Sine, Cosine, and Tangent — AP® Precalculus Notes

In a right triangle, sine and cosine only make sense for angles between $0^\circ$ and $90^\circ$ . To model a turning wheel or a tide you need them for every angle, including angles bigger than a full turn and negative angles. The fix is to put the angle on a circle. This page sets up the three ideas the rest of the unit is built on: radians, the unit circle, and sine, cosine and tangent as coordinates and a slope.

Radians: measuring angles by arcs

An angle at the center of a circle cuts off an arc. The bigger the circle, the longer the arc, but the ratio arc length over radius depends only on the angle. That ratio is the angle's radian measure:

$\theta = \frac{s}{r}$ (radians), so $s = r\theta$ .

A full turn cuts off the whole circumference: $\frac{2\pi r}{r} = 2\pi$ radians. So $360^\circ = 2\pi$ and

$\pi \text{ radians} = 180^\circ$ .

One radian is the angle whose arc is exactly one radius long, about $57.3^\circ$ . Calculus and this course use radians by default: when an angle has no degree sign, it is in radians.

Worked example 1. Convert $150^\circ$ to radians and $\frac{7\pi}{4}$ to degrees.

Multiply degrees by $\frac{\pi}{180}$ : $150 \cdot \frac{\pi}{180} = \frac{5\pi}{6}$ .

Multiply radians by $\frac{180}{\pi}$ : $\frac{7\pi}{4} \cdot \frac{180}{\pi} = 315^\circ$ .

Degrees	$0^\circ$	$30^\circ$	$45^\circ$	$60^\circ$	$90^\circ$	$180^\circ$	$270^\circ$	$360^\circ$
Radians	$0$	$\frac{\pi}{6}$	$\frac{\pi}{4}$	$\frac{\pi}{3}$	$\frac{\pi}{2}$	$\pi$	$\frac{3\pi}{2}$	$2\pi$

Angles in standard position

An angle is in standard position when its vertex is at the origin and its initial ray lies along the positive $x$ -axis. It turns to its terminal ray: counterclockwise for a positive angle, clockwise for a negative one. Turning more than $2\pi$ just goes around again, so $\theta$ and $\theta + 2\pi$ have the same terminal ray. So do $\theta$ and $\theta - 2\pi$ , $\theta + 4\pi$ , and so on.

Sine and cosine on the unit circle

The unit circle is the circle of radius 1 centered at the origin, $x^2 + y^2 = 1$ . The terminal ray of $\theta$ meets it at exactly one point $P$ . By definition,

$\cos\theta = x$ -coordinate of $P$ , $\sin\theta = y$ -coordinate of $P$ , so $P = (\cos\theta, \sin\theta)$ .

A circle of radius 1 centered at the origin. An angle theta in standard position starts on the positive x-axis (the initial ray) and turns counterclockwise to the terminal ray, which meets the circle at P. The arc from (1, 0) to P, drawn in green, has length theta, because the radius is 1. A dashed vertical segment from P down to the x-axis has length sine of theta; the horizontal distance from the origin is cosine of theta. So P is (cos theta, sin theta). The slope of the terminal ray is sin theta over cos theta, which is tangent of theta. — **Figure 1.** The point at angle $\theta$ , its coordinates, and the arc of length $\theta$ . LevlPrep original diagram.

For an acute angle this matches the triangle definition: drop a segment from $P$ to the $x$ -axis and you get a right triangle with hypotenuse 1, so opposite $= \sin\theta$ and adjacent $= \cos\theta$ . The circle version also works for angles the triangle cannot handle. Because the radius is 1, the angle $\theta$ in radians is also the length of the arc from $(1, 0)$ to $P$ : sine and cosine turn an arc length into a coordinate.

The four points on the axes give values you should know at a glance:

$\theta$	Point $P$	$\cos\theta$	$\sin\theta$
$0$	$(1, 0)$	$1$	$0$
$\frac{\pi}{2}$	$(0, 1)$	$0$	$1$
$\pi$	$(-1, 0)$	$-1$	$0$
$\frac{3\pi}{2}$	$(0, -1)$	$0$	$-1$

The quadrant of $P$ sets the signs. In Quadrant I both are positive; in Quadrant II, $\cos\theta < 0$ and $\sin\theta > 0$ ; in Quadrant III both are negative; in Quadrant IV, $\cos\theta > 0$ and $\sin\theta < 0$ .

Tangent as a slope

The terminal ray goes through the origin and $P$ , so its slope is rise over run, $\frac{\sin\theta}{\cos\theta}$ . That slope is $\tan\theta$ :

$\tan\theta = \frac{\sin\theta}{\cos\theta}$ , defined when $\cos\theta \ne 0$ .

At $\theta = \frac{\pi}{2}$ the ray is vertical, $\cos\theta = 0$ , and $\tan\theta$ is undefined. Any point on the ray gives the same slope, not only $P$ .

Worked example 2. The terminal ray of $\theta$ meets the unit circle at $P\left(-\frac{3}{5}, \frac{4}{5}\right)$ . Find $\sin\theta$ , $\cos\theta$ and $\tan\theta$ , and say where $\theta$ lies if $0 < \theta < 2\pi$ .

Read the coordinates: $\cos\theta = -\frac{3}{5}$ and $\sin\theta = \frac{4}{5}$ . Then $\tan\theta$ $= \frac{4/5}{-3/5}$ $= -\frac{4}{3}$ , the slope of the ray from the origin to $P$ .

$P$ is in Quadrant II, so $\frac{\pi}{2} < \theta < \pi$ .

Worked example 3. A point on the rim of a wheel of radius 30 cm moves 90 cm along the rim. Through what angle has the wheel turned, and where is the point if it started on the positive $x$ -axis and turned counterclockwise?

$\theta = \frac{s}{r} = \frac{90}{30} = 3$ radians. Since $\frac{\pi}{2}$ $\approx 1.57 < 3 < \pi$ $\approx 3.14$ , the point is in Quadrant II: left of the center and above it.

Why this makes sine and cosine periodic

Adding $2\pi$ to an angle brings its terminal ray back to the same place, so

$\sin(\theta + 2\pi) = \sin\theta$ and $\cos(\theta + 2\pi) = \cos\theta$ .

So sine and cosine are periodic, with period $2\pi$ . The next topic finds their exact values at the angles you meet most, and the one after draws their graphs.