Math Core

In a right triangle, sine, cosine and tangent are ratios of sides, so they only make sense for acute angles. But angles like $150^\circ$ , $270^\circ$ or $-45^\circ$ show up all the time in rotation, waves and circular motion. In this lesson you'll redefine the trig functions using coordinates, so they work for any angle.

From triangles to coordinates

Put an angle $\theta$ in standard position: vertex at the origin, initial side along the positive $x$ -axis, and terminal side rotated counterclockwise (or clockwise, for a negative angle). Pick any point $P(x, y)$ on the terminal side, other than the origin. Its distance from the origin is

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

The distance $r$ is always positive. The coordinates $x$ and $y$ can be positive, negative or zero, depending on where the terminal side points.

The terminal side of θ passes through P(-3, 4), which is r = 5 units from the origin.

If $\theta$ is acute, $P$ is in Quadrant I, and dropping a perpendicular to the $x$ -axis makes a right triangle with legs $x$ and $y$ and hypotenuse $r$ . Then "opposite over hypotenuse" is $\dfrac{y}{r}$ , "adjacent over hypotenuse" is $\dfrac{x}{r}$ , and "opposite over adjacent" is $\dfrac{y}{x}$ . The new definitions simply keep these formulas and let $x$ and $y$ carry signs.

Definition

Trig functions of any angle

Let $\theta$ be an angle in standard position, and let $P(x, y)$ be any point on its terminal side other than the origin, with $r = \sqrt{x^2 + y^2}$ . Then

\sin\theta = \frac{y}{r} \qquad \cos\theta = \frac{x}{r} \qquad \tan\theta = \frac{y}{x}\ (x \ne 0)

\csc\theta = \frac{r}{y}\ (y \ne 0) \qquad \sec\theta = \frac{r}{x}\ (x \ne 0) \qquad \cot\theta = \frac{x}{y}\ (y \ne 0)

Does it matter which point you pick? No. Any other point on the same terminal side is $(kx, ky)$ for some $k > 0$ , and its distance is $kr$ . Every ratio has a factor of $k$ on top and bottom, which cancels. So the values depend only on the angle.

Worked example: A point in Quadrant II

The terminal side of $\theta$ passes through $(-3, 4)$ . Find all six trig functions of $\theta$ .

First find $r$ :

r = \sqrt{(-3)^2 + 4^2} = \sqrt{25} = 5.

Now use $x = -3$ , $y = 4$ , $r = 5$ :

\sin\theta = \frac{4}{5}, \quad \cos\theta = -\frac{3}{5}, \quad \tan\theta = -\frac{4}{3}, \quad \csc\theta = \frac{5}{4}, \quad \sec\theta = -\frac{5}{3}, \quad \cot\theta = -\frac{3}{4}.

Notice that cosine is negative here. That's impossible in a right triangle, but perfectly normal for an angle that opens past $90^\circ$ .

Worked example: When r is not a whole number

The terminal side of $\theta$ passes through $(2, -5)$ . Find $\sin\theta$ , $\cos\theta$ and $\tan\theta$ .

r = \sqrt{2^2 + (-5)^2} = \sqrt{29}.

\sin\theta = \frac{-5}{\sqrt{29}} = -\frac{5\sqrt{29}}{29}, \qquad \cos\theta = \frac{2}{\sqrt{29}} = \frac{2\sqrt{29}}{29}, \qquad \tan\theta = \frac{-5}{2} = -\frac{5}{2}.

Rationalizing the denominator is the usual final form, but $-\dfrac{5}{\sqrt{29}}$ is the same number.

Common mistake

$r$ is a distance, so it is never negative, even when both coordinates are negative. For the point $(-6, -8)$ , $r = \sqrt{36 + 64} = 10$ , not $-10$ . All the signs in the answers come from $x$ and $y$ .

The unit circle

Since any point on the terminal side works, choose the one that makes the arithmetic easiest: the point where the terminal side crosses the circle of radius $1$ centered at the origin. This circle, $x^2 + y^2 = 1$ , is the unit circle.

On the unit circle, $r = 1$ , so the formulas collapse:

\sin\theta = \frac{y}{1} = y, \qquad \cos\theta = \frac{x}{1} = x.

On the unit circle, the point on the terminal side of θ is (cos θ, sin θ).

Coordinates on the unit circle

If the terminal side of $\theta$ meets the unit circle at $P$ , then

P = (\cos\theta, \sin\theta), \qquad \tan\theta = \frac{\sin\theta}{\cos\theta}.

Cosine is the $x$ -coordinate and sine is the $y$ -coordinate. Since every point on the unit circle has coordinates between $-1$ and $1$ , so do sine and cosine.

This picture is the heart of the rest of the course. As $\theta$ grows, $P$ travels around the circle, and $\cos\theta$ and $\sin\theta$ are just its shadows on the two axes.

Quadrantal angles

An angle whose terminal side lies on an axis is called a quadrantal angle: $0^\circ$ , $90^\circ$ , $180^\circ$ , $270^\circ$ and any angle coterminal with them. Their unit circle points are easy to read off, and some of their trig values are undefined because a denominator is $0$ .

$\theta$	point	$\sin\theta$	$\cos\theta$	$\tan\theta$
$0^\circ$ or $0$	$(1, 0)$	$0$	$1$	$0$
$90^\circ$ or $\frac{\pi}{2}$	$(0, 1)$	$1$	$0$	undefined
$180^\circ$ or $\pi$	$(-1, 0)$	$0$	$-1$	$0$
$270^\circ$ or $\frac{3\pi}{2}$	$(0, -1)$	$-1$	$0$	undefined

You don't need to memorize the table. Picture the point, then apply the definitions.

Worked example: A quadrantal angle

Find all six trig functions of $\theta = 270^\circ$ .

The terminal side points straight down, so it meets the unit circle at $(0, -1)$ . With $x = 0$ , $y = -1$ , $r = 1$ :

\sin 270^\circ = -1, \qquad \cos 270^\circ = 0, \qquad \csc 270^\circ = \frac{1}{-1} = -1, \qquad \cot 270^\circ = \frac{0}{-1} = 0.

Tangent and secant both divide by $x = 0$ , so $\tan 270^\circ$ and $\sec 270^\circ$ are undefined.

Tip

"Undefined" and "zero" are different. A trig function is $0$ when its numerator is $0$ , and undefined when its denominator is $0$ . For $\cot 270^\circ = \dfrac{x}{y} = \dfrac{0}{-1}$ , the zero is on top, so the value is $0$ .

Worked example: Finding a missing coordinate

The terminal side of $\theta$ meets the unit circle at a point in Quadrant III with $y$ -coordinate $-\dfrac{5}{13}$ . Find $\cos\theta$ .

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

x^2 + \frac{25}{169} = 1 \quad\Rightarrow\quad x^2 = \frac{144}{169} \quad\Rightarrow\quad x = \pm\frac{12}{13}.

In Quadrant III, $x$ is negative, so $\cos\theta = x = -\dfrac{12}{13}$ .