---
title: Review Ib
tags: teaching, review, calculus
---
# Review I (Trig., exponential, and logarithmic functions)
## Trig functions
### Meaning of sine and cosine functions
Setup: Say that $P$ is a point on the unit circle, so that going a distance $t$ along the circle, starting from $(1,0)$ and going counter-clockwise, puts you at $P$ (going to $t$ radians). Then $\cos(t)$ is defined to be the $x$-coordinate and $\sin(t)$ is the $y$-coordinate. See the image below.
When $t<0$ then you travel clockwise along the circle instead of counter-clockwise. Since reflecting about the $x$-axis won't change an $x$-coordinate, we have
> $$\cos(-t) = \cos(t);\quad\text{and }\sin(-t)=-\sin(t).$$
Also, since points on the circle satisfy $x^2+y^2=1$,
::: info
The Main Trig identity $$\cos^2(t) + \sin^2(t) = 1$$
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2. Other Trig Functions
> $$\tan(t) = \frac{\sin(t)}{\cos(t)};\qquad \cot(t) = \frac{\cos(t)}{\sin(t)};$$
> $$\sec(t) = \frac{1}{\cos(t)};\qquad \csc(t) = \frac{1}{\sin(t)}.$$
### Exercises: Other identities
Use the Main Trig identity to verify the following.
1. $\tan^2(t) = \sec^2(t)-1$
▷ Solution is [here](https://i.imgur.com/NokM6ad.mp4).
2. $\cot^2(t) + 1 = \csc^2(t)$
### Values of sine and cosine
*You should know the following values*
| $t$ | $\cos(t)$ | $\sin(t)$ | $t$ | $\cos(t)$ | $\sin(t)$|
| -------- | -------- | -------- | -------- | -------- | -------- |
| $0$ | $1$ | $0$ | $3\pi/2$ | $0$ | $-1$ |
| $2\pi$ | $1$ | $0$ | $\pi/4$ | $\frac{1}{\sqrt2}$ | $\frac{1}{\sqrt2}$ |
| $\pi/2$ | $0$ | $1$ | $\pi/6$ | $\frac{\sqrt3}{2}$ | $\frac{1}{2}$ |
| $\pi$ | $-1$ | $0$ | $\pi/3$ | $\frac{1}{2}$ | $\frac{\sqrt3}{2}$ |
*Reasoning for the values in the table* :
::: spoiler
Since $t=0$ means that the point $P$ is $(1,0)$ get $\cos(0)=1$ and $\sin(0)=0$ immediately from the definition. Also, the unit circle has circumference $2\pi$, making a full revolution around the circle, so $\cos(2\pi)=1$ and $\sin(2\pi)=0$.
Since $\pi/2$ is one-fourth of the way around the circle, this puts you at the point $(0,1)$, and so $\cos(\pi/2)=0$ and $\sin(\pi/2)=1$. Similar thinking gives the values at $t=\pi$ and $t=3\pi/2$.
Since $t=\pi/4$ radians puts $P$ on the line $y=x$, you must have $\cos(\pi/4) = \sin(\pi/4)$. Then, $$1 = \cos^2(\pi/4)+\sin^2(\pi/4)=2\cos^2(\pi/4).$$ Solve for $\cos(\pi/4)$ (using that $x$-coordinates are positive in the 1^st^ quadrant), and you get it.
For $t=\pi/3$, an equilateral triangle, with side lengths equal to 1, has each angle equal to 60^o^ ($\pi/3$ radians), and if you bisect one angle (getting a perpendicular to the opposite side) then you cut the opposite side in half. So, the hypotenuse of each of the remaining 30-60-90 right triangles is twice as long as the side opposite the 30^o^ angle ($\pi/6$ radians). Placing such a triangle (a length 1 hypotenuse) with the $\pi/3$ radians angle at the origin, the other side lengths are $\cos(\pi/3)$ and $\sin(\pi/3)$ -- with $\cos(\pi/3)$ being the one with length $1/2$, as the hypotenuse is twice its length. Using the Main Trig identity, you get that $\frac{1}{4}+\sin^2(\pi/3)=1$. Since this is in the 1^st^ quadrant, $\sin(\pi/3)=\sqrt3/2$.
By reflecting about $y=x$ you get the values at $t=\pi/6$.
:::
### Exercises
3. Find all solutions to $2\tan(t)+\sec(t)=0$ in the interval $[0,2\pi]$.
:::spoiler
There are two solutions, $t=7\pi/6$ and $t=11\pi/6$.
:::
4. Find all solutions to $\cos(2t)=\cos^2(t)$ in the interval $[0,2\pi]$.
5. Verify the identity $$\tan(t)+\cot(t) = \sec(t)\csc(t).$$
::: spoiler
[Video solution](https://i.imgur.com/UVP7hIx.mp4).
:::
The definitions of $\sin(t), \cos(t),$ and $\tan(t)$ above agree with definitions based on the ratios of side lengths in a right triangle, when $0<t<\pi/2$ (by dropping a vertical line from $P$ to the $x$-axis; then that $x$-axis point, $P$, and the origin make a right triangle that has hypotenuse length 1).
>Since you get a hypotenuse with length 1, the ratio (Opposite/Hypotenuse) equals the $y$-coordinate of $P$; the ratio (Adjacent/Hypotenuse) equals the $x$-coordinate of $P$, and (Opposite/Adjacent) is then the same as $\sin(t)/\cos(t) = \tan(t)$.
### Exercises
6. Given a number $x$ and an angle $t$, suppose that $\sin(t)=x$ (or $\frac{x}{1}$, if you want). Assuming $\cos(t)\ge 0$, find an expression in $x$ that is equal to $\cos(t)$. (Note: you could do this without thinking about triangles, but trig identities instead.)
> [Video](https://youtu.be/bSM7RNSbWhM) on trig functions from triangles.
### Sum (and double) angle formulas
For two angles $a$ and $b$,
> $$\sin(a+b) = \sin(a)\cos(b) + \sin(b)\cos(a);$$
> $$\cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b).$$
Use what has been discussed already to verify
- $\sin(a+\pi/2) = \cos(a)$
- $\sin(2a) = 2\sin(a)\cos(a)$
- $\cos(2a) = 2\cos^2(a)-1 = 1-2\sin^2(a)$ (Called the double-angle formula)
### More exercises for practice
There are examples in Section 1.3 of the [textbook](https://openstax.org/books/calculus-volume-1/pages/1-introduction). Consider trying:
| Section 1.3 | Task | Exercises |
| -------- | -------- | -------- |
| | Simplify the expression | 144, 145 |
| | Verify the identity | 147,149,150,153 |
| | Solve in the interval | 155,156 |
| | "Word" problem | 181(b-c) |
For more examples of working out identities with trig functions, check out the following two videos.
> [Video 1](https://youtu.be/UIOu4BBwIps) (Two examples in the second half of video.)
> [Video 2]()
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## Exponential functions and logarithms
### A variable as an exponent
A function $f(x)$ is an exponential function if has the form $f(x)=b^x$ for some number $b>0$ (called the *base*). For example, if $b$ is the number $e \approx 2.718\ldots$ then you have $f(x) =e^x$, one example of an exponential function.
The values of such a function are easy to make sense of when $x$ is an [integer](https://en.wikipedia.org/wiki/Integer) or [rational number](https://en.wikipedia.org/wiki/Rational_number) ...
::: spoiler
Note that positive integer exponents are simply repeated multiplication of the base, and for negative integers it means that you find 1 divided by the base raised to the absolute value of the power. If the exponent is rational ($p/q$ for integers $p,q$, where $q$ is not zero) then you take the $q^{th}$ root, and then raise that to the $p^{th}$ power.
:::
However, what is meant by raising a number to an irrational number? What do you *do* in order to raise $e$ to the $\pi$ power? The answer requires **limits**. Sorting this out is necessary in order to have the domain of a function like $e^x$ be all real numbers[^first].
[^first]:Sorting out what such powers means is a necessity to make sure all the mathematics works out as we think it should. However, in Calculus I, we do not get into how to sort it out.
### Inverses of exponential functions: logarithms
The *inverse function* of $e^x$ is $\ln(x)$. By definition, this means
:::info
- For any number $x$, $\qquad\ln(e^x) = x$,
and
- For any positive number $x$, $\qquad e^{\ln(x)}=x$.
<span style="color:lightslategray;"> having $x$ be *positive* in bullet 2 is needed since the range of $e^x$ is $(0,\infty)$ </span>
:::
For a different base $b$, the inverse of the function defined by $b^x$ is the *base-$b$* logarithm, $\log_b(x)$. Note, $\log_e(x)$ is the same as $\ln(x)$, and this is often simply written as $\log(x)$.
> When simply $\log(x)$ is written, the intended base $b$ differs depending on the field of study. In **most** of the sciences and math & stats, *professionals and academics* use $\log(x)$ for base $e$ (so the same as $\ln(x)$).
> In high school, on calculators, and in a few engineering and biology settings, $\log(x)$ means base 10. In computer science, $\log(x)$ can often mean base 2.
> See the [table here](https://en.wikipedia.org/wiki/Logarithm#Particular_bases).
The function $\ln(x)$ has domain $(0,\infty)$, since this is the range of $e^x$. The same is true for other bases. Additionally, using that the natural log and base $e$ exponential are inverses, for $b>0$ and for all $x$ we have $$b^x = (e^{\ln(b)})^x = e^{\ln(b)x}.$$ Say that, for a particular $x$ and some $y$, we have $x = \log_b(y)$. Then $y = b^x = e^{\ln(b)\log_b(y)},$ and so
:::info
$$\frac{\ln(y)}{\ln(b)} = \log_b(y).$$
:::
### Graph of $y=e^x$ and $y=\ln(x)$
Any time you have a pair of functions that are inverses of each other, you can get the graph of one of them from the graph of the other by *reflecting* about $y=x$. That means: take every point $(x_0,y_0)$ on the one graph and switch it for the point $(y_0,x_0)$; then doing that results in the graph of the inverse.
To check this is true (for $e^x$ and $\ln(x)$), remember that on a graph of a function $f$, a point with $x$-coordinate $a$ has $y$-coordinate $f(a)$. So, in the case of $e^x$ and $\ln(x)$, if $(x_0,y_0)$ is on the graph of $e^x$, then $y_0=e^{x_0}$. But that means $x_0 = \ln(y_0)$, which says $(y_0,x_0)$ is on the graph of $\ln(x)$. This all works in the opposite direction too, so the graph of $\ln(x)$ is what you get by reflecting the graph of $e^x$ about $y=x$.
> I didn't use anything special about $e^x$ and $\ln(x)$ *except* the fact that they are inverse functions of each other. So, the same holds for any pair of inverse functions.
### Properties of logarithms
For all positive numbers $p$ and $q$, and any real number $r$, the following properties hold for a logarithm. (I will write it with the natural log $\ln$, but the same properties work for *any* base $b>0$.
::: info
- $\ln(pq) = \ln(p) + \ln(q)$;
- $\ln(p/q) = \ln(p) - \ln(q)$;
- $\ln(p^r) = r\ln(p)$.
:::
*Reasoning*
> Every positive number equals $e$ raised to something (the domain of $\ln(x)$ is $(0,\infty)$). Say that $p = e^s$ and $q = e^t$. Then:
> $$pq = e^se^t = e^{s+t}.$$
> So, $\ln(pq) = \ln(e^{s+t}) = s+t = \ln(p)+\ln(q)$.
> Also, $\ln(p^r) = \ln((e^s)^r) = \ln(e^{rs}) = rs$. Since $s = \ln(p)$, this says that $\ln(p^r) = r\ln(p)$.
> Use the fact that $p/q = (p)(q^{-1})$, and the 1^st^ and 3^rd^ properties with $r=-1$, to get the second property.
These properties of logarithms are useful when solving equations that involve exponentials and logarithms.
#### Examples
1. Find all solutions to $\ln(x^2) = 2$.
:::spoiler
Putting both sides into the inverse function $e^x$, this is the same as $x^2 = e^2$, which has solutions $x=e$ and $x=-e$.
:::
2. Solve $e^{x^2} = 3e^{2-x}$.
▷ Solution. Put both sides into $\ln(x)$. You get $x^2 = \ln(3e^{2-x})$. Using the first property above, the equation is the same as $x^2 = \ln(3)+\ln(e^{2-x})$, which is $$x^2 = \ln(3)+2-x.$$ Now, $\ln(3)$ is just a constant (approximately 1.1). So, we collect terms and solve with the quadratic formula: $$x = \frac{-1\pm\sqrt{1+4(2+\ln(3))}}{2}$$
::: danger
**Some craziness!** Of the two solutions to $\ln(x^2)=2$, only *one* of them ($x=e$) is a solution to $\ln(x)+\ln(x) = 2$, even though one of the properties of logs tells us $\ln(x^2) = \ln(x)+\ln(x)$. It is important to remember the assumption of those properties, that the number being put into $\ln$ is **positive**.
:::
#### Some more examples
> See this [Video](https://youtu.be/rBnQiLa2TYo).
### Solving equations through substitution
At times, an equation will arise that invovles some of the functions we've seen here, and in order to solve it helps to replace the trig, logarithmic, or exponential function by a new variable. Here are a few examples.
#### Examples
3. Find all solutions, for $t$ in $[-\pi,\pi]$, to $\cos(t)^3-\cos(t) = 0$.
▷ Solution. If we substitute $\cos(t) = x$, then the equation is $x^3-x=0$. The left side factors $$x(x^2-1) = x(x-1)(x+1),$$
and so solutions are when $\cos(t) = x$ is $0, 1,$ or $-1$. In the interval, that means $t = -\pi, -\pi/2, 0, \pi/2,$ or $\pi$.
4. Find all solutions, for $t$ in $[0,2\pi]$, to $\sin^2(t)-3\sin(t)+2=0$.
▷ Solution (mostly). Use the same idea as above and notice that $x^2-3x+2 = (x-1)(x-2)$. So, $\sin(t)=1$ or $\sin(t)=2$. The second equation has no solutions, so you just get $t=\pi/2$ from the first one.
5. Find all solutions to $e^x + 2e^{-x} = 3$.
▷ Solution. First, substitute $e^x = y$, so the equation is $y + 2y^{-1} = 3$. It's easier to deal with non-negative powers if possible, so multiply both sides by $y$, then move terms to one side to get $y^2+2-3y=0$ (this is just like finding common denominators; $y^2+2-3y$ is what you would have in the numerator in that case). Now, $$y^2-3y+2 = (y-2)(y-1) = 0.$$ And so, $e^x=2$ or $e^x=1$ (since $y=e^x$). This gives solution $x = \ln(2)$ or $x=0$.
### Exercises
1. Solve the equation $\ln(\sqrt{x-3})=-1$.
:::spoiler
Answer: $x=\frac{1}{e^2}+3$
:::
3. Determine (by hand) where the graph $y=\ln(x-3.15)$ crosses the $x$-axis.
4. Solve $7^{3x-2} = 11$.
5. Find all $x$ such that $e^{2x}+e^x = 6$. (There is only one possible $x$.)
#### More exercises
- Exercise 291 at the end of [Section 1.5](https://openstax.org/books/calculus-volume-1/pages/1-5-exponential-and-logarithmic-functions) of the textbook.
- [Exercises 322 through 325](https://openstax.org/books/calculus-volume-1/pages/1-chapter-review-exercises) in the Chapter 1 review.
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