions of the test. Look at the bottom left corner of the Cover Sheet to see which version yours was.

I have a few general comments about these problems also:

• As usual, please pay attention to the comments I wrote. Very common comment this time was “invalid canceling” or “incorrect canceling”: please make sure you understand why what you did was wrong.

• If I wrote any comment that said that you did not show enough work (however phrased) please understand that this a serious matter which you need to attend to. I will not be lenient on this in future tests.

• A lot of people did well on the first page and then very poorly on page two. One big cause of this was failure to read the instructions: students treated the expressions that they were to simplify as equations to be solved, which they were not. READ THE FRIENDLY INSTRUCTIONS!

• On the third page, where you had to solve equations, many people made the problems much harder than they needed to be, because of not CLEARING the denominators, or because of not using the LCM but rather using the product of all the denominators to clear the denominators. I will put some examples of how this makes things more difficult on this post later on.

• There will be a few problems like problems 2, 3, and 4 on the next Test as well, so make sure that you correct your erroneous methods. The WeBWorKs on Rational Expressions and on Fractional (really rational) Equations are available for you to practice and they offer hints and solutions if you use “show me another”

 

Test 4 is rescheduled to Monday 2 December, and will not be re-scheduled barring emergency situation.

 

Notes, links, homework for Wednesday 20 November

Topics:

More rationalizing  denominators and complex numbers

• Trigonometry topic: special right triangles

The two important right triangles are:
• The isosceles right triangle,
• The half of an equilateral triangle – which I will call the half-equilateral triangle for short.
These are commonly called by other names which refer specifically to the degree measure of their angles, but it is much better to call them by these names for two reasons (at least):
• Later we will use the radian measure of the angles, so we do not want to tie ourselves too much to degree measure
• The names given above remind us of what these triangles really are, so that we can recreate them if we forget the relationships of their sides, for example.

Trigonomety in right triangles:

 

 

Homework:

• Review the isosceles right triangles and how we found the length of the hypotenuse using the Pythagorean Theorem, and also the half-equilateral right triangle and how we found its height using the Pythagorean Theorem. If you do this a few times, you will end up memorizing the triangle! You must learn it by heart as we will use it (and the other special right triangle) a lot in Trigonometry. Here is a good web source that shows how these triangles were developed. (Do not get too attached to the degree-measure names, though!)

Here is another good source on the half-equilateral triangle. (This one was also linked above in the notes.)

 

Here is the chapter from Prof. Africk’s textbook which discusses similar triangles: AfrickGeometryTexctbookSimilarTriangles

In addition to the WeBWorK, you should do problems 7 – 10 from p. 174 there. The answers will be posted here later.

 

Notes and homework Wednesday 13 November

Topics:

Simplifying radicals (including higher roots)

Adding and subtracting radicals

Multiplying radicals

Dividing radicals (simple case)

Here are the notes I handed out in class: OperationsWithRadicals

The problems we worked on in class were (some of) the even-numbered problems from the textbook, page 39

Homework homework yes you have homework

Please make sure to work on the WeBWorK assignments which are due tomorrow night, but start them now! There will be a quiz on adding/subtracting and multiplying radicals tomorrow.

Also, do the following problems on dividing radicals: assume all variables are $\ge 0$

(1) $\frac{\sqrt{57}}{\sqrt{19}}$

(2)  $\frac{\sqrt{2x^{3}}}{\sqrt{18x}}$

(3) $\frac{3\sqrt{2x}}{\sqrt{8x}}$

(4) $\frac{5\sqrt{30}}{10\sqrt{120}}$

 

 

Notes for Thursday 7 November class

Topic: Fractional (Rational) Exponents

Today’s active learning was on extending our work on exponents to include rational exponents. We started out by reviewing how to simplify square roots. Here are the problems:

Simplify, assuming that $x$ and $y$ are both $\ge 0$. [Why do we need to assume this? What difference does it make?]

(1) $\sqrt{9x^{2}}$

(2) $\sqrt{128}$

(3) *Sorry, I’ve forgotten exactly what I wrote for this problem. Can anyone tell us what it was?

(4) $\sqrt{x^{4}y^{9}}$

The active learning problems are here:

MAT1275-F15-Shaver.sshaver.RootsRadicalsShort

For these we used the definition of the fractional (rational) exponents, together with all the previous definitions and properties of exponents. All of the properties continue to work even with rational exponents, as long as we are careful about when the radicand needs to be $\ge 0$.

 

Homework homework yes you have homework!

Recommended order for the WeBWorK:

• First complete the assignments that are due tonight, obviously.

• Then complete the assignments HigherRootsAlgebraic, RationalExponents, and FrractionalEquations, if you have not already done so.

• At the same time you are working on those assignments, start working the Test3Review a few problems at a time. It’s best to schedule time on at least 3 of the 4 days before the Test (including today) and not try to do all the Test Review at once.

Here are some typed notes and a summary of the definitions and properties of exponents, which may be helpful:

MAT1275coRootsNotes

MAT1275coExponentsDefinitionsLaws-Condensed

A more comprehensive post on reviewing for Test 3 will go up soon, but until then, see the Test 2 Review post, which contains much of the same advice.

Notes and homework comments for Wednesday 6 November

See also this post

Topic: Higher Roots (corresponds to Session 4, roughly, in the textbook)

I elaborated on the  Math is Fun page on n-th roots. My comments and summary are below, but please read that page also and follow the link to “fractional exponents” to get more depth.

Definitions: I will use exponential notation rather than the multiplicative notation that is used in the Math is Fun page

In what follows, $n$ represents a natural number greater than 1, and $a$ represents a real number.

An n-th root of $a$ is a number, $b$, such that $b^{n} = a$

I’m not using the notation $\sqrt[n]{a}$ in that definition, because of this
Note: (not mentioned in Math is Fun, but I mentioned) There may be more than one n-th root of $a$ in the real numbers, in which case the notation $\sqrt[n]{a}$ refers to the positive n-th root.

However, it is a general fact that
$\left(\sqrt[n]{a}\right)^{n} = a$
this is another way to say the definition of square root.

Properties of n-th roots:

Roots and multiplication:

nth root ab
(If n is even, a and b must both be ≥ 0)

Roots and division:

nth root a divide b
(If n is even, then a≥0)(and b>0 no matter what n is)
(b cannot be zero, as we can’t divide by zero)

Roots and addition/subtraction: they don’t work well together!

Watch out for the following:

$\sqrt[n]{a + b} \neq \sqrt[n]{a} + \sqrt[n]{b}$

$\sqrt[n]{a – b} \neq \sqrt[n]{a} – \sqrt[n]{b}$

$\sqrt[n]{a^{n}+ b^{n}} \neq {a} + {b}$

$\sqrt[n]{a^{n} – b^{n}} \neq {a} – {b}$

 

Roots of powers:

Here’s the handy table from the Math is Fun page:

n is odd n is even
a ≥ 0 nth root a^n nth root a^n
a < 0 nth root a^n nth root a^n = abs(a)

I discussed at some length why there needs to be an absolute value in that bottom right cell of the table.

Quiz next time: Explain why we need the absolute value in the rule $\sqrt{x} = |x|$. Why do we not need an absolute value in the rule $\sqrt[3]{x} = x$?

More Roots of Powers:

A useful property: $\sqrt[n]{a^{m}} = \left(\sqrt[n]{a}\right)^{m}$ for a ≥ 0

Sometimes this is helpful in simplifying n-th roots. But even more powerful is to use fractional exponents to represent roots:

Definitions:

$\sqrt[n]{a} := a^{\frac{1}{n}}$ for a ≥ 0

$\sqrt[n]{a^{m}} = \left(\sqrt[n]{a}\right)^{m} := a^{\frac{m}{n}}$ for a ≥ 0

 

Homework homework homework yes there is homework!

Before Thursday’s class, please do the following:

• Read and try to understand the notes above, notes you took in class, the Math is Fun page on n-th roots, and also the page on fractional exponents.

• Do the “Your Turn” problems at the bottom of the Math is Fun page on n-th roots

• Do the WeBWorK “HigherRoots”

• Start working on the WeBWorK “HigherRootsAlgebraic” – we will probably pick up on this tomorrow.

A myth debunking and morals from 6 November

First, for the myth debunking, a/k/a “When a student tells you one thing and your teacher tells you a different thing, which one is more likely to be right?’

We worked on this problem in class:
$\left(\frac{5a^{4}z^{−7}}{3a^{−6}z{−4}}\right)^{−3}$
The student who was working it at the board started out by trying to simplify what is inside the parentheses first, which, I would like to point out, is absolutely correct: however, she made a mistake in that simplification. When I stopped her and said there was an error, several students told her that she was “supposed” to change the -3 power into the 3rd power of the reciprocal. (Not necessarily in those words.) I then countered by saying as insistently as I could that no, she did not have to do that first, she could simplify inside the parentheses first, I was only objecting that she had made an error in simplifying.

So what happened? The student listened to the students, who were WRONG in what they said.

It would have been totally correct to simplify inside the parentheses first and then deal with the -3 power. As I will do here:

$\left(\frac{5a^{4}z^{−7}}{3a^{−6}z{−4}}\right)^{−3} = \left(\frac{5a^{4}a^{6}z^{4}}{3z{7}}\right)^{−3}$
$= \left(\frac{5a^{10}z^{4}}{3z{7}}\right)^{−3}$
$= \left(\frac{5a^{10}}{3z{3}}\right)^{−3}$
$= \left(\frac{3z{3}}{5a^{10}}\right)^{3}$
$= \frac{27z{9}}{125a^{30}}$

 

Morals of the story: there are at least two!

• When a student says one thing and the teacher says the opposite, maybe it’s better to listen to the teacher?

• There is not only one correct way to work these problems, and, indeed, that is true of a lot of what we do in this course. The important thing is to know the correct properties and definitions and how to use them correctly: they are your tools. And then, in working a problem, you look at what you have at each step and ask, what can I do with this? What tools can I use that may get me where I want to go? As long as you use mathematically correct “tools” and use them correctly and get to what the problem was asking for, your work is correct.

Thus it is your job to look to the examples to understand why we choose to do what we do at each step: what in the problem itself is suggesting to use this particular tool? Avoid at all costs memorizing “steps” whose purpose you do not understand. That is the opposite of mathematics!

 

Election Day! Don’t forget to vote!

There are important elections happening even though it’s an “off-off” year. And that means that your vote can count more!

If you are registered to vote in NYC, there is an important election for Public Advocate and also there are 5 proposed changes to the City Charter which look very interesting. You can read about them here.

 

Find your polling place here, and congratulations on exercising your precious right to vote!