Note: These are rough notes for the Math 1113 Precalculus course at the University of Georgia
Precalculus
University of Georgia, Spring 2021
D. Zack Garza
University of Georgiadzackgarza@gmail.com
Last updated: 2021-04-27
Note: These are rough notes for the Math 1113 Precalculus course at the University of Georgia
Last updated: 2021-04-27
If \(a,b\) are the legs of a right triangle with hypotenuse \(c\), there is a relation \begin{align*} a^2 + b^2 = c^2 .\end{align*}
If \(p = (x_1, y_1)\) and \(q = (x_2, y_2)\) are points in the Cartesian plane, then there is a distance function \begin{align*} d: \left\{{ \text{Pairs of points } (p, q) }\right\} \to {\mathbb{R}}\\ (p, q) &\mapsto d(p, q) \mathrel{\vcenter{:}}=\sqrt{ (x_2 - x_1)^2 + (y_2 - y_q)^2} .\end{align*}
A function \(f:{\mathbb{R}}\to {\mathbb{R}}\) is linear if and only if \(f\) has a formula of the following form: \begin{align*} f(x) = \alpha x + \beta && \alpha, \beta \in {\mathbb{R}} .\end{align*}
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.
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: \begin{align*} R = \left\{{ (x_0, y_0), (x_1, y_2), \cdots }\right\} .\end{align*}
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.
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 \((x_0, y_0) \in R\).
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.
For domains and ranges:
To determine a domain:
Define \begin{align*} f: {\mathbb{R}}&\to {\mathbb{R}}\\ x &\mapsto {1\over x} .\end{align*}
Then \(\operatorname{dom}(f) = {\mathbb{R}}\setminus\left\{{0}\right\}= (-\infty, 0) \cup(0, \infty)\) and \(\mathop{\mathrm{range}}(f) = {\mathbb{R}}\).
Define \begin{align*} f: {\mathbb{R}}&\to {\mathbb{R}}\\ x &\mapsto \sqrt{x} .\end{align*}
Then \(\operatorname{dom}(f) = {\mathbb{R}}\setminus(-\infty, 0) = [0, \infty)\) and \(\mathop{\mathrm{range}}(f) = [0, \infty)\).
In this section, always draw a picture! Virtually 100% of the time.
Use exact ratios wherever possible.
There are too many details and formulas to just memorize in this unit: focus on the processes.
Some facts to remember:
In geometric terms, an angle in radians in the ratio of the arc length \(s(\theta, R)\) to the radius \(R\), so \begin{align*} \theta_R = {s(\theta, R) \over R} .\end{align*}
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 \(\theta\).
In radians: \begin{align*} \theta_R \simeq \theta_R + k\cdot 2\pi && k\in {\mathbb{Z}} .\end{align*}
In degrees: \begin{align*} \theta_D \simeq \theta_D + k\cdot 360^\circ && k\in {\mathbb{Z}} .\end{align*}
\begin{align*} {\theta \over 1\,\text{rev}} = {\theta_R \over 2\pi\, \text{rad} } = {\theta_D \over 360^\circ} .\end{align*}
\begin{align*} {\theta \over 1\,\text{rev}} = {s(R, \theta) \over 2\pi R} = { A(R, \theta) \over \pi R^2 } .\end{align*}
This implies that \begin{align*} A(R, \theta) &={R^2 \theta \over 2} \\ s(R, \theta) &= R\theta .\end{align*}
There are 6 trigonometric functions defined by the following ratios:
| Function | Domain | Range |
|---|---|---|
| \(\sin\) | \({\mathbb{R}}\) | \([-1, 1]\) |
| \(\cos\) | \({\mathbb{R}}\) | \([-1, 1]\) |
| \(\tan\) | \({\mathbb{R}}\setminus\left\{{\pm {\pi \over 2}, \pm{3\pi \over 2}, \cdots}\right\}\) | ? |
| \(\csc\) | \({\mathbb{R}}\setminus\left\{{0, \pm {\pi}, \pm{2\pi}, \cdots}\right\}\) | ? |
| \(\sec\) | \({\mathbb{R}}\setminus\left\{{\pm {\pi \over 2}, \pm{3\pi \over 2}, \cdots}\right\}\) | ? |
| \(\cot\) | \({\mathbb{R}}\setminus\left\{{0, \pm {\pi}, \pm{2\pi}, \cdots}\right\}\) | ? |
The unit circle is defined as \begin{align*} S^1 \mathrel{\vcenter{:}}=\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\} ,\end{align*} the set of all points in the plane that are distance exactly 1 from the origin.
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 \begin{align*} \mathbf{p} = {\left[ {x, y} \right]} = {\left[ {R\cos(\theta), R\sin(\theta)} \right]} .\end{align*}
Conversely, if \((x, y)\) are known, then the corresponding \(R\) and \(\theta\) are given by \begin{align*} [R, \theta] = {\left[ { \sqrt{x^2 + y^2}, \arctan\qty{y\over x} } \right]} .\end{align*}
If \(R=1\), so \(\mathbf{v}\) is on the unit circle \(S^1\), then \begin{align*} [x, y] = [\cos(\theta), \sin(\theta)] .\end{align*}
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.
Cosines are like \(x\) coordinates, sines are like \(y\) coordinates.
Given \(\theta_R = 4\pi/3\), what is the corresponding point on the unit circle \(S^1\)?
Note that \(\sin(\theta), \cos(\theta)\) work for any \(\theta\) at all. However, \(\cos(\theta) = 0\) sometimes, so \(\tan(\theta) \mathrel{\vcenter{:}}=\sin(\theta) / \cos(\theta)\) will on occasion be problematic. Similar story for the other functions.
For reference: the unit circle.
Idea: we want to partition the circle simultaneously
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.
Reference angles for each quadrant: \begin{align*} \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 .\end{align*}
Given \(\sin(\theta) = 7/25\), what are the five remaining trigonometric functions of \(\theta\)?
Method:
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: \begin{align*} \sin(\theta) &= 7/25 \\ \implies \arcsin( \sin(\theta) ) = \arcsin(7/25) \\ \implies \theta = \arcsin(7/25) \\ .\end{align*}
\begin{align*} (\sin(\theta))^2 + (\cos(\theta))^2 &= 1 \\ 1 + (\cot(\theta))^2 &= (\csc(\theta))^2 \\ (\tan(\theta))^2 + 1 &= (\sec(\theta))^2 .\end{align*}
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 \(\qty{\cos(\theta)}^2\).
Thinking of \(\cos(\theta)\) as a function of \(\theta\), is it
Why do we care? The Fundamental Theorem of Calculus.
Plot vectors for \(\theta, -\theta\) on \(S^1\) and flip over the \(x{\hbox{-}}\)axis.
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?
The standard form of a wave function is given by \begin{align*} f(t) \mathrel{\vcenter{:}}= A \cos(\omega (t - \varphi)) + \delta ,\end{align*}
where
Note that this is nothing more than a usual cosine wave, just translated/dilated in the \(x\) direction and the \(y\) direction.
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: \begin{align*} f(t) &= \cos(Bt+C) \\ &= \cos( B (t + {C\over B}) ) \\ &= \cos(\omega (t - \phi) ) \\ \\ &\implies B = \omega, \phi = -{C\over B} .\end{align*}
Put the following wave in standard form: \begin{align*} f(t) \mathrel{\vcenter{:}}= 4\cos(3t+2) .\end{align*}
Put the following wave in standard form: \begin{align*} f(t) \mathrel{\vcenter{:}}=\alpha \cos(\beta t+\gamma) .\end{align*}
How to plot the graph of a wave equation:
Plot the following function in the \(t\) plane: \begin{align*} f(t) = 2 \cos\qty{5t - {\pi \over 2} } + 7 .\end{align*}
Plot the following: \begin{align*} f(t) = -2\sin(3t-7) .\end{align*}
Given a picture of a graph of a sine wave,
Determine the equation of the following wave function:
\begin{align*} f(t) = 2\sin\qty{4t + {\pi \over 6} } .\end{align*}
Note that we can graph other trigonometric functions: they get pretty wild though.
The goal: reduce a complicated mess of trigonometric functions to something as simple as possible. We’ll use a boxing-up method.
On verifying identities: if you want to show \(f(\theta) = g(\theta)\), start at one and arrive at the other: \begin{align*} f(\theta) &= \text{simplify } f \\ &= \cdots \\ &= \cdots \\ &= \cdots \\ &= g(\theta) \\ .\end{align*}
If you end up with something like \(1=1\) or \(0=0\), this is hinting at a problem with your logic.
Simplify the following: \begin{align*} F(\theta) \mathrel{\vcenter{:}}=\qty{ \sin(\theta) \cos(\theta) \over \cot(\theta)} \cos(\theta)\csc(\theta) .\end{align*}
\begin{align*} F = s \qty{s \over c} .\end{align*}
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.
Show the following identity: \begin{align*} {\sin(-\theta) + \csc(\theta)} = \cot(\theta) \cos(\theta) \end{align*} by showing both sides are separately equal to \(h(\theta) \mathrel{\vcenter{:}}=\csc(\theta) - \sin(\theta)\).
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 \begin{align*} f(\sin(\theta)) &= \operatorname{id}(\theta) = \theta \\ \sin(f(y)) &= \operatorname{id}(y) = y .\end{align*}
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 \(\mathop{\mathrm{range}}(\sin) = [-1, 1]\): 
Similarly, we need \(\mathop{\mathrm{range}}(f) \subset \operatorname{dom}(\sin)\).
Optimistically imagine that we had some such inverse function. Then we could evaluate some expressions without even knowing anything else about it. The trick: \begin{align*} \theta &= \arccos(p/q) \\ \implies \cos(\theta) &= \cos(\arccos(p/q)) \\ \implies \cos(\theta) &= p/q .\end{align*} Now embed this in a triangle. We can’t solve for \(\theta\), but we can solve for other trigonometric functions.
\begin{align*} \cos\qty{ \arccos\qty{ \sqrt 5 \over 5 } } &= {\sqrt 5 \over 5} \\ \arccos\qty{ \cos \qty{ \sqrt 5 \over 5 } } &= {\sqrt 5 \over 5} .\end{align*}
\begin{align*} \tan\qty{ \arcsin\qty{ p \over q } } = {p \over \sqrt{q^2 - p^2} } .\end{align*}
Compute \(\arcsin(3/5)\).
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 \(\csc(3/5)\).
\begin{align*} \theta &= \arcsin(3/5) \\ \implies \sin(\theta) &= (3/5) && \text{roughly by injectivity} \\ \implies &= \cdots ? .\end{align*} 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:
So for example, \(\cos(\arcsin(3/5)) = 4/5\).
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.
Show the following:
The setup: try swapping \(y\) and \(\theta\) in the graph of \(y=\sin(\theta)\):
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:
Back on our original graph, this looks like the following:
Restricting, we get
A similar analysis works for \(\sin(\theta)\):
Restricting, we get
This gives us a new tool to solve equations: \begin{align*} \vdots &= \vdots \\ \implies \cos(x) &= b \\ \implies \arccos(\cos(x)) &= \arccos(b) \\ \implies x &= \arccos(b) ,\end{align*} but only if we know this makes sense based on domain/range issues.
Restrict domains in the following ways:
| Function | Domain | Range |
|---|---|---|
| \(\arcsin\) | \([-1, 1]\) | \([-\pi/2, \pi /2]\) |
| \(\arccos\) | \([-1, 1]\) | \([0, \pi]\) |
| \(\arctan\) | \({\mathbb{R}}\) | \((-\pi/2, \pi/2)\) |
| \(\operatorname{arccsc}\) | \({\mathbb{R}}\setminus\left\{{0, \pm {\pi}, \pm{2\pi}, \cdots}\right\}\) | \([-\pi/2, \pi/2]\setminus\left\{{0}\right\}\) |
| \(\operatorname{arcsec}\) | \({\mathbb{R}}\setminus\left\{{\pm {\pi \over 2}, \pm{3\pi \over 2}, \cdots}\right\}\) | \([0, \pi]\setminus\left\{{\pi/2}\right\}\) |
| \(\operatorname{arccot}\) | \({\mathbb{R}}\) | \((0, \pi)\) |
There is an easy way to remember this:
We have some exact values.
Sines should be in QI or QIV:
Cosines should be in QI or QII:
Tangents should be in QI or QIV:
Note that if \(f, g\) are an inverse pair, we have \begin{align*} f\circ g = \operatorname{id}\quad\iff\quad f(g(x)) = x,\quad g(f(x)) = x .\end{align*} However, we have to be careful with domains for trigonometric functions:
Sometimes we are interested in superposition of waves, see Desmos 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.
We can sometimes rewrite these as a single wave with a phase shift.
Identities: \begin{align*} \sin(\theta + \psi) &= \sin(\theta) \cos(\psi) + \cos(\theta) \sin(\psi) \\ \cos(\theta + \psi) &= \cos(\theta) \cos(\psi) + \sin(\theta) \sin(\psi) .\end{align*} Note that you can divide these to get \begin{align*} \tan(\theta + \psi) &= {\tan(\theta) + \tan(\psi) \over 1 - \tan(\theta) \tan(\psi) } ,\end{align*} and replace \(\psi\) with \(-\psi\) and use even/odd properties to get formulas for \(\sin(\theta - \psi), \cos(\theta - \psi)\)
Sines are friendly and cosines are clique-y!
Taking \(\theta = \psi\) is the above identities yields \begin{align*} \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) .\end{align*}
The latter is not equal to 1! That would be \(\cos^2(\theta) + \sin^2(\theta)\).
Why do we care? We had 16 special angles, this gives a lot more. For example, \begin{align*} \cos(\pi/12) = \cos(\pi/3 - \pi/4) = \cdots \text{ plug in} .\end{align*} By allowing increments of \(\pi/12\), we have 24 total angles.
Starting from the following: \begin{align*} \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 ,\end{align*} one can solve for \begin{align*} \cos^2(\theta) = {1\over 2}\qty{1 + \cos(2\theta) } .\end{align*}
Similarly \begin{align*} \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 ,\end{align*} solving yields \begin{align*} \sin^2(\theta) = {1\over 2} (1 - \cos(2\theta) ) .\end{align*}
These are very important in Calculus! This gives us a way to reduce the exponents on expressions like \(\sin^n(\theta)\).
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.
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 \(\mathbf{v} = a \widehat{\mathbf{x}} + b \widehat{\mathbf{y}}\).
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 \((a, b) \mathrel{\vcenter{:}}= a \widehat{ \mathbf{x} } + b\widehat{ \mathbf{y} }\) by a single number \(a + ib\).
Complex numbers can be added: \begin{align*} (a + bi) + (c + di) = (a + c) + (b + d)i .\end{align*} This is perhaps easier to understand in the ordered pair notation: you just add the components in each component: \begin{align*} [a, b] + [c, d] = [a + c, b + d] .\end{align*}
Complex numbers can be multiplied: \begin{align*} (a +bi)(c+di) &= a(c+di) + bi(c+di) \\ &= ac + adi + bci + bdi^2 \\ &= (ac - bd) + (ad + bc)i .\end{align*} This is harder to see in the ordered pair notation.
We can compare complex numbers: they are equal iff their components are equal: \begin{align*} a + bi = c+di \iff a=c \text{ and } b = d ,\end{align*} or in ordered pair notation, \begin{align*} [a, b] = [c, d] \iff a = c \text{ and } b = d .\end{align*}
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 \({\mathbb{C}}\).
The following is the main link between exponentials and waves:
\begin{align*} e^{i\theta} = \cos(\theta) + i\sin(\theta) .\end{align*}
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.
\begin{align*} e^{i\pi} = -1 .\end{align*}
This is remarkable! It relates some of the most fundamental constant numbers in mathematics:
Proof: just plug \(\pi\) into Euler’s equation. Geometric interpretation: \(\pi\) radians is directly to the left.
An application: proving the angle sum formulas algebraically. We start by considering the angle \(\alpha + \beta\). On one hand, Euler’s formula says \begin{align*} e^{i( \alpha + \beta) } = \cos(\alpha + \beta) + i\sin(\alpha + \beta) = [\cos(\alpha + \beta), \sin(\alpha + \beta)] .\end{align*}
On the other hand, we can use properties of exponentials first and expand: \begin{align*} 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] .\end{align*}
Now we just equate components:
\begin{align*} [\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) .\end{align*}
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: \begin{align*} z \mathrel{\vcenter{:}}= 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} .\end{align*}
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!
