The unit circle

So far, the trigonometric functions have had angles as inputs. In calculus and physics, we also need them for plain real numbers, such as time in seconds. The unit circle turns any real number into a point, and the point gives the function values.

The unit circle and terminal points

The unit circle is the circle with center ${(0,0)}$ and radius ${1}$. Its equation is ${x^2+y^2=1}$.

Take a real number ${t}$. Start at the point ${(1,0)}$, and walk a distance ${|t|}$ along the circle: counterclockwise if ${t}$ is positive, clockwise if ${t}$ is negative. The point where you stop is the terminal point of ${t}$. Call it ${P(x,y)}$.

Go a distance ${t}$ along the unit circle from ${(1,0)}$. You reach the terminal point ${P(x,y)}$.

The arc you walked has length ${t}$, and the radius is ${1}$. By the formula ${s=r\theta}$, the central angle is ${\theta=\dfrac{t}{1}=t}$ radians. So the number ${t}$ is the same as an angle of ${t}$ radians, and ${P}$ is on its terminal side.

The whole circle has length ${2\pi}$. So ${t=\pi}$ takes you halfway around, to ${(-1,0)}$, and ${t=\dfrac{\pi}{2}}$ takes you a quarter of the way, to ${(0,1)}$.

Trigonometric functions of real numbers

On the unit circle, ${r=1}$. So the definitions from the last lesson become very simple. If ${P(x,y)}$ is the terminal point of ${t}$, then

${\sin t=y}$${\cos t=x}$

${\tan t=\dfrac{y}{x}}$${\cot t=\dfrac{x}{y}}$

${\sec t=\dfrac{1}{x}}$${\csc t=\dfrac{1}{y}}$

In words: the cosine is the ${x}$-coordinate, and the sine is the ${y}$-coordinate of the terminal point. The terminal point of ${t}$ is ${(\cos t,\sin t)}$.

These values are the same as the values for an angle of ${t}$ radians. So everything from the last lesson still works: reference angles, signs in each quadrant, and the identities.

Example 1: Find the six trigonometric functions of ${t=\pi}$.

Solution:

Going halfway around the circle from ${(1,0)}$ ends at ${P(-1,0)}$. So ${x=-1}$ and ${y=0}$:

${\sin\pi=0}$${\cos\pi=-1}$

${\tan\pi=\dfrac{0}{-1}=0}$${\sec\pi=\dfrac{1}{-1}=-1}$

${\cot\pi}$ and ${\csc\pi}$ are undefined, because ${y=0}$.

The special values

The special triangles from Right triangle trigonometry give the terminal points for ${\dfrac{\pi}{6}}$, ${\dfrac{\pi}{4}}$, and ${\dfrac{\pi}{3}}$. Each coordinate is a cosine or a sine from the table there.

The terminal points of ${\dfrac{\pi}{6}}$, ${\dfrac{\pi}{4}}$, and ${\dfrac{\pi}{3}}$.

The other special points are mirror images of these three. For example, ${\dfrac{5\pi}{6}}$ has reference angle ${\dfrac{\pi}{6}}$ and is in quadrant II. Its terminal point is $\left(-\dfrac{\sqrt{3}}{2},\dfrac{1}{2}\right)$: the same sizes as for ${\dfrac{\pi}{6}}$, but ${x}$ is negative.

The special values of ${t}$ between ${0}$ and ${2\pi}$.

To find a terminal point on this circle:

Find the reference number. (It is the reference angle, now called a number.)
Take the coordinates of the matching point in quadrant I.
Give each coordinate the sign it has in the quadrant of ${t}$.

Example 2: Find the terminal point of ${t}$. Then find ${\sin t}$, ${\cos t}$, and ${\tan t}$.

(a) ${t=\dfrac{7\pi}{4}}$ (b) ${t=\dfrac{4\pi}{3}}$

Solution:

(a) ${\dfrac{7\pi}{4}}$ is in quadrant IV. Its reference number is

${2\pi-\dfrac{7\pi}{4}=\dfrac{\pi}{4}}$

In quadrant IV, ${x}$ is positive and ${y}$ is negative. So the terminal point is

$P\left(\dfrac{\sqrt{2}}{2},-\dfrac{\sqrt{2}}{2}\right)$

Read off the values:

${\sin\dfrac{7\pi}{4}=-\dfrac{\sqrt{2}}{2}}$${\cos\dfrac{7\pi}{4}=\dfrac{\sqrt{2}}{2}}$

${\tan\dfrac{7\pi}{4}=\dfrac{y}{x}=-1}$

(b) ${\dfrac{4\pi}{3}}$ is in quadrant III. Its reference number is

${\dfrac{4\pi}{3}-\pi=\dfrac{\pi}{3}}$

In quadrant III, both coordinates are negative. So the terminal point is

$P\left(-\dfrac{1}{2},-\dfrac{\sqrt{3}}{2}\right)$

Read off the values:

${\sin\dfrac{4\pi}{3}=-\dfrac{\sqrt{3}}{2}}$${\cos\dfrac{4\pi}{3}=-\dfrac{1}{2}}$

$\tan\dfrac{4\pi}{3}=\dfrac{-\sqrt{3}/2}{-1/2}=\sqrt{3}$

Example 3: The point ${P\left(-\dfrac{3}{5},y\right)}$ is on the unit circle in quadrant III. It is the terminal point of a number ${t}$. Find ${\sin t}$, ${\cos t}$, and ${\tan t}$.

Solution:

The point is on the unit circle, so ${x^2+y^2=1}$:

$\begin{align*}\left(-\dfrac{3}{5}\right)^2+y^2&=1\\y^2&=1-\dfrac{9}{25}\\&=\dfrac{16}{25}\end{align*}$

So ${y=\pm\dfrac{4}{5}}$. In quadrant III, ${y}$ is negative, so ${y=-\dfrac{4}{5}}$. Now

${\sin t=-\dfrac{4}{5}}$${\cos t=-\dfrac{3}{5}}$

${\tan t=\dfrac{-4/5}{-3/5}=\dfrac{4}{3}}$

Domain and range

Every real number ${t}$ has a terminal point, so ${\sin t}$ and ${\cos t}$ are defined for all real numbers. Their domain (the set of allowed inputs) is all real numbers.

Every point on the unit circle has ${x}$ and ${y}$ between ${-1}$ and ${1}$. So

${-1\le\sin t\le 1}$${-1\le\cos t\le 1}$

The range (the set of possible outputs) of sine and cosine is the interval ${[-1,1]}$. For example, no number ${t}$ has ${\sin t=2}$.

Tangent and secant are undefined where ${x=0}$, at the top and bottom of the circle: ${t=\dfrac{\pi}{2}}$, ${\dfrac{3\pi}{2}}$, and so on. Cotangent and cosecant are undefined where ${y=0}$: ${t=0}$, ${\pi}$, ${2\pi}$, and so on.

Even and odd properties

The terminal point of ${-t}$ is found by walking the same distance, but clockwise. So it is the mirror image of the terminal point of ${t}$ across the ${x}$-axis. The ${x}$-coordinate stays the same, and the ${y}$-coordinate changes sign.

The terminal point of ${-t}$ is the mirror image of the terminal point of ${t}$ across the ${x}$-axis.

Since cosine is ${x}$ and sine is ${y}$, this gives

${\cos(-t)=\cos t}$${\sin(-t)=-\sin t}$

${\tan(-t)=-\tan t}$

The tangent rule follows from the other two: ${\tan(-t)=\dfrac{-y}{x}=-\tan t}$. In the same way, secant behaves like cosine, and cosecant and cotangent behave like sine.

A function with ${f(-t)=f(t)}$ is called even, and one with ${f(-t)=-f(t)}$ is called odd. (See Graphs of functions in College Algebra.) So cosine and secant are even. The other four are odd.

Example 4: Find the exact value.

(a) ${\sin\left(-\dfrac{\pi}{6}\right)}$ (b) ${\cos\left(-\dfrac{\pi}{4}\right)}$ (c) ${\tan\left(-\dfrac{\pi}{3}\right)}$

Solution:

(a) Sine is odd, so

$\begin{align*}\sin\left(-\dfrac{\pi}{6}\right)&=-\sin\dfrac{\pi}{6}\\&=-\dfrac{1}{2}\end{align*}$

(b) Cosine is even, so

$\begin{align*}\cos\left(-\dfrac{\pi}{4}\right)&=\cos\dfrac{\pi}{4}\\&=\dfrac{\sqrt{2}}{2}\end{align*}$

$\begin{align*}\tan\left(-\dfrac{\pi}{3}\right)&=-\tan\dfrac{\pi}{3}\\&=-\sqrt{3}\end{align*}$

Periodic properties

Adding ${2\pi}$ to ${t}$ means walking one more full turn around the circle. You end at the same terminal point. So the function values repeat:

${\sin(t+2\pi)=\sin t}$${\cos(t+2\pi)=\cos t}$

A function that repeats like this is called periodic. The smallest positive number ${p}$ with ${f(t+p)=f(t)}$ for every ${t}$ is the period. Sine, cosine, secant, and cosecant have period ${2\pi}$.

Tangent and cotangent repeat sooner. Adding ${\pi}$ takes you halfway around, to the opposite point ${(-x,-y)}$. Then ${\dfrac{-y}{-x}=\dfrac{y}{x}}$, so ${\tan(t+\pi)=\tan t}$. Tangent and cotangent have period ${\pi}$.

Example 5: Find the exact value.

(a) ${\sin\dfrac{13\pi}{6}}$ (b) ${\cos\left(-\dfrac{9\pi}{4}\right)}$

Solution:

(a) Split off the full turns:

$\begin{align*}\dfrac{13\pi}{6}&=\dfrac{\pi}{6}+\dfrac{12\pi}{6}\\&=\dfrac{\pi}{6}+2\pi\end{align*}$

Removing the full turn gives

$\sin\dfrac{13\pi}{6}=\sin\dfrac{\pi}{6}=\dfrac{1}{2}$

(b) Cosine is even, so first drop the minus sign:

$\cos\left(-\dfrac{9\pi}{4}\right)=\cos\dfrac{9\pi}{4}$

Then ${\dfrac{9\pi}{4}=\dfrac{\pi}{4}+2\pi}$, so

$\cos\dfrac{9\pi}{4}=\cos\dfrac{\pi}{4}=\dfrac{\sqrt{2}}{2}$

Using a calculator

For a number ${t}$ that is not special, use a calculator in radian mode. For example, the terminal point of ${t=2}$ is

${(\cos 2,\sin 2)\approx(-0.4161,\ 0.9093)}$

The ${x}$-coordinate is negative and the ${y}$-coordinate is positive, so this point is in quadrant II. That makes sense: ${2}$ is between ${\dfrac{\pi}{2}\approx 1.57}$ and ${\pi\approx 3.14}$.

Watch out: When ${t}$ is a real number, such as ${2}$ or ${\dfrac{\pi}{6}}$, your calculator must be in radian mode. In degree mode, ${\cos 2}$ means the cosine of ${2^\circ}$, which is about ${0.9994}$.

Summary

The unit circle is ${x^2+y^2=1}$. The terminal point of ${t}$ is reached by walking a distance ${t}$ along it from ${(1,0)}$.
The terminal point of ${t}$ is ${(\cos t,\sin t)}$: cosine is the ${x}$-coordinate, and sine is the ${y}$-coordinate.
A real number ${t}$ gives the same values as an angle of ${t}$ radians.
Sine and cosine are defined for all ${t}$, and their values are between ${-1}$ and ${1}$.
Cosine is even: ${\cos(-t)=\cos t}$. Sine and tangent are odd: ${\sin(-t)=-\sin t}$.
Sine and cosine have period ${2\pi}$. Tangent has period ${\pi}$.

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