Tutorial Week 3

In this week’s tutorial, we’ll be going over inverse trigonometric functions and limits.

Trigonometric and inverse trigonometric functions

Let’s start this class with a (not so) quick review of the trigonometric functions and their inverses. Try to name each of the following functions:


  1. cos(x)

  2. cot(x)

  3. sin(x)

  4. sec(x)

  5. tan(x)

  6. csc(x)

Now that we have reviewed the 6 trigonometric functions, we can start looking at their inverses. Most of these inverses can’t be defined on the entire domain of their original functions because those functions aren’t all one-to-one (think about what the inverse of a non one-to-one function would look like).

The functions are as follows:

\(sin^{-1}(x)\) with domain and range \(x \in [-1, 1], sin^{-1}(x) \in [-\frac{\pi}{2}, \frac{\pi}{2}]\)

\(cos^{-1}(x)\) with domain and range \(x \in [-1, 1], cos^{-1}(x) \in [0, \pi]\)

\(tan^{-1}(x)\) with domain and range \(x \in \mathbb{R}, tan^{-1}(x) \in [-\frac{\pi}{2}, \frac{\pi}{2}]\)

\(csc^{-1}(x)\) with domain and range \(x \in (-\infty, 1] \cup [1, \infty), csc^{-1}(x) \in [-\frac{\pi}{2}, \frac{\pi}{2}] \setminus \{ 0 \}\)

\(sec^{-1}(x)\) with domain and range \(x \in [-1, 1], sec^{-1}(x) \in [0, \pi] \setminus \{ \frac{\pi}{2} \}\)

\(cot^-1(x)\) with domain and range \(x \in \mathbb{R}, cot^{-1}(x) \in (0, \pi)\)

Inverse trigonometric practice

Now that we have all the functions down, let’s do a few practice questions with these functions.

Q1: What is the value of \(sin^{-1}(1)\)?

This means that you need to find the value of \(\theta\) that gives you \(sin(\theta) = 1\) and \(-\dfrac{\pi}{2} \le \theta \le \dfrac{\pi}{2}\).

\(\theta = \dfrac{\pi}{2}\) works in this case. (Be familiar with the values of the trig functions on a unit circle)

Q2: What is the value of \(cos^{-1}(cos(\dfrac{56\pi}{5}))\)?

This question requires knowing 3 properties:

  1. \(cos(x) = cos(x + 2\pi)\)

and

  1. \(cos^{-1}(cos(x)) = x \;\;\) if \(0 \le x \lt \pi\;\;\) (Cancellation rules)

and

  1. \(cos(x) = cos(-x)\)

First, use property 1 to get \(\dfrac{56\pi}{5}\) within \([-\pi, \pi]\), giving us:

\(cos(\frac{56\pi}{5} - \frac{10\pi}{5}) = cos(\frac{46\pi}{5})\)

\(cos(\frac{46\pi}{5} - 5 \cdot \frac{10\pi}{5}) = cos(-\frac{4\pi}{5})\)

If the value is negative, use property 3 to make it positive.

So \(cos(-\frac{4\pi}{5}) = cos(\frac{4\pi}{5})\)

Then use cancellation rules (property 2) to get

\(cos^{-1}(cos(\frac{4\pi}{5})) = \frac{4\pi}{5}\)

Q3: Find the inverse of the function \(f(x) = cos^{-1}(x^3 + 2)\).

\[\begin{split}\begin{aligned} x &= cos^{-1}(y^3 + 2 )\\ cos(x) &= cos(cos^{-1}(y^3 + 2)) \\ cos(x) &= (y^3 + 2) \\ \sqrt[3]{cos(x) - 2} &= y \\ \end{aligned}\end{split}\]

with a domain of \(x \in [0, \pi]\) since that’s the range of \(f(x)\).

Finite Limits

Sometimes, we want to be able to find the values of a function as we get close (and in most cases, infinitely close!) to some point that may be undefined on the function.

To formally define this notion of a limit, we would need to use the delta-epsilon definition of a limit, which is way out of scope of this course, so just use your intuition here (but remember to adequetly justify your steps!). If you’re really curious, the definition is \(\lim_{x \to a} f(x) = L \Leftrightarrow (\forall \epsilon > 0, \exists \delta > 0 \text{ st } 0 < |x - c| < \delta \Rightarrow |f(x) - L| < \epsilon)\)

One example of this is at x = 1 for the function \(f(x) = \frac{(x - 1)x}{x-1}\). Even though x - 1 cancels out, the function is still left undefined at x = 1, yet we can easily compute the limit \(\lim_{x \to 1} \frac{(x-1)x}{x-1} = \lim_{x \to 1} x = 1\).

Sometimes you may want to evaluate a limit from only one side of the point, so we have the notion of a left-hand limit \(\lim_{x \to a^-}\) and a right-hand limit \(\lim_{x \to a^+}\). One definition to keep in mind here is that \(\lim_{x \to a} f(x) = L \Leftrightarrow \lim_{x \to a^+} = L\) and \(\lim_{x \to a^-}f(x) = L\) (this is useful for showing that a limit does not exist).

Hopefully this isn’t the first time you’ve encountered limits, since this next question will be a bit tricky to do…

Q4: Find the limit if it exists: \(\lim_{x \to 3} \dfrac{\sqrt{x + 13} - 4}{x - 3}\).

Multiply by conjugate, simplify, then directly subtitute.

\[\begin{split}\begin{aligned} \lim_{x \to 3} \frac{\sqrt{x+13}-4}{x-3} &= \lim_{x \to 3} \frac{\left(\sqrt{x+13}-4\right)\left(\sqrt{x+13}+4\right)}{\left(x-3\right)\left(\sqrt{x+13}+4\right)}\\ &= \lim_{x \to 3} \frac{\left(\sqrt{x+13}\right)^{2}-16}{\left(x-3\right)\left(\sqrt{x+13}+4\right)} \\ &= \lim_{x \to 3} \frac{x+13-16}{\left(x-3\right)\left(\sqrt{x+13}+4\right)} \\ &= \lim_{x \to 3} \frac{x-3}{\left(x-3\right)\left(\sqrt{x+13}+4\right)} \\ &= \lim_{x \to 3} \frac{1}{\left(\sqrt{x+13}+4\right)} \\ &= \frac{1}{\left(\sqrt{3+13}+4\right)} \\ &= \frac{1}{8} \end{aligned}\end{split}\]

Q5: Let \(f(x) = \begin{cases} x < 1 & x^3 + xk \\ x > 1& \frac{1 + x^2k}{xk}\end{cases}\). For what values of \(k\) does \(\lim_{x \to 1} f(x)\) exist?

For the limit to exist, we need \(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x)\).

\(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x^3 + xk\)

and

\(\lim{x \to 1^+} f(x) = \lim_{x \to 1^+} \frac{1 + x^2k}{xk}\).

So we need to solve for k in \(\lim_{x \to 1^-} x^3 + xk = \lim_{x \to 1^+} \frac{1 + x^2k}{xk}\)

\[\begin{split}\begin{aligned} \lim_{x \to 1^-} x^3 + xk &= \lim_{x \to 1^+} \frac{1 + x^2k}{xk} \\ 1^3 + (1)k &= \frac{1 + (1^2)k}{(1)k} \\ k + k^2 &= 1 + k \\ k^2 &= 1 \\ k = \pm 1 \end{aligned}\end{split}\]