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\newcommand{\cores}[0]{\operatorname{cores}} \newcommand{\res}[0]{\operatorname{res}} \newcommand{\Res}[0]{\operatorname{Res}} \newcommand{\infl}[0]{\operatorname{inf}} \newcommand{\coinfl}[0]{\operatorname{coinf}} \newcommand{\ind}[0]{\operatorname{ind}} \newcommand{\Ind}[0]{\operatorname{Ind}} \newcommand{\Coind}[0]{\operatorname{coInd}} \newcommand{\submfds}{\operatorname{SubMfds}} \newcommand{\quotright}[2]{ {}^{#1}\mkern-2mu/\mkern-2mu_{#2} } \newcommand{\quotleft}[2]{ {}_{#2}\mkern-.5mu\backslash\mkern-2mu^{#1} } \newcommand\B{{ \mathsf{B}}} \newcommand\D{{ \mathsf{D}}} \newcommand\Rd{{ \mathsf{R}}} \newcommand\Ld{{ \mathsf{R}}} # 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** \[ d: \ts{ \text{Pairs of points } (p, q) } \to \RR \\ (p, q) &\mapsto d(p, q) \da \sqrt{ (x_2 - x_1)^2 + (y_2 - y_q)^2} .\] ::: \todo[inline]{Law of cosines} :::{.definition title="Linear Functions"} A function $f:\RR \to \RR$ is **linear** if and only if $f$ has a formula of the following form: \[ f(x) = \alpha x + \beta && \alpha, \beta \in \RR .\] ::: :::{.definition title="Intercepts"} Given a function $f: \RR\to \RR$, an **$x\dash$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 $x\dash$axis. \ A **$y\dash$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 $y\dash$axis. ::: :::{.definition title="Relation"} A **relation** on two sets $X$ and $Y$ is a set of ordered pairs $(x, y) \in X \cross Y$, so $R$ can be described as a set: \[ R = \ts{ (x_0, y_0), (x_1, y_2), \cdots } .\] 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 $x_0\in \dom(R)$, 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 $\RR$. 2. Throw out "problematic" points. 3. Draw a number line and write out what you are left with in interval notation. ::: :::{.example title="?"} Define \[ f: \RR &\to \RR \\ x &\mapsto {1\over x} .\] Then $\dom(f) = \RR \smz = (-\infty, 0) \union (0, \infty)$ and $\range(f) = \RR$. ::: :::{.example title="?"} Define \[ f: \RR &\to \RR \\ x &\mapsto \sqrt{x} .\] Then $\dom(f) = \RR \sm (-\infty, 0) = [0, \infty)$ and $\range(f) = [0, \infty)$. ::: # 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: - $\sin\inv(\theta) \neq 1/\sin(\theta)$. Mnemonic: reciprocals of trigonometric functions already have a better name, here $\csc(\theta)$. ## Basic Trigonometric Functions \todo[inline]{Sin/cos/etc as ratios} ## Proportionality Relationships :::{.definition title="Radian"} \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 $k\in \ZZ$. Any such angle is said to be **coterminal** to $\theta$. ::: :::{.remark} In radians: \[ \theta_R \simeq \theta_R + k\cdot 2\pi && k\in \ZZ .\] In degrees: \[ \theta_D \simeq \theta_D + k\cdot 360^\circ && k\in \ZZ .\] ::: :::{.proposition title="Degrees are related to radians"} \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"} \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: \todo[inline]{soh-cah-toa, cho-sha-cao} ::: :::{.proposition title="Domains of trigonometric functions"} | Function | Domain | Range | |----------|--------|-------| |$\sin$ | $\RR$ | $[-1, 1]$ | |$\cos$ | $\RR$ | $[-1, 1]$ | |$\tan$ | $\RR \smts{\pm {\pi \over 2}, \pm{3\pi \over 2}, \cdots}$ | ? | |$\csc$ | $\RR \smts{0, \pm {\pi}, \pm{2\pi}, \cdots}$ | ? | |$\sec$ | $\RR \smts{\pm {\pi \over 2}, \pm{3\pi \over 2}, \cdots}$ | ? | |$\cot$ | $\RR \smts{0, \pm {\pi}, \pm{2\pi}, \cdots}$ | ? | ::: ## Polar Coordinates :::{.definition title="Unit Circle"} The **unit circle** is defined as \[ S^1 \da \ts{ \vector p = (x, y) \in \RR^2 \st d(\vector p, \vector 0) = 1 } = \ts{ (x, y) \in \RR^2 \st x^2 + y^2 = 1 } ,\] the set of all points in the plane that are distance exactly 1 from the origin. ::: :::{.theorem title="Polar Coordinates"} If a vector $\vector v$ has at an angle of $\theta$ in radians and has length $R$, the corresponding point $\vector p$ at the end of $\vector v$ is given by \[ \vector p = \tv{x, y} = \tv{R\cos(\theta), R\sin(\theta)} .\] Conversely, if $(x, y)$ are known, then the corresponding $R$ and $\theta$ are given by \[ [R, \theta] = \tv{ \sqrt{x^2 + y^2}, \arctan\qty{y\over x} } .\] ::: :::{.corollary title="Polar Coordinates on $S^1$"} If $R=1$, so $\vector v$ 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 $\tan(\theta) \da \sin(\theta) / \cos(\theta)$ 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"} \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** $\thetaref$ is the acute angle in the triangle formed by dropping a perpendicular to the nearest horizontal axis. ::: :::{.proposition title="?"} Reference angles for each quadrant: \[ \text{Quadrant II}: && \theta + \thetaref = \pi \\ \text{Quadrant III}: && \pi + \thetaref = \theta \\ \text{Quadrant IV}: && \theta + \thetaref = 2\pi .\] ::: :::{.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="?"} \envlist - $f(\theta) \da \cos(\theta)$ is an even function. - $g(\theta) \da \sin(\theta)$ is an odd function. ::: :::{.proof title="?"} Plot vectors for $\theta, -\theta$ on $S^1$ and flip over the $x\dash$axis. ::: :::{.corollary title="?"} \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 \[ f(t) \da A \cos(\omega (t - \varphi)) + \delta ,\] where - $A$ is the **amplitude**, - $\omega$ is the **frequency**, - $\phi$ is the **phase shift**, and - $\delta$ is the **vertical shift**. - $P \da 2\pi / \omega$ is the **period**, so $f(t+kP) = f(t)$ for all $k\in \ZZ$. \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: \[ f(t) \da 4\cos(3t+2) .\] ::: :::{.example title="?"} Put the following wave in standard form: \[ f(t) \da \alpha \cos(\beta t+\gamma) .\] ::: :::{.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 $2\abs{A}$ 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 $\abs{A}$. 2. 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 $y\dash$axis. 3. Measure the period $P$ and reverse-engineer it to get $\omega$: $P = 2\pi/\omega \implies \omega = 2\pi/P$. 4. Measure the distance from the starting point to the $y\dash$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: \[ F(\theta) \da \qty{ \sin(\theta) \cos(\theta) \over \cot(\theta)} \cos(\theta)\csc(\theta) .\] ::: :::{.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 $h(\theta) \da \csc(\theta) - \sin(\theta)$. ::: ## 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 $\sin: \RR_A \to \RR_B$, this would be some function $f: \RR_B \to \RR_A$ such that \[ f(\sin(\theta)) &= \id(\theta) = \theta \\ \sin(f(y)) &= \id(y) = y .\] ![Input-Output perspective: important!](figures/image_2021-04-18-22-24-55.png) Note that we only ever have to define $f$ on $\range(\sin)$, since we're only ever sending outputs of $f$ in as the inputs of $\sin$. So we need $\range(\sin) \subset \dom(f)$, noting that $\range(\sin) = [-1, 1]$: ![image_2021-04-18-22-26-56](figures/image_2021-04-18-22-26-56.png) Similarly, we need $\range(f) \subset \dom(\sin)$. ::: ### 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 $\sin\inv(3/5)$, 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 $[-\pi/2, \pi/2] = \range(\arcsin)$. - $\tan(\arccos(-10/26)) = 10/26$ - Write $\theta = \arccos(-10/26)$, note $\theta$ is in $[0, \pi] = \range(\arccos)$ ::: ### 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 - $\dom(\arccos) \da \range({ \color{green} \cos} ) = [-1, 1]$. - $\range(\arccos) \da \dom( {\color{green} \cos} ) = [0, \pi]$. ::: :::{.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 - $\dom(\arcsin) \da \range({ \color{green} \sin} ) = [-1, 1]$. - $\range(\arcsin) \da \dom( {\color{green} \sin }) = [-\pi/2, \pi/2]$. ::: :::{.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 | |----------|--------|-------| |$\arcsin$ | $[-1, 1]$ | $[-\pi/2, \pi /2]$ | |$\arccos$ | $[-1, 1]$ | $[0, \pi]$ | |$\arctan$ | $\RR$ | $(-\pi/2, \pi/2)$ | |$\arccsc$ | $\RR \smts{0, \pm {\pi}, \pm{2\pi}, \cdots}$ | $[-\pi/2, \pi/2]\smts{0}$ | |$\arcsec$ | $\RR \smts{\pm {\pi \over 2}, \pm{3\pi \over 2}, \cdots}$ | $[0, \pi]\smts{\pi/2}$ | |$\arccot$ | $\RR$ | $(0, \pi)$ | ::: :::{.slogan} There is an easy way to remember this: - Cosines are $x\dash$values, pick the upper (or lower) half of the circle to make them unique. - Sines are $y\dash$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 \[ f\circ g = \id \quad\iff\quad f(g(x)) = x,\quad g(f(x)) = x .\] 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 $\tan$) - $\tan(\arctan(x)) = x \iff x\in \RR$ - 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 $\vector v\in \RR^2$ breaks up as the sum of two vectors, i.e. $\vector v = \vector v_x + \vector v_y$. In coordinates, if $\vector v = (a, b)$, we have $\vector v_x = (a, 0)$ and $\vector v_y = (0, b)$. Alternatively, we can drop the ordered pair notation and write $\vector v = a \hat{\vector x} + b \hat{\vector y}$. ::: :::{.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 $\vhat{y}$ direction with the $i$ direction -- this amounts to replacing ordered pairs $(a, b) \da a \vhat{x} + b\vhat{y}$ 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 $\RR$. 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 $\RR$, but *will* have a solution in $\CC$. ::: :::{.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 $\vector v[r\cos(\theta), r\sin(\theta)]$, note that Euler's formula gives us a way to get a complex number $z$ that corresponds to it: \[ z \da re^{i\theta} = r(\cos(\theta) + i\sin(\theta)) = r\cos(\theta) + i\cdot r\sin(\theta) = [r\cos(\theta), r\sin(\theta)] = \vector v .\] ::: :::{.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) :::