Lesson 3 (08/28/26)
Today:
- Measuring Angle
- Review of Trig. Functions
- Inverse Trig. Functions
- Examples
Office Hours: MWF 1:30pm – 2:30pm, MATH 842
Announcements:
- MRR opens on Monday (MATH G175)
- Quiz 1 on Tuesday (Lesson 1, 2)
Office Hours: MWF 1:30pm – 2:30pm, MATH 842
\[ \sin\theta = \frac{\text{opp.}}{\text{hyp.}}, \qquad \operatorname{cosec}\theta = \frac{1}{\sin\theta} \]
\[ \cos\theta = \frac{\text{adj.}}{\text{hyp.}}, \qquad \sec\theta = \frac{1}{\cos\theta} \]
\[ \tan\theta = \frac{\text{opp.}}{\text{adj.}} = \frac{\sin\theta}{\cos\theta}, \qquad \cot\theta = \frac{1}{\tan\theta} \]
Reciprocal
these are all periodic function
there is a \(p\) such that \(f(x+p) = f(x)\)
\(\cos\theta = x\) is Horizontal distance
\(\sin\theta = y\) is Vertical distance
\(\theta\) is +ve if measured \(\text{CCW}\) w.r.t +ve x-axis
-ve if measured clock wise
\[ 1\text{ full circle} = 360\text{ Degrees} = 2\pi\text{ Radian} \]
Unless specified, angles in calculus always measured in Radians.
\[ 1\text{ Radian} = \text{Angle measure when radius equals Arc length.} \]
Conversion:
\[ 1^\circ = \frac{2\pi}{360}\text{ Radian} \] \[ 1\text{ Radian} = \frac{360^\circ}{2\pi} \]
\[ f(\theta) = a \sin(b(\theta - c)) + d \]
Not invertible on \((-\infty, \infty)\)
Restrict the Domain to \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \leadsto\) it is invertible
\[ \arcsin(x) = \sin^{-1}(x) = \theta \leadsto \text{then } \theta \text{ lies in } \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \text{ such that } \sin\theta = x \]
\(D: [-1, 1]\)
\(R: \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)
Find \(\theta\) in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) such that \(\sin\theta = \frac{1}{2}\)
\(\left[-\frac{\pi}{2}, 0\right]\) is IV Q
\(\left[0, \frac{\pi}{2}\right]\) is I Q
want \(\theta\) in \(\left[0, \frac{\pi}{2}\right]\) such that \(\sin\theta = \frac{1}{2}\)
\[\theta = \pi/6\]
NOT \(5\pi/6\), bcz its not in \(\left[0, \frac{\pi}{2}\right] = \text{Q}_1\)
Not invertible on \((-\infty, \infty)\)
Restrict Domain to \([0, \pi]\) we can define \(\arccos(x) = \cos^{-1}(x)\)
\[ \arccos(x) = \cos^{-1}(x) = \theta \iff \text{then } \cos\theta = x, \quad \theta \underset{\text{is in}}{\in} [0, \pi] \]
\(\text{D: } [-1, 1]\)
\(\text{R: } [0, \pi]\)
eg: Evaluate \(\cos^{-1}\left(\cos\left(\frac{4\pi}{3}\right)\right)\)
\[\neq \frac{4\pi}{3}\]
Find \(\theta\) in \([0, \pi]\) such that
\[\cos^{-1}\left(\cos\left(\frac{4\pi}{3}\right)\right) = \theta\] \[\cos\theta = \cos\left(\frac{4\pi}{3}\right)\]
\[\cos\left(\frac{2\pi}{3}\right) = \cos\left(\frac{4\pi}{3}\right)\] \[\cos^{-1}\left(\cos\left(\frac{4\pi}{3}\right)\right) = \frac{2\pi}{3}\]
\[ \cos \theta = x \]
\[ \cos(-\theta) = x \]
\[ \cos(\pi - \theta) = \cos(\pi + \theta) \]
Restrict \(D: \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), \(R: (-\infty, \infty) \leadsto \tan x\) is invertible
\[ \arctan(x) = \tan^{-1}(x) = \theta \leadsto \tan\theta = x, \quad \theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \]
\(D: (-\infty, \infty)\)
\(R: \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)