MATH 221

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 28 July 2014

Lecture 2

Angles

$$\begin{array}{c} y x 1 -1 1 -1 π is the distance half way around a circle of radius 1 \end{array}$$ Measure angles according to the distance traveled on a circle of radius 1. $$\begin{array}{c} y x 1 -1 1 -1 ←θ The angle θ is measured by travelling a distance θ on a circle of radius 1. \end{array}$$ Sketch both $x$ and $y$ to get a circle of radius $r\text{.}$ $$\begin{array}{c} y x r -r r -r 1 -1 ← distance r⁢θ ← distance θ The distance θ stretches to r⁢θ. \end{array}$$ The distance $2\pi$ around a circle of radius 1 stretches to $2\pi r$ around a circle of radius $r\text{.}$ So the circumference of a circle is $2\pi r$ if the circle is radius $r\text{.}$

To find the area of a circle first approximate with a polygon inscribed in the circle. The eight triangles form an octagon $P_8$ in the circle. The area of the octagon $P_8$ is almost the same as the area of the circle. Unwrap the octagon. $$\begin{array}{c} { { b B → h y x \end{array}$$ The area of the octagon is the area of the 8 triangles. The area of each triangle is $\frac{1}{2}bh\text{.}$ So the area of the octagon is $\frac{1}{2}Bh\text{.}$

Take the limit as the number of triangles in the interior polygon gets larger and larger (the polygon gets closer and closer to being the circle). Then $$\begin{array}{ccc}{\displaystyle \text{Area of the circle}}& =& {\displaystyle \underset{n\to \infty }{\lim }\left(\text{area of an}\hspace{0.17em}n\text{-sided polygon}\hspace{0.17em}{P}_n\right)}\\ & =& {\displaystyle \underset{n\to \infty }{\lim }\left(\frac{1}{2}Bh\right)}\\ & =& {\displaystyle \frac{1}{2}\left(2\pi r\right)\left(r\right)}\\ & =& {\displaystyle \pi r^2}\end{array}$$ Where $B$ is the total base, $h$ is the height of the triangle, $2\pi$ is the length of an unwrapped circle and $r$ is the radius of the circle.

So the area of a circle is $\pi r^2$ if the circle is radius $r\text{.}$

Trigonometric functions

$$\begin{array}{c} { } } θ cos⁡θ sin⁡θ ← \end{array}$$ $\sin \theta$ is the $y$-coordinate of a point at distance $\theta$ on a circle of radius 1. $\cos \theta$ is the $x$-coordinate of a point at distance $\theta$ on a circle of radius 1. $$\text{tan}\hspace{0.17em}\theta =\frac{\text{sin}\hspace{0.17em}\theta }{\text{cos}\hspace{0.17em}\theta },\text{cot}\hspace{0.17em}\theta =\frac{\text{cos}\hspace{0.17em}\theta }{\text{sin}\hspace{0.17em}\theta },\text{sec}\hspace{0.17em}\theta =\frac{1}{\text{cos}\hspace{0.17em}\theta },\text{csc}\hspace{0.17em}\theta =\frac{1}{\text{sin}\hspace{0.17em}\theta }\text{.}$$ Since the equation of a circle of radius 1 is $x^2+y^2=1$ this forces $${\text{sin}}^2\theta +{\text{cos}}^2\theta =1\text{.}$$ The pictures $$\begin{array}{c} y x } θ cos⁡θ sin⁡θ and y x } -θ -cos⁡θ -sin⁡θ \end{array}$$ show that $$\text{sin}(-\theta )=-\text{sin}\hspace{0.17em}\theta \text{and}\text{cos}(-\theta )=\text{cos}\hspace{0.17em}\theta \text{.}$$ Also $$\begin{array}{c} y x -1 1 and y x } π2 -1 1 \end{array}$$ show that $$\begin{array}{ccc}{\displaystyle \text{sin}\hspace{0.17em}0}& =& {\displaystyle 0,}\\ {\displaystyle \text{cos}\hspace{0.17em}0}& =& {\displaystyle 1,}\end{array}\text{and}\begin{array}{ccc}{\displaystyle \text{sin}\hspace{0.17em}\frac{\pi }{2}}& =& {\displaystyle 1,}\\ {\displaystyle \text{cos}\hspace{0.17em}\frac{\pi }{2}=0\text{.}}& =& {\displaystyle }\end{array}$$ Draw the graphs $y=\text{sin}\hspace{0.17em}\theta$ and $y=\text{cos}\hspace{0.17em}\theta$ by seeing how the $x$ and $y$ coordinates change as you walk around the circle. $$\begin{array}{c} y θ -2π -3π2 -π -π2 π2 π 3π2 2π 1 -1 \end{array}$$ and $$\begin{array}{c} y θ -2π -3π2 -π -π2 π2 π 3π2 2π 1 -1 \end{array}$$

Verify $\frac{\text{sec}\hspace{0.17em}B}{\text{cos}\hspace{0.17em}B}-\frac{\text{tan}\hspace{0.17em}B}{\text{cot}\hspace{0.17em}B}=1\text{.}$ $$\begin{array}{ccc}{\displaystyle \frac{\text{sec}\hspace{0.17em}B}{\text{cos}\hspace{0.17em}B}-\frac{\text{tan}\hspace{0.17em}B}{\text{cot}\hspace{0.17em}B}}& =& {\displaystyle \frac{\frac{1}{\text{cos}\hspace{0.17em}B}}{\text{cos}\hspace{0.17em}B}-\frac{\frac{\text{sin}\hspace{0.17em}B}{\text{cos}\hspace{0.17em}B}}{\frac{\text{cos}\hspace{0.17em}B}{\text{sin}\hspace{0.17em}B}}}\\ & =& {\displaystyle \frac{1}{{\text{cos}}^{2}B}-\frac{{\text{sin}}^{2}B}{{\text{cos}}^{2}B}}\\ & =& {\displaystyle \frac{1-{\text{sin}}^{2}B}{{\text{cos}}^{2}B}}\\ & =& {\displaystyle \frac{{\text{cos}}^{2}B}{{\text{cos}}^{2}B}}\\ & =& {\displaystyle 1\text{.}}\end{array}$$

Verify $\text{cot}\hspace{0.17em}\alpha -\text{cot}\hspace{0.17em}\beta =\frac{\text{sin}(\beta -\alpha )}{\text{sin}\hspace{0.17em}\alpha \hspace{0.17em}\text{sin}\hspace{0.17em}\beta }\text{.}$ $$\begin{array}{ccc}{\displaystyle \text{Left Hand Side}}& =& {\displaystyle \text{cot}\hspace{0.17em}\alpha -\text{cot}\hspace{0.17em}\beta }\\ & =& {\displaystyle \frac{\text{cos}\hspace{0.17em}\alpha }{\text{sin}\hspace{0.17em}\alpha }-\frac{\text{cos}\hspace{0.17em}\beta }{\text{sin}\hspace{0.17em}\beta }}\\ & =& {\displaystyle \frac{\text{cos}\hspace{0.17em}\alpha \hspace{0.17em}\text{sin}\hspace{0.17em}\beta -\text{cos}\hspace{0.17em}\beta \hspace{0.17em}\text{sin}\hspace{0.17em}\alpha }{\text{sin}\hspace{0.17em}\alpha \hspace{0.17em}\text{sin}\hspace{0.17em}\beta }}\\ {\displaystyle \text{Right Hand Side}}& =& {\displaystyle \frac{\text{sin}(\beta -\alpha )}{\text{sin}\hspace{0.17em}\alpha \hspace{0.17em}\text{sin}\hspace{0.17em}\beta }}\\ & =& {\displaystyle \frac{\text{sin}\hspace{0.17em}\beta \hspace{0.17em}\text{cos}(-\alpha )+\text{cos}\hspace{0.17em}\beta \hspace{0.17em}\text{sin}(-\alpha )}{\text{sin}\hspace{0.17em}\alpha \hspace{0.17em}\text{sin}\hspace{0.17em}\beta }}\\ & =& {\displaystyle \frac{\text{sin}\hspace{0.17em}\beta \hspace{0.17em}\text{cos}\hspace{0.17em}\alpha +\text{cos}\hspace{0.17em}\beta (-\text{sin}\hspace{0.17em}\alpha )}{\text{sin}\hspace{0.17em}\alpha \hspace{0.17em}\text{sin}\hspace{0.17em}\beta }}\\ & =& {\displaystyle \frac{\text{sin}\hspace{0.17em}\beta \hspace{0.17em}\text{cos}\hspace{0.17em}\alpha -\text{cos}\hspace{0.17em}\beta \hspace{0.17em}\text{sin}\hspace{0.17em}\alpha }{\text{sin}\hspace{0.17em}\alpha \hspace{0.17em}\text{sin}\beta }\text{.}}\end{array}$$ So $$\begin{array}{ccc}{\displaystyle \text{Left Hand Side}}& =& {\displaystyle \text{Right Hand Side.}}\end{array}$$

Verify $\frac{\text{tan}\hspace{0.17em}A0\text{sin}\hspace{0.17em}A}{\text{sec}\hspace{0.17em}A}=\frac{{\text{sin}}^{3}A}{1+\text{cos}\hspace{0.17em}A}\text{.}$ $$\frac{\text{tan}\hspace{0.17em}A0\text{sin}\hspace{0.17em}A}{\text{sec}\hspace{0.17em}A}\stackrel{?}{=}\frac{{\text{sin}}^{3}A}{1+\text{cos}\hspace{0.17em}A}\text{.}$$ So So $$\text{tan}\hspace{0.17em}A+\text{cos}\hspace{0.17em}A\hspace{0.17em}\text{tan}\hspace{0.17em}A-\text{sin}\hspace{0.17em}A-\text{sin}\hspace{0.17em}A\hspace{0.17em}\text{cos}\hspace{0.17em}A\stackrel{?}{=}{\text{sin}}^{3}A\hspace{0.17em}\text{sec}\hspace{0.17em}A\text{.}$$ So $$\frac{\text{sin}\hspace{0.17em}A}{\text{cos}\hspace{0.17em}A}+\text{sin}\hspace{0.17em}A-\text{sin}\hspace{0.17em}A-\text{sin}\hspace{0.17em}A\hspace{0.17em}\text{cos}\hspace{0.17em}A\stackrel{?}{=}\frac{{\text{sin}}^{3}A}{\text{cos}\hspace{0.17em}A}\text{.}$$ So $$\frac{\text{sin}\hspace{0.17em}A-\text{sin}\hspace{0.17em}A\hspace{0.17em}{\text{cos}}^{2}A}{\text{cos}\hspace{0.17em}A}\stackrel{?}{=}\frac{{\text{sin}}^{3}A}{\text{cos}\hspace{0.17em}A}\text{.}$$ So $$\text{sin}\hspace{0.17em}A-\text{sin}\hspace{0.17em}A\hspace{0.17em}{\text{cos}}^{2}A\stackrel{?}{=}{\text{sin}}^{3}A\text{.}$$ So $$1-{\text{cos}}^{2}A\stackrel{?}{=}{\text{sin}}^{2}A\text{.}$$ YES because ${\text{sin}}^{2}A+{\text{cos}}^{2}A=1\text{.}$

Notes and References

These are a typed copy of Lecture 2 from a series of handwritten lecture notes for the class MATH 221 given on September 8, 2000.

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