MATH 221 Lecture 2

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

Last updated: 10 July 2012

Angles

Measure angles according to the distance traveled on a circle of radius 1.

Sketch both $x$ and $y$ to get a circle of radius $r$.

The distance $2\pi$ around a circle of radius 1 stretches to $2\pi r$ around a circle of radius $r$. So the circumference of a circle is $2\pi r$ if the circle is radius $r$.

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.

The area of the octagon is the area of the 8 triangles. The area of each triangle is $\frac{1}{2}bh$. So the area of the octagon is $\frac{1}{2}Bh$.

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

$$\begin{array}{ccc}\text{Area of the circle}& =& {\displaystyle \underset{n\to \infty }{\lim }}\left(\text{area of an }n\text{-sided polygon }{P}_n\right)\\ & =& {\displaystyle \underset{n\to \infty }{\lim }}\left(\frac{1}{2}Bh\right)\\ & =& \frac{1}{2}\left(2\pi r\right)\left(r\right)\\ & =& \pi r^2\end{array}$$

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 $r$.

Trigonometric functions

$\sin \theta$ 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

${\displaystyle \begin{array}{cccc}\tan \theta =\frac{\sin \theta }{\cos \theta }\text{,}& \cot \theta =\frac{\cos \theta }{\sin \theta }\text{,}& \sec \theta =\frac{1}{\cos \theta }\text{,}& \csc \theta =\frac{1}{\sin \theta }\end{array}}$

Since the equation of a circle of radius 1 is $x^2+y^2=1$ this forces ${\sin }^2\theta +{\cos }^2\theta =1$.

The pictures

show that

${\displaystyle \begin{array}{ccc}\sin \left(-\theta \right)=-\sin \theta & \text{and}& \cos \left(-\theta \right)=\cos \theta \end{array}}$

Also

show that

${\displaystyle \begin{array}{ccc}\sin 0=0& \text{and}& \sin \frac{\pi }{2}=1\\ \cos 0=1& \cos \frac{\pi }{2}=0\end{array}}$

Draw the graphs

$y=\sin \theta$

$y=\cos \theta$

by seeing how the $x$ and $y$ coordinates change as you walk around the circle.

Example: Verify ${\displaystyle \frac{\sec B}{\cos B}-\frac{\tan B}{\cot B}=1}$

${\displaystyle \begin{array}{ccccc}\frac{\sec B}{\cos B}-\frac{\tan B}{\cot B}& =& \frac{\frac{1}{\cos B}}{\cos B}-\frac{\frac{\sin B}{\cos B}}{\frac{\cos B}{\sin B}}& =& \frac{1}{{\cos }^{2}B}-\frac{{\sin }^{2}B}{{\cos }^{2}B}\\ & =& \frac{1-{\sin }^{2}B}{{\cos }^{2}B}& =& \frac{{\cos }^{2}B}{{\cos }^{2}B}\\ & =& 1\end{array}}$

Example: Verify ${\displaystyle \cot \alpha -\cot \beta =\frac{\sin (\beta -\alpha )}{\sin \alpha \sin \beta }}$

${\displaystyle \begin{array}{ccccc}\text{Left hand side}& =& \cot \alpha -\cot \beta & =& \frac{\cos \alpha }{\sin \alpha }-\frac{\cos \beta }{\sin \beta }\\ & =& \frac{\cos \alpha \sin \beta -\cos \beta \sin \alpha }{\sin \alpha \sin \beta }\\ \\ \text{Right hand side}& =& \frac{\sin (\beta -\alpha )}{\sin \alpha \sin \beta }& =& \frac{\sin \beta \cos (-\alpha )+\cos \beta \sin (-\alpha )}{\sin \alpha \sin \beta }\\ & =& \frac{\sin \beta \cos \alpha +\cos \beta (-\sin \alpha )}{\sin \alpha \sin \beta }& =& \frac{\sin \beta \cos \alpha -\cos \beta \sin \alpha }{\sin \alpha \sin \beta }\end{array}}$

So

$\text{Left hand side}=\text{Right hand side}$

Example: Verify ${\displaystyle \frac{\tan A-\sin A}{\sec A}=\frac{{\sin }^{3}A}{1+\cos A}}$

${\displaystyle \begin{array}{cccc}& \frac{\tan A-\sin A}{\sec A}& \stackrel{?}{=}& \frac{{\sin }^{3}A}{1+\cos A}\\ \text{So}& (1+\cos A)(\tan A-\sin A)& \stackrel{?}{=}& {\sin }^{3}A\sec A\\ \text{So}& \tan A+\cos A\tan A-\sin A-\sin A\cos A& \stackrel{?}{=}& {\sin }^{3}A\sec A\\ \text{So}& \frac{\sin A}{\cos A}+\cos A\frac{\sin A}{\cos A}-\sin A-\sin A\cos A& \stackrel{?}{=}& {\sin }^{3}A\frac{1}{\cos A}\\ \text{So}& \frac{\sin A}{\cos A}+\sin A-\sin A-\sin A\cos A& \stackrel{?}{=}& \frac{{\sin }^{3}A}{\cos A}\\ \text{So}& \frac{\sin A-\sin A{\cos }^{2}A}{\cos A}& \stackrel{?}{=}& \frac{{\sin }^{3}A}{\cos A}\\ \text{So}& \sin A-\sin A{\cos }^{2}A& \stackrel{?}{=}& {\sin }^{3}A\\ \text{So}& 1-{\cos }^{2}A& \stackrel{?}{=}& {\sin }^{2}A\end{array}}$

References

[Ram] A. Ram, MATH 221 Lecture 2, September 8, 2000, University of Wisconsin.

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