General Angles and Reference Angles, precalculus lesson

Right triangles only give trigonometric ratios for angles between $0^\circ$ and $90^\circ$ . A wheel, though, turns through $150^\circ$ or $480^\circ$ , and a bolt can be turned backwards. To handle any angle, place it on coordinate axes and read the ratios from a point on its terminal side. The signs of that point’s coordinates carry the extra information.

Key terms

General angle

Definition A general angle is an angle in standard position, with its vertex at the origin and its initial side along the positive $x$ -axis, whose measure may be any real number of degrees or radians. A positive measure means a counterclockwise rotation from the initial side to the terminal side, a negative measure means a clockwise rotation, and a measure larger than $360^\circ$ or $2\pi$ means more than one full turn. Angles with the same terminal side are coterminal and differ by an integer multiple of $360^\circ$ or $2\pi$ . An angle lies in a quadrant when its terminal side does. When the terminal side lies on an axis the angle is quadrantal, its measure is an integer multiple of $90^\circ$ or $\frac{\pi}{2}$ , and it lies in no quadrant.

In plain words An angle is now a rotation, not just a corner of a triangle. It can be large or negative or wind around several times, and for trigonometry all that matters is where the terminal side ends up.

Example The angle $480^\circ$ is one full turn plus $120^\circ$ , so it is coterminal with $120^\circ$ and lies in Quadrant II. The angle $-45^\circ$ turns clockwise, is coterminal with $315^\circ$ and lies in Quadrant IV. The angle $-270^\circ$ is quadrantal, with the same terminal side as $90^\circ$ .

Terminal point

Definition For an angle $\theta$ in standard position, a terminal point is any point $P = (x, y)$ other than the origin that lies on the terminal side of $\theta$ . The terminal side is a ray starting at the origin, so if $(x_0, y_0)$ is one terminal point, the terminal points are exactly the points $(t x_0, t y_0)$ with $t > 0$ . Some books use the name only for the one terminal point at distance 1 from the origin, the point on the unit circle.

In plain words Pick any point on the terminal side except the origin. Its coordinates tell you in which direction the side points.

Example The terminal side of $120^\circ$ makes a $60^\circ$ angle with the negative $x$ -axis and passes through $(-1, \sqrt{3})$ . Doubling gives another terminal point, $(-2, 2\sqrt{3})$ . The point $(1, -\sqrt{3})$ is on the same line but on the opposite ray, since it comes from $t = -1$ , so it is not a terminal point of $120^\circ$ .

Radius

Definition The radius of a terminal point $P = (x, y)$ is its distance from the origin,

r = \sqrt{x^2 + y^2}.

Because $P$ is not the origin, $r > 0$ . The point $P$ lies on the circle of radius $r$ centered at the origin.

In plain words The radius is how far the chosen point is from the origin. It is always positive, even when $x$ or $y$ is negative.

Example For $P = (-3, 4)$ the radius is $r = \sqrt{9 + 16} = 5$ . For the terminal point $(-1, \sqrt{3})$ of $120^\circ$ it is $r = \sqrt{1 + 3} = 2$ .

Six trigonometric functions

Definition Let $\theta$ be an angle in standard position, let $(x, y)$ be a terminal point of $\theta$ , and let $r = \sqrt{x^2 + y^2}$ . Then

\begin{gathered} \sin \theta = \frac{y}{r}, \qquad \cos \theta = \frac{x}{r}, \\[0.8em] \tan \theta = \frac{y}{x}, \end{gathered}

\begin{gathered} \csc \theta = \frac{r}{y}, \qquad \sec \theta = \frac{r}{x}, \\[0.8em] \cot \theta = \frac{x}{y}, \end{gathered}

each defined whenever its denominator is not zero. The values do not depend on which terminal point you choose. Replacing $(x, y)$ by $(tx, ty)$ with $t > 0$ also replaces $r$ by $tr$ , and the factor $t$ cancels in every ratio. For an acute angle the terminal point, its foot on the $x$ -axis and the origin form a right triangle with legs $x$ and $y$ and hypotenuse $r$ , so these definitions agree with the right triangle ratios.

In plain words These are the right triangle ratios with $x$ in place of the adjacent side, $y$ in place of the opposite side and $r$ in place of the hypotenuse. The difference is that $x$ and $y$ can now be negative.

Example For the terminal point $(-3, 4)$ with $r = 5$ ,

$\sin \theta$	$\cos \theta$	$\tan \theta$	$\csc \theta$	$\sec \theta$	$\cot \theta$
$\frac{4}{5}$	$-\frac{3}{5}$	$-\frac{4}{3}$	$\frac{5}{4}$	$-\frac{5}{3}$	$-\frac{3}{4}$

With the point $(-6, 8)$ on the same ray, $r = 10$ and every ratio comes out the same.

Excluded denominators

Definition Since $r > 0$ , sine and cosine are defined for every angle. Tangent and secant have $x$ in the denominator, so they are undefined exactly when $x = 0$ , which happens when the terminal side lies on the $y$ -axis, at $\theta = 90^\circ + 180^\circ k$ or $\frac{\pi}{2} + k\pi$ for an integer $k$ . Cotangent and cosecant have $y$ in the denominator, so they are undefined exactly when $y = 0$ , which happens when the terminal side lies on the $x$ -axis, at $\theta = 180^\circ k$ or $k\pi$ for an integer $k$ .

In plain words A ratio with zero in the denominator has no value. At every quadrantal angle exactly two of the six functions fail in this way.

Example For $\theta = 270^\circ$ a terminal point is $(0, -1)$ with $r = 1$ . So $\sin 270^\circ = -1$ , $\cos 270^\circ = 0$ , $\csc 270^\circ = -1$ and $\cot 270^\circ = \frac{0}{-1} = 0$ . Tangent and secant would need division by $x = 0$ , so $\tan 270^\circ$ and $\sec 270^\circ$ are undefined.

Quadrant signs

Definition Because $r > 0$ , the sign of each function comes from the signs of $x$ and $y$ . Sine and cosecant have the sign of $y$ , cosine and secant have the sign of $x$ , and tangent and cotangent have the sign of $xy$ .

Quadrant	Sign of $x$	Sign of $y$	Positive functions
I	$+$	$+$	all six
II	$-$	$+$	sine and cosecant
III	$-$	$-$	tangent and cotangent
IV	$+$	$-$	cosine and secant

The functions not listed are negative in that quadrant.

In plain words The quadrant decides the signs. A common memory aid is All Students Take Calculus, which lists all, sine, tangent and cosine as the positive ones, going counterclockwise from Quadrant I.

Example Suppose $\sin \theta < 0$ and $\tan \theta > 0$ . The first tells you $y < 0$ . The second tells you $x$ and $y$ have the same sign, so $x < 0$ too, and $\theta$ lies in Quadrant III.

Reference angle

Definition Let $\theta$ be an angle in standard position that is not quadrantal. Its reference angle $\theta'$ is the acute angle between the terminal side of $\theta$ and the $x$ -axis, so $0^\circ < \theta' < 90^\circ$ . A quadrantal angle has no reference angle, and its values are read directly from a terminal point on the axis. For $0^\circ < \theta < 360^\circ$ it is found as follows.

Quadrant of $\theta$	I	II	III	IV
Reference angle $\theta'$	$\theta$	$180^\circ - \theta$	$\theta - 180^\circ$	$360^\circ - \theta$

For any other angle, first replace it by the coterminal angle between $0^\circ$ and $360^\circ$ . In radians use $\pi$ for $180^\circ$ and $2\pi$ for $360^\circ$ . If $(x, y)$ is a terminal point of $\theta$ , then $(\lvert x \rvert, \lvert y \rvert)$ is a terminal point of $\theta'$ with the same $r$ , so each function of $\theta$ has the same absolute value as that function of $\theta'$ ,

\begin{aligned} \lvert \sin \theta \rvert &= \sin \theta', \\[0.4em] \lvert \cos \theta \rvert &= \cos \theta', \\[0.4em] \lvert \tan \theta \rvert &= \tan \theta', \end{aligned}

and the same holds for cosecant, secant and cotangent wherever they are defined.

In plain words The reference angle measures how far the terminal side is from the nearest part of the $x$ -axis. It gives the size of each value, and the quadrant gives the sign.

Example A few angles and their reference angles.

Angle $\theta$	Coterminal angle used	Quadrant	Reference angle $\theta'$
$150^\circ$	$150^\circ$	II	$30^\circ$
$225^\circ$	$225^\circ$	III	$45^\circ$
$300^\circ$	$300^\circ$	IV	$60^\circ$
$-200^\circ$	$160^\circ$	II	$20^\circ$
$480^\circ$	$120^\circ$	II	$60^\circ$
$\frac{5\pi}{6}$	$\frac{5\pi}{6}$	II	$\frac{\pi}{6}$

Exact values

Definition Let $\theta$ be an angle that is not quadrantal, with reference angle $\theta'$ . For each of the six functions $f$ that is defined at $\theta$ ,

f(\theta) = \pm f(\theta'),

with the sign that $f$ has in the quadrant of $\theta$ . When $\theta'$ is $30^\circ$ , $45^\circ$ or $60^\circ$ , the value $f(\theta')$ is known exactly from the special right triangles, so $f(\theta)$ is known exactly too. For a quadrantal angle use the terminal point $(1, 0)$ , $(0, 1)$ , $(-1, 0)$ or $(0, -1)$ with $r = 1$ .

In plain words The size comes from the reference angle and the sign comes from the quadrant. Find each separately, then put them together.

Example Each row finds the quadrant, the reference angle and the sign.

Angle	Quadrant	Reference angle	Exact value
$210^\circ$	III	$30^\circ$	$\cos 210^\circ = -\cos 30^\circ = -\frac{\sqrt{3}}{2}$
$300^\circ$	IV	$60^\circ$	$\tan 300^\circ = -\tan 60^\circ = -\sqrt{3}$
$\frac{3\pi}{4}$	II	$\frac{\pi}{4}$	$\sin \frac{3\pi}{4} = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$
$-120^\circ$	III	$60^\circ$	$\sec(-120^\circ) = -\sec 60^\circ = -2$

For the last row, $-120^\circ$ is coterminal with $240^\circ$ , which lies in Quadrant III, where secant is negative.

Check yourself

The point $(5, -12)$ lies on the terminal side of an angle $\theta$ in standard position. Find all six trigonometric values of $\theta$ .

Show the answer

The radius is $r = \sqrt{25 + 144} = 13$ . With $x = 5$ and $y = -12$ ,

\begin{gathered} \sin \theta = -\frac{12}{13}, \qquad \cos \theta = \frac{5}{13}, \\[0.8em] \tan \theta = -\frac{12}{5}, \end{gathered}

\begin{gathered} \csc \theta = -\frac{13}{12}, \qquad \sec \theta = \frac{13}{5}, \\[0.8em] \cot \theta = -\frac{5}{12}. \end{gathered}

The angle lies in Quadrant IV, where only cosine and secant are positive, which matches.

Find the exact values of $\sin 330^\circ$ and $\tan\left(-\frac{3\pi}{4}\right)$ .

Show the answer

The angle $330^\circ$ lies in Quadrant IV with reference angle $30^\circ$ , and sine is negative there, so $\sin 330^\circ = -\frac{1}{2}$ . The angle $-\frac{3\pi}{4}$ is coterminal with $\frac{5\pi}{4}$ in Quadrant III, with reference angle $\frac{\pi}{4}$ , and tangent is positive there, so $\tan\left(-\frac{3\pi}{4}\right) = 1$ .

Suppose $\cos \theta = -\frac{2}{3}$ and $\sin \theta > 0$ . Find $\sin \theta$ and $\tan \theta$ .

Show the answer

Any terminal point gives the same values, so use the one at distance $r = 3$ from the origin. Since $\cos \theta = \frac{x}{r}$ , its first coordinate is $x = -2$ . The condition $\sin \theta > 0$ makes $y$ positive, so $y = \sqrt{9 - 4} = \sqrt{5}$ and $\theta$ lies in Quadrant II. Then

\begin{gathered} \sin \theta = \frac{\sqrt{5}}{3} \\[0.8em] \text{and} \\[0.4em] \tan \theta = \frac{\sqrt{5}}{-2} = -\frac{\sqrt{5}}{2}. \end{gathered}

General Angles and Reference Angles

Key terms

General angle

Terminal point

Radius

Six trigonometric functions

Excluded denominators

Quadrant signs

Reference angle

Exact values

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus

Resources for this lesson

Unit circle explorer