Math Core

Right-triangle trigonometry only works for acute angles, but the things trigonometry models (rotations, sound, tides, alternating current) keep going around forever. The unit circle turns sine and cosine into functions of any real number, and it does so in radians, the unit that makes calculus with these functions clean. This lesson builds that definition and the tools you need to evaluate trig functions exactly.

Radian measure

An angle in standard position has its vertex at the origin and its initial side along the positive $x$ -axis. Counterclockwise rotation is positive; clockwise is negative.

Definition

Radian

On a circle of radius $r$ , a central angle that cuts off an arc of length $s$ has measure

\theta = \frac{s}{r} \text{ radians}.

One radian is the angle whose arc is exactly one radius long.

A full turn cuts off the whole circumference $2\pi r$ , so a full turn is $\dfrac{2\pi r}{r} = 2\pi$ radians. That single fact gives every conversion:

360^\circ = 2\pi \quad\Longrightarrow\quad 180^\circ = \pi, \qquad 1^\circ = \frac{\pi}{180}, \qquad 1 \text{ radian} = \frac{180^\circ}{\pi} \approx 57.3^\circ.

To convert degrees to radians multiply by $\dfrac{\pi}{180}$ ; to go back multiply by $\dfrac{180}{\pi}$ . For example, $135^\circ = \dfrac{135\pi}{180} = \dfrac{3\pi}{4}$ and $\dfrac{7\pi}{6} = \dfrac{7 \cdot 180^\circ}{6} = 210^\circ$ .

Because a radian is a ratio of two lengths, it has no units. That is why the definition rearranges so neatly into the arc length formula $s = r\theta$ (with $\theta$ in radians). A wheel of radius $0.4$ m turning through $5$ radians rolls $0.4 \cdot 5 = 2$ m.

Angles that differ by a full turn, like $\dfrac{\pi}{3}$ and $\dfrac{\pi}{3} + 2\pi = \dfrac{7\pi}{3}$ , share a terminal side. They are called coterminal.

Sine and cosine from the unit circle

The unit circle is $x^2 + y^2 = 1$ : radius $1$ , centered at the origin. On it, arc length and radian measure are the same number, since $s = 1 \cdot \theta$ . So you can think of a real number $t$ as a distance walked along the circle from $(1, 0)$ : counterclockwise if $t > 0$ , clockwise if $t < 0$ .

Unit circle definitions

Let $P(x, y)$ be the point on the unit circle reached by the angle $t$ . Then

\cos t = x, \qquad \sin t = y, \qquad \tan t = \frac{y}{x}\ (x \ne 0).

The reciprocal functions are $\sec t = \dfrac{1}{x}$ , $\csc t = \dfrac{1}{y}$ and $\cot t = \dfrac{x}{y}$ , each defined when its denominator is not $0$ .

For an acute angle this agrees with SOH-CAH-TOA: drop a perpendicular from $P$ to the $x$ -axis and you get a right triangle with hypotenuse $1$ , adjacent side $x$ and opposite side $y$ . The difference is that the circle definition keeps working past $90^\circ$ , for negative angles, and for angles larger than a full turn.

The unit circle with several special angles. The point at π/6 is (√3/2, 1/2); the others are reflections of first-quadrant points.Open in grapher →

The special values

The $30$ - $60$ - $90$ and $45$ - $45$ - $90$ triangles with hypotenuse $1$ give the first-quadrant points you should know cold:

$t$	$0$	$\frac{\pi}{6}$	$\frac{\pi}{4}$	$\frac{\pi}{3}$	$\frac{\pi}{2}$
$\cos t$	$1$	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{1}{2}$	$0$
$\sin t$	$0$	$\frac{1}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{3}}{2}$	$1$
$\tan t$	$0$	$\frac{\sqrt{3}}{3}$	$1$	$\sqrt{3}$	undefined

Every other special angle is a reflection of one of these. The circle is symmetric across both axes, so the point for any angle has the same coordinates (up to sign) as the point for its reference angle, the acute angle between the terminal side and the $x$ -axis.

Quadrant II: reference angle $\pi - t$ .
Quadrant III: reference angle $t - \pi$ .
Quadrant IV: reference angle $2\pi - t$ .

The signs come straight from the coordinates. Cosine is $x$ , so it is positive on the right half of the circle; sine is $y$ , positive on the top half; tangent is $y/x$ , positive where they agree (Quadrants I and III).

Evaluating any special angle

Replace $t$ by a coterminal angle in $[0, 2\pi)$ if needed.
Find the quadrant and the reference angle.
Take the value at the reference angle and attach the sign for that quadrant.

Worked example: Reference angles in three quadrants

Find $\cos\dfrac{5\pi}{6}$ , $\sin\dfrac{4\pi}{3}$ and $\tan\left(-\dfrac{\pi}{4}\right)$ .

Solution. $\dfrac{5\pi}{6}$ is in Quadrant II with reference angle $\dfrac{\pi}{6}$ . Cosine is negative there, so $\cos\dfrac{5\pi}{6} = -\dfrac{\sqrt{3}}{2}$ .

$\dfrac{4\pi}{3}$ is in Quadrant III with reference angle $\dfrac{\pi}{3}$ . Sine is negative there, so $\sin\dfrac{4\pi}{3} = -\dfrac{\sqrt{3}}{2}$ .

$-\dfrac{\pi}{4}$ is an eighth of a turn clockwise, landing in Quadrant IV at $\left(\dfrac{\sqrt{2}}{2}, -\dfrac{\sqrt{2}}{2}\right)$ . So $\tan\left(-\dfrac{\pi}{4}\right) = \dfrac{-\sqrt{2}/2}{\sqrt{2}/2} = -1$ .

Worked example: Large angles and reciprocals

Find $\sec\dfrac{11\pi}{3}$ and $\csc\dfrac{19\pi}{6}$ .

Solution. Subtract $2\pi = \dfrac{6\pi}{3}$ : $\dfrac{11\pi}{3}$ is coterminal with $\dfrac{5\pi}{3}$ , in Quadrant IV with reference angle $\dfrac{\pi}{3}$ . So $\cos\dfrac{11\pi}{3} = \dfrac{1}{2}$ and $\sec\dfrac{11\pi}{3} = 2$ .

Subtract $2\pi = \dfrac{12\pi}{6}$ : $\dfrac{19\pi}{6}$ is coterminal with $\dfrac{7\pi}{6}$ , in Quadrant III with reference angle $\dfrac{\pi}{6}$ . So $\sin\dfrac{19\pi}{6} = -\dfrac{1}{2}$ and $\csc\dfrac{19\pi}{6} = -2$ .

Common mistake

The reference angle is always measured to the $x$ -axis, never the $y$ -axis. For $\dfrac{2\pi}{3}$ the reference angle is $\pi - \dfrac{2\pi}{3} = \dfrac{\pi}{3}$ , not $\dfrac{\pi}{6}$ . Mixing these up swaps the sine and cosine values.

Properties that come free with the circle

The circle picture explains the basic behavior of sine and cosine as functions.

Domain and range. Every real $t$ gives a point, so sine and cosine have domain $(-\infty, \infty)$ . Coordinates on the unit circle lie between $-1$ and $1$ , so both have range $[-1, 1]$ .
Periodicity. Adding $2\pi$ brings you back to the same point: $\sin(t + 2\pi) = \sin t$ and $\cos(t + 2\pi) = \cos t$ . Tangent repeats sooner. Adding $\pi$ sends $(x, y)$ to $(-x, -y)$ , and $\dfrac{-y}{-x} = \dfrac{y}{x}$ , so $\tan(t + \pi) = \tan t$ .
Even and odd. The angle $-t$ reflects $P$ across the $x$ -axis, from $(x, y)$ to $(x, -y)$ . So $\cos(-t) = \cos t$ (cosine is even) and $\sin(-t) = -\sin t$ (sine is odd). Tangent is odd too.
The Pythagorean identity. Since $x^2 + y^2 = 1$ for every point on the circle,

\sin^2 t + \cos^2 t = 1.

The identity lets you find all six values from just one, provided you know the quadrant.

Worked example: One value and a quadrant

Suppose $\cos t = -\dfrac{5}{13}$ and $\sin t > 0$ . Find $\sin t$ , $\tan t$ and $\csc t$ .

Solution. Cosine negative and sine positive puts $t$ in Quadrant II. From the identity,

\sin^2 t = 1 - \left(-\frac{5}{13}\right)^2 = 1 - \frac{25}{169} = \frac{144}{169}, \qquad \sin t = \frac{12}{13}

(positive, because of the quadrant). Then $\tan t = \dfrac{12/13}{-5/13} = -\dfrac{12}{5}$ and $\csc t = \dfrac{13}{12}$ .

Tip

Before you trust an answer, picture the point on the circle. $\sin\dfrac{5\pi}{4}$ should be negative (the point is below the axis), and $\left|\cos\dfrac{\pi}{6}\right|$ should be bigger than $\left|\sin\dfrac{\pi}{6}\right|$ (the point is closer to the $x$ -axis than to the $y$ -axis).