audio read-through The Unit Circle Approach to Trigonometry

There are two basic approaches to trigonometry:

Both approaches were introduced in Introduction to Trigonometry. Be sure to read this prior section, since it covers important notation and conventions.

The Unit Circle Approach to Trigonometry

Start with a circle, centered at the origin, of radius $\,1\,.$ In trigonometry, this is called the ‘unit circle’. (‘Unit’ refers to the radius being equal to $\,1\,.$ ) The equation of this circle is $\,x^2 + y^2 = 1\,.$

In the unit circle approach to trigonometry, angles can have any real number measure. The angle must be ‘laid out’ in the circle in a standard way:

Let $\,\theta\,$ indicate the desired angle.

the unit circle approach to trigonometry

Focus attention on the point where the terminal (ending) side of the angle intersects the unit circle. This point is called the terminal point for the angle.

Once you have the terminal point, you can forget how you got there! The only thing that matters, for determining the values of the trigonometric functions, is the location of the terminal point!

The terminal point has coordinates: an $x$-value, and a $y$-value. Notice immediately that both coordinates are always numbers between $\,-1\,$ and $\,1\,$ (inclusive), since it's a point on the unit circle.

By definition:

$\sin\theta := \text{the $y$-value of the terminal point}$

$\cos\theta := \text{the $x$-value of the terminal point}$

Thus, cosine and sine are just the $\,x\,$ and $\,y\,$ values (respectively) of points on the unit circle!

Infinitely many angles can get you to the same terminal point. You might get there as efficiently as possible. You might sweep around counterclockwise many times. You might sweep around clockwise many times. It doesn't matter! All that matters is where you end up (the terminal point). This is a really important thing to remember!

The sketch below shows two ways to get to the same terminal point:

$30^\circ\,$
(Positive angles are swept out counterclockwise; up)

$-330^\circ\,$
(Negative angles are swept out clockwise; down)

two ways to get to the same terminal point

It doesn't matter how we get there! All that matters are the coordinates of the terminal point:

$\displaystyle\cos 30^\circ = \cos(-330^\circ) = \frac{\sqrt 3}{2}$

$\displaystyle \sin 30^\circ = \sin(-330^\circ) = \frac{1}{2}$

Example: Find Terminal Point, Sine, Cosine, Tangent for ‘Easy’ Angles

What is the terminal point for $\,90^\circ\,$? From this information, find (where possible) the sine, cosine, and tangent of $\,90^\circ\,.$

Solution:

As shown below, the terminal point for $\,90^\circ\,$ is $\,(0,1)\,.$

the terminal point for 90 degrees

Sine is the $y$-value of the terminal point:

$$\cssId{s48}{\sin 90^\circ = 1}$$

Cosine is the $x$-value of the terminal point:

$$\cssId{s50}{\cos 90^\circ = 0}$$

Tangent is defined as sine divided by cosine, when the cosine is nonzero. Since division by zero is not allowed, $\,\tan 90^\circ\,$ is not defined.

Example: Given a Uniquely Identified Terminal Point, Find Trig Functions

Let $\,P(x,y)\,$ be a point on the unit circle. Let $\,\theta\,$ be any angle that has $\,P\,$ as its terminal point.

Solution:

finding trig functions for a uniquely defined point on the unit circle

Since $\,P(x,y)\,$ is on the unit circle $\,x^2 + y^2 = 1\,,$ we have:

$$\begin{gather} \cssId{s60}{x^2 + (-\frac 13)^2 = 1}\cr\cr \cssId{s61}{x^2 + \frac 19 = 1}\cr\cr \cssId{s62}{x^2 = 1 - \frac 19 = \frac 89} \end{gather} $$
$$ \begin{align} x\ &=\ \pm\sqrt{\frac 89}\cr\cr &=\ \pm \frac{\sqrt{4\cdot 2}}{3}\cr\cr &=\ \pm \frac{2\sqrt{2}}{3} \end{align} $$
Thus (since $\,x\lt 0\,$): $$P(x,y) = P\bigl(-\frac{2\sqrt{2}}{3},-\frac 13\bigr)$$

Sine is the $y$-value of the terminal point:

$$ \cssId{s66}{\sin \theta = -\frac 13} $$

Cosine is the $x$-value of the terminal point:

$$ \cssId{s68}{\cos \theta = -\frac{2\sqrt{2}}{3}} $$

Tangent is sine divided by cosine:

$$ \begin{align} \tan \theta &:= \frac{\sin\theta}{\cos\theta}\cr\cr &= \frac{-1/3}{-2\sqrt{2}/3} \cdot \frac{3}{3}\cr\cr &= \frac{1}{2\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}\cr\cr &= \frac{\sqrt{2}}{4} \end{align} $$

Concept Practice