\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus - Mathematical Proofs and Logic

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Although you may not have had a class in mathematical logic yet, there are a few concepts that we will cover here that should help you understand calculus.
First, we recommend that you read the proofs in your textbook when you come across a new theorem. Even if you don't understand each step, starting to read them now will help you understand them much more easily later on, when you DO need to understand them. We recommend this even if you are not planning on taking any more math in the future. Just the effort of trying to understand the proofs will change the chemistry of your brain now and will help you in your current course.

Before we get started, here is a great video.

PatrickJMT - Introduction to Fundamental Math Proof Techniques [14mins]

video by PatrickJMT

Okay, now for a few logic concepts. We hope you have noticed in the statement of some theorems that the language has an if-then structure (even if they don't explictly state the words 'if' and 'then'). Other theorems use an 'if and only if' phrase. It is important that you understand the difference and how the wording affects your use of the theorem.

One Way Implication; if A, then B

Let's start with 'if-then'. This is called an implication. It may be written as, 'if A, then B' and symbolically we usually write it as A → B. This means that if we know that A is true, then, based on that information alone, we can correctly conclude that B is also true. But you need to be careful here. The other direction cannot be assumed, i.e. you cannot automatically say B → A.

The character '~' is called a tilde. You may have seen it on your keyboard and used it in an emoticon or something. In mathematics, we use it to mean 'not' or 'negation'. So if I write ~A, I mean 'not A'. It turns out that if you know that A → B, then, based on that information alone, can also correctly say ~B → ~A. Notice the direction of the arrow. It may be easier for you to see if you write ~A ← ~B. Basically the idea is that if we have A → B and we know A is true, then B is also true. You can also be guaranteed that by negating both and changing the direction of the arrow, you get ~A ← ~B, which is also true. However, mathematicians don't really like to have arrows going backwards like this, so they usually leave the arrow and switch A and B to get ~B → ~A.

Now, I know this is a lot to take in and when you get into a mathematical logic course, you will learn why these work this way. For now, let me just give you an example.

A: I am a college math teacher.
B: I have a degree in mathematics.
Now, since most schools, including where I have taught calculus, require college math teachers to have a degree in mathematics, we know that A → B, i.e. if I am a college math teacher, then I have a degree in mathematics. However, you cannot assume B → A, if I have a degree in mathematics, then I am a college math teacher. There are a lot of people with math degrees who are not college math teachers. Some of them engineers. So based on the information that I am a college math teacher, which is true, then you know I have a degree in mathematics.

Now, let's look at ~B → ~A ( called the contrapositive of A → B ). Using this same example, if I do NOT have a degree in mathematics, then I am NOT a college math teacher. This is also true because, as I said above, the schools where I have taught calculus all require a math degree. Do you see how this works? In mathematical logic, if the implication A → B is true, then the contrapositive ~B → ~A is also true. This can be shown using truth tables.

Two Way Implication; A if and only if B

You can say that 'if-then' is a part or a subset of 'if and only if'. Because what 'if and only if' (sometimes shortened to iff) says is that the other direction is true also. What I mean is if you have 'A iff B', then you know A → B and B → A. Is that cool or what?!

Why is this important? Because starting with calculus, theorems become increasing important as you go on in mathematics. And you need to be able to use theorems correctly, even at the beginning calculus level. If you are studying infinite series, you have to understand these ideas in order to know how to determine convergence or divergence of an infinite series.

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Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)


\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)


\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)


\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)


\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)


\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)


\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)


\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)


\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)


\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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One Way Implication

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