Tutorial Week 3

This week’s tutorial will be based on the content from the second week of lectures.

Be sure to be familiar with the material from Week 1 (https://q.utoronto.ca/courses/267509/pages/week-1-sept-5-9?module_item_id=3863942).

Inverse Trig

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 \(0 \le x \lt \pi\)

and

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

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

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

Then use cancellation laws (property 2).

Q3: Rewrite \(tan(cos^{-1}(3x))\) without trigonometric functions.

Relate to SOHCAHTOA, so \(cos^{-1}(3x)\) means the triangle with \(\dfrac{A}{H} = 3x\).

One such triangle would have side length 3x and hypotenuse length 1,

Use pythagorean theorem to find the length of the remaining side, then use SOHCAHTOA to find the tangent of that angle.

Inverse Trig Functions

Q4: Find the domain of the function \(f(x) = cos^{-1}(x^3 + 2)\).

\(cos^{-1}(x)\) has domain \([-1, 1]\)

so let

\(-1 \le x^3 + 2 \le 1\)

\(-3 \le x^3 \le -1\)

\(\sqrt[3]{-3} \le x \le -1\)

which is the domain.

Q5: 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 restriction of \([0, \pi]\)

Q6: For what values of \(x\) satisfy \(tan^2(x) = tan(x)\) if \(-\dfrac{\pi}{2} \lt x \lt \dfrac{\pi}{2}\)?

\[\begin{split}\begin{aligned} tan^2(x) &= tan(x)\\ tan^2(x) - tan(x) &= 0 \\ tan(x)(1 - tan(x)) &= 0 \\ \end{aligned}\end{split}\]

Use inverse trig to find the values of tan(x).

Q7: What is the domain of \(tan^{-1}(e^x)\)?

Domain of \(tan^{-1}(x)\) and \(e^x\) is \((-\infty, \infty)\), so domain is \((-\infty, \infty)\).

Inverse Trig Cancellation Rules

Q8: For what values of \(x\) is \(sin^{-1}(cos(x)) = x + \dfrac{\pi}{2}\)?

Recall the property \(sin^{-1}(sin(x)) = x\) if \(x \in [-\dfrac{\pi}{2}, \dfrac{\pi}{2}]\).

Also note that \(cos(x) = sin(x + \dfrac{\pi}{2})\).

Substitute in the above and use cancellation laws.

Graphs of Limits

Q9: Find the following values of \(f(x)\):

[Placeholder for graph to be drawn during tutorial]

Q10: Draw \(f(x)\) given the following requirements:

  1. \(\lim_{x \to 2} f(x) = 3\)

  2. \(\lim_{x \to 3^+} f(x) = 4\)

  3. \(\lim_{x \to 3^-} f(x) = 6\)

  4. \(\lim_{x \to 0} f(x) = 0\)

  5. \(0\) is not in the domain of \(f(x)\)

  6. \(\lim_{x \to \infty} f(x) = -\infty\)

  7. \(\lim_{x \to -\infty} f(x) = 0\)

Calculating Limits

Q11: Find \(\lim_{x \to 2} \dfrac{x^2 - x - 2}{x - 2}\).

Simplify to remove the \(x - 2\) term from both the numerator and denominator, then directly subtitute.

Q12: Find the limit if it exists: \(\lim_{x \to 3} \dfrac{\sqrt{x + 13} - 4}{x - 3}\).

Multiply by conjugate, simplify, then directly subtitute.

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