3. Trigonometry

c. Circle Definitions

Recall: \[\begin{aligned} \sin\theta&=\dfrac{y}{r}\quad &\tan\theta&=\dfrac{y}{x}\quad &\sec\theta&=\dfrac{r}{x} \\[8pt] \cos\theta&=\dfrac{x}{r}\quad &\cot\theta&=\dfrac{x}{y}\quad &\csc\theta&=\dfrac{r}{y} \end{aligned}\]

3. Unit Circle

Now that we know how to find the trig functions using the circle definitions, we want to understand what this says when the circle is a unit circle, i.e. a circle of radius \(1\).

The fact that the radius is \(1\) allows us to write the trig functions as follows: \[\begin{aligned} \sin\theta&=y\quad &\tan\theta&=\dfrac{y}{x}\quad &\sec\theta&=\dfrac{1}{x} \\[8pt] \cos\theta&=x\quad &\cot\theta&=\dfrac{x}{y}\quad &\csc\theta&=\dfrac{1}{y} \end{aligned}\]

In particular:

If a point on the unit circle has coordinates \((x,y)\) then: \[ \sin\theta=y \qquad \text{and} \qquad \cos\theta=x \] Conversely, the point on the unit circle at angle \(\theta\) is: \[ (x,y)=(\cos\theta,\sin\theta) \]

As an example, let's find the coordinates of a point on the unit circle at an angle of \(30^\circ\) from the \(x\)-axis. We know the coordinates are \((x,y)=(\cos30^\circ,\sin30^\circ)\). To find \(\sin\) and \(\cos\), we look at the \(30^\circ - 60^\circ - 90^\circ\) triangle with sides \(1, 2, \sqrt{3}\). So \[ \sin30^\circ=\dfrac{1}{2} \quad \text{and} \quad \cos30^\circ=\dfrac{\sqrt{3}}{2} \] So the coordinates of the point are: \[ (x,y)=\left(\dfrac{\sqrt{3}}{2},\dfrac{1}{2}\right) \]

Using the unit circle, find the \(\sin135^\circ\) and \(\cos135^\circ\).

How are \(\sin135^\circ\) and \(\cos135^\circ\) related to \(\sin45^\circ\) and \(\cos45^\circ\)?

\(\begin{aligned} \sin135^\circ&=\dfrac{\sqrt{2}}{2} \\ \cos135^\circ&=-\,\dfrac{\sqrt{2}}{2} \end{aligned}\)

Look at the point on the unit circle at \(135^\circ\): \[ (x,y)=(\cos135^\circ,\sin135^\circ) \] The coordinates are the same as those at \(45^\circ\): \[ (x,y)=(\cos45^\circ,\sin45^\circ) =\left(\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}\right) \] except there needs to be a minus on the \(x\) component. So \[ (x,y)=(\cos135^\circ,\sin135^\circ) =\left(-\,\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}\right) \]

We conclude: \[ \sin135^\circ=\dfrac{\sqrt{2}}{2} \quad \text{and} \quad \cos135^\circ=-\,\dfrac{\sqrt{2}}{2} \]

aas

Consider a unit circle and the ray at the angle \(\theta\) which passes through the point \((0.8,-0.6)\), identify the six trig functions for the angle \(\theta\).

\(\begin{aligned} \sin\theta&=-\,\dfrac{3}{5}\quad &\tan\theta&=-\,\dfrac{3}{4}\quad &\sec\theta&=\dfrac{5}{4} \\ \cos\theta&=\dfrac{4}{5}\quad &\cot\theta&=-\,\dfrac{4}{3}\quad &\csc\theta&=-\,\dfrac{5}{3} \end{aligned}\)

Given that this is a unit circle, we know that it has a radius of \(r=1\), and we are given that the ray passes through the point \((x,y)=(0.8,-0.6)\). Therefore, the trig functions are: \[\begin{aligned} \sin\theta&=\dfrac{y}{r}=\dfrac{-0.6}{1}=-\,\dfrac{3}{5}\quad &\cos\theta&=\dfrac{x}{r}=\dfrac{0.8}{1}=\dfrac{4}{5} \\[8pt] \tan\theta&=\dfrac{y}{x}=\dfrac{-0.6}{0.8}=-\,\dfrac{3}{4}\quad &\cot\theta&=\dfrac{x}{y}=\dfrac{0.8}{-0.6}=-\,\dfrac{4}{3} \\[8pt] \sec\theta&=\dfrac{r}{x}=\dfrac{1}{0.8}=\dfrac{5}{4}\quad &\csc\theta&=\dfrac{r}{y}=\dfrac{1}{-0.6}=-\,\dfrac{5}{3} \end{aligned}\]

3. Trigonometry

c. Circle Definitions

3. Unit Circle

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