# Preface # Unit 1: Functions ::: {.theorem title="The Pythagorean Theorem"} If \( a,b \) are the legs of a right triangle with hypotenuse \( c \), there is a relation \[ a^2 + b^2 = c^2 .\] ::: ::: {.theorem title="The Distance Formula"} If \( p = (x_1, y_1) \) and \( q = (x_2, y_2) \) are points in the Cartesian plane, then there is a **distance function** \[ d: \left\{{ \text{Pairs of points } (p, q) }\right\} \to {\mathbb{R}}\\ (p, q) &\mapsto d(p, q) \coloneqq\sqrt{ (x_2 - x_1)^2 + (y_2 - y_q)^2} .\] ::: ```{=tex} \todo[inline]{Law of cosines} ``` ::: {.definition title="Linear Functions"} A function \( f:{\mathbb{R}}\to {\mathbb{R}} \) is **linear** if and only if \( f \) has a formula of the following form: \[ f(x) = \alpha x + \beta && \alpha, \beta \in {\mathbb{R}} .\] ::: ::: {.definition title="Intercepts"} Given a function \( f: {\mathbb{R}}\to {\mathbb{R}} \), an **\( x{\hbox{-}} \)intercept** of \( f \) is a point \( (x_0, 0) \) on the graph of \( f \), so \( f(x_0) = 0 \). Equivalently, it is a point on the intersection of the graph and the \( x{\hbox{-}} \)axis. \ A **\( y{\hbox{-}} \)intercept** of \( f \) is a point \( (0, y_0) \) on the graph of \( f \), so \( f(0) = y_0 \). Equivalently, it is a point on the intersection of the graph and the \( y{\hbox{-}} \)axis. ::: ::: {.definition title="Relation"} A **relation** on two sets \( X \) and \( Y \) is a set of ordered pairs \( (x, y) \in X \times Y \), so \( R \) can be described as a set: \[ R = \left\{{ (x_0, y_0), (x_1, y_2), \cdots }\right\} .\] The **domain** of the relation is the set of all \( x\in X \) that occur in the first slot of these pairs, and the **range** is the set of all \( y\in Y \) that occur in the second slot. ::: ::: {.definition title="Function"} A relation \( R \) is a **function** if it satisfies the following *deterministic property*: for every \( x_0\in \operatorname{dom}(R) \), there is exactly *one* pair of the form \( (x_0, y_0) \in R \). ::: ::: {.remark} This says we can think of \( X \) as "inputs" and \( Y \) as "output", and a function is a way to unambiguously assign inputs to outputs. It can be useful to think of functions like programs: if I send in an \( x \), what \( y \) should the program return to me? If I run this program today, tomorrow, and 100 years from now, sending in the same \( x \) every time, we might want it to give the same output every time, which is the *deterministic* property: I can *determine* a single unique output if I know what the input is. If my program tells me that \( 2+2=4 \) today but \( 2+2=5 \) tomorrow, who knows what it will return in 100 years! We can't "determine" it. ::: ::: {.slogan} For domains and ranges: - Domains: the set of *meaningful* inputs that the function "knows" how to handle. - Ranges: the set of *attainable* outputs that we can expect. ::: ::: {.remark} To determine a domain: 1. Naively hope it is *all* of \( {\mathbb{R}} \). 2. Throw out "problematic" points. 3. Draw a number line and write out what you are left with in interval notation. ::: ::: {.example title="?"} Define \[ f: {\mathbb{R}}&\to {\mathbb{R}}\\ x &\mapsto {1\over x} .\] Then \( \operatorname{dom}(f) = {\mathbb{R}}\setminus\left\{{0}\right\}= (-\infty, 0) \cup(0, \infty) \) and \( \mathop{\mathrm{range}}(f) = {\mathbb{R}} \). ::: ::: {.example title="?"} Define \[ f: {\mathbb{R}}&\to {\mathbb{R}}\\ x &\mapsto \sqrt{x} .\] Then \( \operatorname{dom}(f) = {\mathbb{R}}\setminus(-\infty, 0) = [0, \infty) \) and \( \mathop{\mathrm{range}}(f) = [0, \infty) \). ::: # Unit 2: Exponential and Logarithmic Functions # Unit 3: Trigonometric Functions ## General Notes - In this section, always draw a picture! Virtually 100% of the time. - In particular, a unit circle should almost always show up. - Use exact ratios wherever possible. - There are too many details and formulas to just memorize in this unit: focus on the **processes**. ## Common Mistakes Some facts to remember: - \( \sin^{-1}(\theta) \neq 1/\sin(\theta) \). Mnemonic: reciprocals of trigonometric functions already have a better name, here \( \csc(\theta) \). ## Basic Trigonometric Functions ```{=tex} \todo[inline]{Sin/cos/etc as ratios} ``` ## Proportionality Relationships ::: {.definition title="Radian"} ```{=tex} \todo[inline]{What is a 1 radian?} ``` ![image_2021-04-18-21-51-59](figures/image_2021-04-18-21-51-59.png) ::: ::: {.remark} In geometric terms, an angle in radians in the ratio of the arc length \( s(\theta, R) \) to the radius \( R \), so \[ \theta_R = {s(\theta, R) \over R} .\] ::: ::: {.definition title="Coterminal Angles"} If \( \theta \) is an abstract angle, we will say \( \theta + k\,\text{rev} \simeq \theta \) for any integer \( k\in {\mathbb{Z}} \). Any such angle is said to be **coterminal** to \( \theta \). ::: ::: {.remark} In radians: \[ \theta_R \simeq \theta_R + k\cdot 2\pi && k\in {\mathbb{Z}} .\] In degrees: \[ \theta_D \simeq \theta_D + k\cdot 360^\circ && k\in {\mathbb{Z}} .\] ::: ::: {.proposition title="Degrees are related to radians"} ```{=tex} \todo[inline]{todo} ``` \[ {\theta \over 1\,\text{rev}} = {\theta_R \over 2\pi\, \text{rad} } = {\theta_D \over 360^\circ} .\] ::: ::: {.proposition title="Arc length and sector area are related to radians"} ```{=tex} \todo[inline]{todo} ``` \[ {\theta \over 1\,\text{rev}} = {s(R, \theta) \over 2\pi R} = { A(R, \theta) \over \pi R^2 } .\] This implies that \[ A(R, \theta) &={R^2 \theta \over 2} \\ s(R, \theta) &= R\theta .\] ::: ## Trigonometric Functions as Ratios ::: {.definition title="?"} There are 6 trigonometric functions defined by the following ratios: ```{=tex} \todo[inline]{soh-cah-toa, cho-sha-cao} ``` ::: ::: {.proposition title="Domains of trigonometric functions"} ------------------------------------------------------------------------------------------------------------------------------------------- Function Domain Range ---------------------------- ------------------------------------------------------------------------------------------ ------------------- \( \sin \) \( {\mathbb{R}} \) \( [-1, 1] \) \( \cos \) \( {\mathbb{R}} \) \( [-1, 1] \) \( \tan \) \( {\mathbb{R}}\setminus\left\{{\pm {\pi \over 2}, \pm{3\pi \over 2}, \cdots}\right\} \) ? \( \csc \) \( {\mathbb{R}}\setminus\left\{{0, \pm {\pi}, \pm{2\pi}, \cdots}\right\} \) ? \( \sec \) \( {\mathbb{R}}\setminus\left\{{\pm {\pi \over 2}, \pm{3\pi \over 2}, \cdots}\right\} \) ? \( \cot \) \( {\mathbb{R}}\setminus\left\{{0, \pm {\pi}, \pm{2\pi}, \cdots}\right\} \) ? ------------------------------------------------------------------------------------------------------------------------------------------- ::: ## Polar Coordinates ::: {.definition title="Unit Circle"} The **unit circle** is defined as \[ S^1 \coloneqq\left\{{ \mathbf{p} = (x, y) \in {\mathbb{R}}^2 {~\mathrel{\Big|}~}d(\mathbf{p}, \mathbf{0}) = 1 }\right\} = \left\{{ (x, y) \in {\mathbb{R}}^2 {~\mathrel{\Big|}~}x^2 + y^2 = 1 }\right\} ,\] the set of all points in the plane that are distance exactly 1 from the origin. ::: ::: {.theorem title="Polar Coordinates"} If a vector \( \mathbf{v} \) has at an angle of \( \theta \) in radians and has length \( R \), the corresponding point \( \mathbf{p} \) at the end of \( \mathbf{v} \) is given by \[ \mathbf{p} = {\left[ {x, y} \right]} = {\left[ {R\cos(\theta), R\sin(\theta)} \right]} .\] Conversely, if \( (x, y) \) are known, then the corresponding \( R \) and \( \theta \) are given by \[ [R, \theta] = {\left[ { \sqrt{x^2 + y^2}, \arctan\qty{y\over x} } \right]} .\] ::: ::: {.corollary title="Polar Coordinates on $S^1$"} If \( R=1 \), so \( \mathbf{v} \) is on the unit circle \( S^1 \), then \[ [x, y] = [\cos(\theta), \sin(\theta)] .\] ::: ::: {.remark} This is a very important fact! The \( x, y \) coordinates on the unit circle *literally* corresponding to cosines and sines of subtended angles will be used frequently. ::: ::: {.slogan} Cosines are like \( x \) coordinates, sines are like \( y \) coordinates. ::: ::: {.example title="?"} Given \( \theta_R = 4\pi/3 \), what is the corresponding point on the unit circle \( S^1 \)? ::: ::: {.warnings} Note that \( \sin(\theta), \cos(\theta) \) work for any \( \theta \) at all. However, \( \cos(\theta) = 0 \) sometimes, so \( \tan(\theta) \coloneqq\sin(\theta) / \cos(\theta) \) will on occasion be problematic. Similar story for the other functions. ::: ## Special Angles For reference: the unit circle. ![image_2021-04-18-21-06-45](figures/image_2021-04-18-21-06-45.png) ::: {.remark} Idea: we want to partition the circle simultaneously - Into 8 pieces, so we increment by \( 2\pi/8 = \pi/4 \) - Into 12 pieces, so we increment by \( 2\pi/12 = \pi/6 \). ::: ::: {.proposition title="Trick to memorize special angles"} ```{=tex} \todo[inline]{Table of special angles, increasing/decreasing} ``` ::: ## Reference Angles and the Flipping Method ::: {.definition title="Reference Angle"} Given a vector at of length \( R \) and angle \( \theta \), the **reference angle** \( { \theta_{\mathrm{Ref} } } \) is the acute angle in the triangle formed by dropping a perpendicular to the nearest horizontal axis. ::: ::: {.proposition title="?"} Reference angles for each quadrant: \[ \text{Quadrant II}: && \theta + { \theta_{\mathrm{Ref} } }= \pi \\ \text{Quadrant III}: && \pi + { \theta_{\mathrm{Ref} } }= \theta \\ \text{Quadrant IV}: && \theta + { \theta_{\mathrm{Ref} } }= 2\pi .\] ::: ::: {.example title="?"} Given \( \sin(\theta) = 7/25 \), what are the five remaining trigonometric functions of \( \theta \)? Method: 1. Draw a picture! Embed \( \theta \) into a right triangle. 2. Find the missing side using the Pythagorean theorem. 3. Use definition of trigonometric functions are ratios. ::: ::: {.remark} Note that you can not necessarily find the angle \( \theta \) here, but we didn't need it. If we *did* want \( \theta \), we would need an inverse function to free the argument: \[ \sin(\theta) &= 7/25 \\ \implies \arcsin( \sin(\theta) ) = \arcsin(7/25) \\ \implies \theta = \arcsin(7/25) \\ .\] ::: ## Identities Using Pythagoras ::: {.proposition title="?"} \[ (\sin(\theta))^2 + (\cos(\theta))^2 &= 1 \\ 1 + (\cot(\theta))^2 &= (\csc(\theta))^2 \\ (\tan(\theta))^2 + 1 &= (\sec(\theta))^2 .\] ::: ::: {.proof title="?"} Derive first from Pythagorean theorem in \( S^1 \). Obtain the second by dividing through by \( \qty{\sin(\theta)}^2 \). Obtain the third by dividing through by \( \qty{\cos(\theta)}^2 \). ::: ## Even/Odd Properties ::: {.question} Thinking of \( \cos(\theta) \) as a function of \( \theta \), is it - Even? - Odd? - Neither? ::: ::: {.remark} Why do we care? The Fundamental Theorem of Calculus. ![image_2021-04-18-22-39-08](figures/image_2021-04-18-22-39-08.png) ::: ::: {.proposition title="?"} ```{=tex} \envlist ``` - \( f(\theta) \coloneqq\cos(\theta) \) is an even function. - \( g(\theta) \coloneqq\sin(\theta) \) is an odd function. ::: ::: {.proof title="?"} Plot vectors for \( \theta, -\theta \) on \( S^1 \) and flip over the \( x{\hbox{-}} \)axis. ::: ::: {.corollary title="?"} ```{=tex} \envlist ``` - \( \cos(t), \sec(t) \) are even. - \( \sin(t), \csc(t), \tan(t), \cot(t) \) are odd. ::: ## Wave Function ::: {.remark} Motivation: let a vector run around the unit circle, where we think of \( \theta \) as a time parameter. What are its \( x \) and \( y \) coordinates? What happens if we plot \( x(t) \) in a new \( \theta \) plane? ::: ::: {.definition title="Standard Form of a Wave Function"} The **standard form** of a wave function is given by \[ f(t) \coloneqq A \cos(\omega (t - \varphi)) + \delta ,\] where - \( A \) is the **amplitude**, - \( \omega \) is the **frequency**, - \( \phi \) is the **phase shift**, and - \( \delta \) is the **vertical shift**. - \( P \coloneqq 2\pi / \omega \) is the **period**, so \( f(t+kP) = f(t) \) for all \( k\in {\mathbb{Z}} \). ```{=tex} \todo[inline]{Insert plot} ``` ::: ::: {.remark} Note that this is nothing more than a usual cosine wave, just translated/dilated in the \( x \) direction and the \( y \) direction. ::: ::: {.warnings} Don't memorize equations like \( y=\sin(Bt+C) \) and e.g. the phase shift if \( \phi = -C/B \). Instead, use a process: always put your equation in standard form, then you can just read off the parameters. For example: \[ f(t) &= \cos(Bt+C) \\ &= \cos( B (t + {C\over B}) ) \\ &= \cos(\omega (t - \phi) ) \\ \\ &\implies B = \omega, \phi = -{C\over B} .\] ::: ::: {.example title="?"} Put the following wave in standard form: \[ f(t) \coloneqq 4\cos(3t+2) .\] ::: ::: {.example title="?"} Put the following wave in standard form: \[ f(t) \coloneqq\alpha \cos(\beta t+\gamma) .\] ::: ::: {.proposition title="?"} How to plot the graph of a wave equation: 1. Put in standard form. 2. Read off the parameters to build a rectangular box of width \( P \) and height \( 2{\left\lvert {A} \right\rvert} \) about the line \( y=\delta \). 3. Break the box into 4 pieces using the key points \( t = \phi + {k\over 4}P \) for \( k=0,1,2,3,4 \). ::: ::: {.example title="Plotting"} Plot the following function in the \( t \) plane: \[ f(t) = 2 \cos\qty{5t - {\pi \over 2} } + 7 .\] ::: ::: {.example title="?"} Plot the following: \[ f(t) = -2\sin(3t-7) .\] ::: ::: {.proposition title="Determining the equation of a sine wave"} Given a picture of a graph of a sine wave, 1. Draw a horizontal line cutting the wave in half. This will be \( \delta \). 2. Measure the distance from this midline to a peak. This will be \( {\left\lvert {A} \right\rvert} \). 3. Restrict to one full period, starting either at a peak (if you want to match \( \cos(t) \)) or a zero (if you want to match \( \sin(t) \)). Pick the period starting as close as possible to the \( y{\hbox{-}} \)axis. 4. Measure the period \( P \) and reverse-engineer it to get \( \omega \): \( P = 2\pi/\omega \implies \omega = 2\pi/P \). 5. Measure the distance from the starting point to the \( y{\hbox{-}} \)axis: this is \( \phi \). ::: ::: {.example title="?"} Determine the equation of the following wave function: ![image_2021-04-18-20-51-34](figures/image_2021-04-18-20-51-34.png) ::: {.solution} \[ f(t) = 2\sin\qty{4t + {\pi \over 6} } .\] ::: ::: ::: {.remark} Note that we can graph other trigonometric functions: they get pretty wild though. ![Tangent](figures/image_2021-04-18-21-01-06.png) ![](figures/image_2021-04-18-21-01-38.png) ::: ## Simplifying Identities ::: {.remark} The goal: reduce a complicated mess of trigonometric functions to something as simple as possible. We'll use a **boxing-up method**. ::: ::: {.remark} On verifying identities: if you want to show \( f(\theta) = g(\theta) \), start at one and arrive at the other: \[ f(\theta) &= \text{simplify } f \\ &= \cdots \\ &= \cdots \\ &= \cdots \\ &= g(\theta) \\ .\] ::: ::: {.warnings} If you end up with something like \( 1=1 \) or \( 0=0 \), this is hinting at a problem with your logic. ::: ::: {.exercise title="?"} Simplify the following: \[ F(\theta) \coloneqq\qty{ \sin(\theta) \cos(\theta) \over \cot(\theta)} \cos(\theta)\csc(\theta) .\] ::: ::: {.solution} \[ F = s \qty{s \over c} .\] ::: ::: {.remark} As an alternative, you can use the **transitivity of equality**: show that \( f(\theta) = h(\theta) \) for some totally different function \( h \), and then show \( g(\theta) = h(\theta) \) as well. ![image_2021-04-18-21-58-52](figures/image_2021-04-18-21-58-52.png) ::: ::: {.exercise title="Reducing both sides to a common expression"} Show the following identity: \[ {\sin(-\theta) + \csc(\theta)} = \cot(\theta) \cos(\theta) \] by showing both sides are separately equal to \( h(\theta) \coloneqq\csc(\theta) - \sin(\theta) \). ::: ## Inverse Functions ### Motivation ::: {.remark} Motivation: we want a way to solve equations where the unknown \( \theta \) is stuck in the argument of a trigonometric function. For example, for \( \sin: {\mathbb{R}}_A \to {\mathbb{R}}_B \), this would be some function \( f: {\mathbb{R}}_B \to {\mathbb{R}}_A \) such that \[ f(\sin(\theta)) &= \operatorname{id}(\theta) = \theta \\ \sin(f(y)) &= \operatorname{id}(y) = y .\] ![Input-Output perspective: important!](figures/image_2021-04-18-22-24-55.png) Note that we only ever have to define \( f \) on \( \mathop{\mathrm{range}}(\sin) \), since we're only ever sending outputs of \( f \) in as the inputs of \( \sin \). So we need \( \mathop{\mathrm{range}}(\sin) \subset \operatorname{dom}(f) \), noting that \( \mathop{\mathrm{range}}(\sin) = [-1, 1] \): ![image_2021-04-18-22-26-56](figures/image_2021-04-18-22-26-56.png) Similarly, we need \( \mathop{\mathrm{range}}(f) \subset \operatorname{dom}(\sin) \). ::: ### Using Triangles ::: {.remark} Optimistically imagine that we had some such inverse function. Then we could evaluate some expressions without even knowing anything else about it. The trick: \[ \theta &= \arccos(p/q) \\ \implies \cos(\theta) &= \cos(\arccos(p/q)) \\ \implies \cos(\theta) &= p/q .\] Now embed this in a triangle. We can't solve for \( \theta \), but we can solve for other trigonometric functions. ::: ::: {.exercise title="Using functional inverse property"} \[ \cos\qty{ \arccos\qty{ \sqrt 5 \over 5 } } &= {\sqrt 5 \over 5} \\ \arccos\qty{ \cos \qty{ \sqrt 5 \over 5 } } &= {\sqrt 5 \over 5} .\] ::: ::: {.exercise title="Using a triangle"} \[ \tan\qty{ \arcsin\qty{ p \over q } } = {p \over \sqrt{q^2 - p^2} } .\] ![image_2021-04-22-22-14-13](figures/image_2021-04-22-22-14-13.png) ::: ::: {.exercise title="Can't extract angles"} Compute \( \arcsin(3/5) \). ::: {.warnings} This is equal to \( \sin^{-1}(3/5) \), which is *not* equal to \( {1\over \sin(3/5)} \)! One way to remember this is that we have another name for reciprocals, here \( \csc(3/5) \). ::: ::: ::: {.solution} \[ \theta &= \arcsin(3/5) \\ \implies \sin(\theta) &= (3/5) && \text{roughly by injectivity} \\ \implies &= \cdots ? .\] We are out of luck, since this isn't a special angle. So we can't find a numerical value of \( \theta \). We can find other trig functions of \( \theta \) though: ![image_2021-04-18-22-30-09](figures/image_2021-04-18-22-30-09.png) So for example, \( \cos(\arcsin(3/5)) = 4/5 \). ::: ::: {.remark} Most inverse trigonometric functions can *not* be exactly solved! We'll have to approximate by calculator if we want the actual angle. If we just want *other* trigonometric functions though, we can always embed in a triangle. ::: ::: {.example title="Using triangles"} Show the following: - \( \cos(\arcsin(24/26)) = 10/26 \) - Write \( \theta = \arcsin(24/26) \), note \( \theta \) is in \( [-\pi/2, \pi/2] = \mathop{\mathrm{range}}(\arcsin) \). - \( \tan(\arccos(-10/26)) = 10/26 \) - Write \( \theta = \arccos(-10/26) \), note \( \theta \) is in \( [0, \pi] = \mathop{\mathrm{range}}(\arccos) \) ::: ### Defining Inverses ::: {.remark} The setup: try swapping \( y \) and \( \theta \) in the graph of \( y=\sin(\theta) \): ![image_2021-04-18-22-32-36](figures/image_2021-04-18-22-32-36.png) Note that the latter is a function (vertical line test) iff the former is injective (horizontal line test). So we take the largest branch where the inverse is a function: ![image_2021-04-18-22-33-27](figures/image_2021-04-18-22-33-27.png) Back on our original graph, this looks like the following: ![image_2021-04-18-20-53-25](figures/image_2021-04-18-20-53-25.png) Restricting, we get - \( \operatorname{dom}(\arccos) \coloneqq\mathop{\mathrm{range}}({ \color{green} \cos} ) = [-1, 1] \). - \( \mathop{\mathrm{range}}(\arccos) \coloneqq\operatorname{dom}( {\color{green} \cos} ) = [0, \pi] \). ::: ::: {.remark} A similar analysis works for \( \sin(\theta) \): ![image_2021-04-18-22-34-21](figures/image_2021-04-18-22-34-21.png) Restricting, we get - \( \operatorname{dom}(\arcsin) \coloneqq\mathop{\mathrm{range}}({ \color{green} \sin} ) = [-1, 1] \). - \( \mathop{\mathrm{range}}(\arcsin) \coloneqq\operatorname{dom}( {\color{green} \sin }) = [-\pi/2, \pi/2] \). ::: ::: {.remark} This gives us a new tool to solve equations: \[ \vdots &= \vdots \\ \implies \cos(x) &= b \\ \implies \arccos(\cos(x)) &= \arccos(b) \\ \implies x &= \arccos(b) ,\] but only if we know this makes sense based on domain/range issues. ::: ::: {.proposition title="Domains of inverse trigonometric functions"} Restrict domains in the following ways: - \( \sin \): \( [-\pi/2, \pi/2] \) - \( \cos: [0, \pi] \) - \( \tan: [-\pi/2, \pi/2] \) --------------------------------------------------------------------------------------------------------------------------------------------------------------------------- Function Domain Range ----------------------------- ------------------------------------------------------------------------------------------ -------------------------------------------------- \( \arcsin \) \( [-1, 1] \) \( [-\pi/2, \pi /2] \) \( \arccos \) \( [-1, 1] \) \( [0, \pi] \) \( \arctan \) \( {\mathbb{R}} \) \( (-\pi/2, \pi/2) \) \( \operatorname{arccsc} \) \( {\mathbb{R}}\setminus\left\{{0, \pm {\pi}, \pm{2\pi}, \cdots}\right\} \) \( [-\pi/2, \pi/2]\setminus\left\{{0}\right\} \) \( \operatorname{arcsec} \) \( {\mathbb{R}}\setminus\left\{{\pm {\pi \over 2}, \pm{3\pi \over 2}, \cdots}\right\} \) \( [0, \pi]\setminus\left\{{\pi/2}\right\} \) \( \operatorname{arccot} \) \( {\mathbb{R}} \) \( (0, \pi) \) --------------------------------------------------------------------------------------------------------------------------------------------------------------------------- ::: ::: {.slogan} There is an easy way to remember this: - Cosines are \( x{\hbox{-}} \)values, pick the upper (or lower) half of the circle to make them unique. - Sines are \( y{\hbox{-}} \)values, pick the right (or left) half of the circle to make them unique. ![image_2021-04-22-22-00-04](figures/image_2021-04-22-22-00-04.png) ::: ::: {.example title="Using special angles"} ![Unit Circle](figures/image_2021-04-18-21-06-45.png) We have some exact values. Sines should be in QI or QIV: - \( \arcsin(1/2) = \pi/6 \) - \( \arcsin(\sqrt{3}/2) = \pi/3 \) - \( \arcsin(-1/2) = -\pi/6 \) Cosines should be in QI or QII: - \( \arccos(\sqrt{3}/2) = \pi/6 \) - \( \arccos(-\sqrt{2}/2) = 3\pi/4 \) - \( \arccos(1/2) = \pi/3 \) Tangents should be in QI or QIV: - \( \arctan(\sqrt{3}/3) = \pi/6 \) - \( \arctan(0) = 0 \) - \( \arctan(1) = \pi/4 \) ::: ::: {.warnings} Note that if \( f, g \) are an inverse pair, we have \[ f\circ g = \operatorname{id}\quad\iff\quad f(g(x)) = x,\quad g(f(x)) = x .\] However, we have to be careful with domains for trigonometric functions: - \( \arcsin(\sin(x)) = x \iff x\in [-\pi/2, \pi/2] \) (restricted domain of \( \sin \)) - \( \sin(\arcsin(x)) = x \iff x\in [-1, 1] \) (domain of \( \arcsin \)) - \( \arccos(\cos(x)) = x \iff x\in [0, \pi] \) (restricted domain of \( \cos \)) - \( \cos(\arccos(x)) = x \iff x\in [-1, 1] \) (domain of \( \arccos \)) - \( \arctan(\tan(x)) = x \iff x\in [0] \) (restricted domain of \( \tan \)) - \( \tan(\arctan(x)) = x \iff x\in {\mathbb{R}} \) - Domain of \( \arctan \), then range is \( [-\pi/2, \pi/2] \), which is in the domain of \( \tan \). ::: ## Double/Half-Angle Identities ::: {.remark} Sometimes we are interested in **superposition** of waves, see [Desmos](https://www.desmos.com/calculator/rhliflgwmv) for an example. Mathematically this is modeled by adding wave functions together. Similarly, we are sometimes interested in **modulating** or **enveloping** waves, which is modeled by multiplying a wave with another function: see [Desmos](https://www.desmos.com/calculator/wjgcl2xfbb). ![image_2021-04-18-22-06-08](figures/image_2021-04-18-22-06-08.png) We can sometimes rewrite these as a *single* wave with a phase shift. ::: ::: {.proposition title="Angle Sum Identities"} Identities: \[ \sin(\theta + \psi) &= \sin(\theta) \cos(\psi) + \cos(\theta) \sin(\psi) \\ \cos(\theta + \psi) &= \cos(\theta) \cos(\psi) + \sin(\theta) \sin(\psi) .\] Note that you can divide these to get \[ \tan(\theta + \psi) &= {\tan(\theta) + \tan(\psi) \over 1 - \tan(\theta) \tan(\psi) } ,\] and replace \( \psi \) with \( -\psi \) and use even/odd properties to get formulas for \( \sin(\theta - \psi), \cos(\theta - \psi) \) ::: ::: {.slogan} Sines are friendly and cosines are clique-y! ::: ::: {.corollary title="Double angle identities"} Taking \( \theta = \psi \) is the above identities yields \[ \sin(2\theta ) &= \sin(\theta) \cos(\theta) + \cos(\theta) \sin(\theta) \\ &= 2\sin(\theta)\cos(\theta) \\ \\ \cos(2\theta) &= \cos(\theta) \cos(\theta) + \sin(\theta) \sin(\theta) \\ &= \cos^2(\theta) - \sin^2(\theta) .\] ::: ::: {.warnings} The latter is not equal to 1! That would be \( \cos^2(\theta) + \sin^2(\theta) \). ::: ::: {.remark} Why do we care? We had 16 special angles, this gives a lot more. For example, \[ \cos(\pi/12) = \cos(\pi/3 - \pi/4) = \cdots \text{ plug in} .\] By allowing increments of \( \pi/12 \), we have 24 total angles. ::: ::: {.corollary title="?"} Starting from the following: \[ \cos(2\theta) &= \cos^2(\theta) - \sin^2(\theta) \\ &= \cos^2(\theta) - \qty{1 - \cos^2(\theta) } \\ &= 2\cos^2(\theta) -1 && \text{using } s^2 + c^2 = 1 ,\] one can solve for \[ \cos^2(\theta) = {1\over 2}\qty{1 + \cos(2\theta) } .\] Similarly \[ \cos(2\theta) &= \cos^2(\theta) - \sin^2(\theta) \\ &= \qty{1 - \sin^2(\theta) } - \sin^2(\theta) \\ &= 1-2\sin^2(\theta) && \text{using } s^2 + c^2 = 1 ,\] solving yields \[ \sin^2(\theta) = {1\over 2} (1 - \cos(2\theta) ) .\] ::: ::: {.remark} These are very important in Calculus! This gives us a way to reduce the exponents on expressions like \( \sin^n(\theta) \). ::: ## Bonus: Complex Exponentials ::: {.question} We spent one entire unit studying the function \( f(x) = e^x \), and another studying the functions \( g(x) = \cos(x), h(x) = \sin(x) \). They seem completely unrelated, but miraculously they are both just shadows of of unifying concept. ::: ::: {.remark} Components of vectors: every \( \mathbf{v}\in {\mathbb{R}}^2 \) breaks up as the sum of two vectors, i.e. \( \mathbf{v} = \mathbf{v}_x + \mathbf{v}_y \). In coordinates, if \( \mathbf{v} = (a, b) \), we have \( \mathbf{v}_x = (a, 0) \) and \( \mathbf{v}_y = (0, b) \). Alternatively, we can drop the ordered pair notation and write \( \mathbf{v} = a \widehat{\mathbf{x}} + b \widehat{\mathbf{y}} \). ::: ::: {.remark} We've worked with the *Cartesian plane* all semester. One powerful tool is replacing this with the *complex* plane. We formally define a new symbol \( i \) and replace the \( \widehat{ \mathbf{y} } \) direction with the \( i \) direction -- this amounts to replacing ordered pairs \( (a, b) \coloneqq a \widehat{ \mathbf{x} } + b\widehat{ \mathbf{y} } \) by a single number \( a + ib \). ::: ::: {.example title="How to work with complex numbers"} Complex numbers can be added: \[ (a + bi) + (c + di) = (a + c) + (b + d)i .\] This is perhaps easier to understand in the ordered pair notation: you just add the components in each component: \[ [a, b] + [c, d] = [a + c, b + d] .\] Complex numbers can be multiplied: \[ (a +bi)(c+di) &= a(c+di) + bi(c+di) \\ &= ac + adi + bci + bdi^2 \\ &= (ac - bd) + (ad + bc)i .\] This is harder to see in the ordered pair notation. We can compare complex numbers: they are equal iff their components are equal: \[ a + bi = c+di \iff a=c \text{ and } b = d ,\] or in ordered pair notation, \[ [a, b] = [c, d] \iff a = c \text{ and } b = d .\] ::: ::: {.remark} The symbol \( i \) happens to have another algebraic property. Consider the family of equations \( f(x, t) = x^2 + t \), and think about finding the roots. Finding a root is solving \( f(x, t) = 0 \), which is the exact same thing as finding the intersection points with the graph of \( g(x) = 0 \). Taking \( t=0 \) yields \( f(x) = x^2 \), which has a root at zero. Taking \( t<0 \) yields two roots. However, taking \( t>0 \) yields no roots -- at least not in \( {\mathbb{R}} \). As it turns out, the function \( f_1(x) = x^2 + 1 \) and \( g(x) = 0 \) *do* intersect in some other, bigger space, and we're only seeing a shadow of this! In other words, \( x^2+1=0 \) didn't have solutions in \( {\mathbb{R}} \), but *will* have a solution in \( {\mathbb{C}} \). ::: ::: {.remark} The following is the main link between exponentials and waves: ::: ::: {.proposition title="Euler's Formula"} \[ e^{i\theta} = \cos(\theta) + i\sin(\theta) .\] ::: ::: {.remark} Really, this is just polar coordinates on the unit circle: if we go back to ordered pair notation, this is just giving a point \( (\cos(\theta), \sin(\theta)) \in S^1 \). So the *complex number* \( e^{i\theta} \) is also a *vector* pointing at an angle \( \theta \) from the origin and landing on the unit circle. ::: ::: {.proposition title="Euler's Identity"} \[ e^{i\pi} = -1 .\] ::: ::: {.remark} This is remarkable! It relates some of the most fundamental constant numbers in mathematics: - \( e = 2.718\ldots \) - \( \pi = 3.14159\ldots \) - \( -1 \) Proof: just plug \( \pi \) into Euler's equation. Geometric interpretation: \( \pi \) radians is directly to the left. ::: ::: {.example title="?"} An application: proving the angle sum formulas algebraically. We start by considering the angle \( \alpha + \beta \). On one hand, Euler's formula says \[ e^{i( \alpha + \beta) } = \cos(\alpha + \beta) + i\sin(\alpha + \beta) = [\cos(\alpha + \beta), \sin(\alpha + \beta)] .\] On the other hand, we can use properties of exponentials first and expand: \[ e^{i(\alpha + \beta)} &= e^{i\alpha} e^{i\beta} \\ &= \qty{ \cos(\alpha) + i\sin(\alpha)} \cdot \qty{ \cos(\beta) + i\sin(\beta) } \\ &= \cos(\alpha) \qty{ \cos(\beta) + i\sin(\beta) } + i\sin(\alpha) \qty{ \cos(\beta) + i\sin(\beta) } \\ &= \cos(\alpha)\cos(\beta) + i \cos(\alpha)\sin(\beta) + i\sin(\alpha)\cos(\beta) + i^2 \sin(\alpha)\sin(\beta) \\ &= \qty{ \cos(\alpha)\cos( \beta) - \sin(\alpha)\sin(\beta) } + i \qty{\cos(\alpha) \sin(\beta) + \sin(\alpha)\cos(\beta) } \\ &= \left[ \cos(\alpha)\cos( \beta) - \sin(\alpha)\sin(\beta),\quad \cos(\alpha) \sin(\beta) + \sin(\alpha)\cos(\beta) \right] .\] Now we just equate components: \[ [\cos(\alpha + \beta), \sin(\alpha + \beta)] &= \left[ \cos(\alpha)\cos( \beta) - \sin(\alpha)\sin(\beta),\quad \cos(\alpha) \sin(\beta) + \sin(\alpha)\cos(\beta) \right] \\ \\ \implies \cos(\alpha + \beta) &= \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta) \\ \implies \sin(\alpha + \beta) &= \cos(\alpha)\sin(\beta) + \sin(\alpha)\cos(\beta) .\] ::: ::: {.remark} The analogy goes farther: polar coordinates are essentially just a shadow of complex numbers. Since \( e^{i\theta} \in S^1 \), we can scale by a radius \( r \) to write \( z = re^{i\theta} \) and get any point in the plane. If we just draw a vector \( \mathbf{v}[r\cos(\theta), r\sin(\theta)] \), note that Euler's formula gives us a way to get a complex number \( z \) that corresponds to it: \[ z \coloneqq re^{i\theta} = r(\cos(\theta) + i\sin(\theta)) = r\cos(\theta) + i\cdot r\sin(\theta) = [r\cos(\theta), r\sin(\theta)] = \mathbf{v} .\] ::: ::: {.remark} Results like these are at the heart of mathematics: having a bunch of equations, seeing patterns, and trying to find some common, unifying, and hopefully simpler structure that underlies all of it. An example you'll see in Calculus: all of the graphs we've been looking at in this class are "shadows" of intersecting shapes in some higher dimensional space! ![Conic Sections](figures/ConicSections.png) :::