Topic 3.2B

Sine, Cosine, and Tangent

In 3.2A we placed angles in standard position and measured them in radians. Now we attach numbers to an angle. The terminal ray of every angle crosses a circle centered at the origin at exactly one point, and the coordinates of that point give three important values: the sine, cosine, and tangent of the angle.

9 MIN READ5 IDEAS32 PROBLEMS5 flashcards

Read this first

30 sec

01
$\cos$ follows $x$ . $\sin$ follows $y$ . $\tan$ is positive when $x$ and $y$ share a sign (Quadrants I and iii).
02
Related angles have the same-sized sine and cosine — only the signs change, based on the quadrant.

Practice 32 problems Next: Chapter 3.3 Sine and Cosine Function Values

Three Ratios from One Point

CONCEPT

Sine, cosine, and tangent

Place an angle $\theta$ in standard position and draw a circle of radius $r$ centered at the origin. The terminal ray meets the circle at a point $P(x, y)$ .

$\sin \theta$ is the vertical displacement of $P$ from the x-axis divided by the distance from the origin to $P$ .

$\cos \theta$ is the horizontal displacement of $P$ from the y-axis divided by the distance from the origin to $P$ .

$\tan \theta$ is the slope of the terminal ray: the vertical displacement divided by the horizontal displacement.

$x$ and $y$ are signed displacements. $r$ is a distance, so it is always positive.

\sin\theta=\frac{y}{r}\qquad\cos\theta=\frac{x}{r}\qquad\tan\theta=\frac{y}{x}\quad(x\ne0)

In Quadrant I, $x$ , $y$ , and $r$ are the sides of a right triangle, so these are the same ratios you may know as SOH-CAH-TOA. The circle version is more powerful: it works in every quadrant, because $x$ and $y$ are allowed to be negative.

Why doesn't the size of the circle matter? Just like radian measure in 3.2A: a bigger circle stretches $x$ , $y$ , and $r$ by the same factor, so every ratio stays the same.

Worked example

Example 1. The terminal ray of an angle $\theta$ in standard position passes through the point $P(-8, 15)$ . Find $\sin \theta$ , $\cos \theta$ , and $\tan \theta$ .

01
Find $r$ , the distance from the origin to $P$ , with the Pythagorean theorem: $r = \sqrt{(-8)^{2} + 15^{2}} = \sqrt{64 + 225} = \sqrt{289} = 17$ .
02
$\sin \theta = y \div r = 15/17$ .
03
$\cos \theta = x \div r = -8/17$ . The negative sign stays: $P$ is 8 units to the LEFT of the y-axis.
04
$\tan \theta = y \div x = 15 \div (-8) = -15/8$ . Check: the terminal ray rises as it goes left, so its slope should be negative. ✓
Try a bigger circle: the same ray also passes through $(-16, 30)$ , which is 34 units from the origin. Then $\sin \theta = 30/34 = 15/17$ and $\cos \theta = -16/34 = -8/17$ — exactly the same values.

COMMON MISTAKE

" $\cos \theta = 8/17$ , because lengths can't be negative."

$r$ is a distance, so it is always positive. But $x$ and $y$ are displacements — they carry a direction. $P$ is to the left of the y-axis, so its horizontal displacement is −8, and $\cos \theta = -8/17$ .

Dropping the sign would describe a point in Quadrant I, which belongs to a completely different angle.

Quick check

The terminal ray of $\theta$ meets the circle of radius $13$ at $(-5, 12)$ . What is $\cos\theta$ ?

The Unit Circle

The simplest circle to use is the unit circle: the circle of radius 1 centered at the origin. With $r = 1$ , the division disappears.

CONCEPT

Sine and cosine on the unit circle

If the terminal ray of $\theta$ meets the unit circle at $P(x, y)$ , then $\cos \theta = x$ and $\sin \theta = y$ . In other words, the point itself is

P=(\cos\theta,\ \sin\theta)

Cosine comes first, just as $x$ comes first in $(x, y)$ . Since $\tan \theta$ is y ÷ x, it can also be written using sine and cosine:

\tan\theta=\frac{y}{x}=\frac{\sin\theta}{\cos\theta}

On the unit circle, the coordinates of $P$ are the cosine and the sine of $\theta$ .

Worked example

Example 2. The terminal ray of $\theta$ meets the unit circle at $P(0.96, -0.28)$ . Find $\sin \theta$ , $\cos \theta$ , and $\tan \theta$ , and name the quadrant of $\theta$ .

01
On the unit circle, read the coordinates directly: $\cos \theta = 0.96$ and $\sin \theta = -0.28$ .
02
$\tan \theta = \sin \theta \div \cos \theta = -0.28 \div 0.96 = -7/24 \approx -0.292$ .
03
$x > 0$ and $y < 0$ , so the terminal ray lies in Quadrant IV.
Every point on the unit circle is exactly 1 unit from the origin, so neither coordinate can be bigger than 1 or smaller than −1. That means $-1 \le \sin \theta \le 1$ and $-1 \le \cos \theta \le 1$ for every angle $\theta$ . Tangent has no such limit: a steep terminal ray can have a very large slope.

The quadrantal angles

When the terminal ray lies on an axis, $P$ is one of the four points where the unit circle crosses the axes:

θ	0	π/2	π	3π/2
P	(1, 0)	(0, 1)	(−1, 0)	(0, −1)
cos θ	1	0	−1	0
sin θ	0	1	0	−1
tan θ	0	undefined	0	undefined

COMMON MISTAKE

" $\tan (\pi /2) = 0$ ."

At $\theta = \pi /2$ , $P = (0, 1)$ , so $\tan \theta = 1 \div 0$ — division by zero. The terminal ray is vertical, and a vertical line has no slope, so $\tan (\pi /2)$ is undefined.

Compare $\theta = \pi$ : $P = (-1, 0)$ , so $\tan \theta = 0 \div (-1) = 0$ . A horizontal ray has slope 0. Zero on top gives 0; zero on the bottom gives undefined.

Signs in Each Quadrant

On the unit circle, $\cos \theta$ is $x$ and $\sin \theta$ is $y$ , so their signs come straight from the quadrant. And $\tan \theta = y \div x$ is positive exactly when $x$ and $y$ have the same sign.

Signs of sine, cosine, and tangent in each quadrant.

KEY RULE

$\cos$ follows $x$ . $\sin$ follows $y$ . $\tan$ is positive when $x$ and $y$ share a sign (Quadrants I and III).

Worked example

Example 3. An angle $\theta$ satisfies $\sin \theta < 0$ and $\tan \theta > 0$ . In which quadrant is the terminal ray of $\theta$ ?

01
$\sin \theta < 0$ means $y < 0$ , so $\theta$ is in Quadrant III or Quadrant IV.
02
$\tan \theta > 0$ means $x$ and $y$ have the same sign. Since $y < 0$ , $x$ must also be negative.
03
$x < 0$ and $y < 0$ , so the terminal ray lies in Quadrant III.

Quick check

In which quadrant is $\sin\theta < 0$ and $\tan\theta > 0$ ?

Reflecting a Point on the Unit Circle

The symmetry of the circle turns one point into four. Suppose the terminal ray of $\theta$ meets the unit circle at P( $20/29$ , $21/29$ ) in Quadrant I. Reflecting $P$ across the axes produces three more points on the unit circle, each belonging to a related angle.

Reflections change the signs of the coordinates, never their sizes.

Point	How it is made	Angle	cos	sin
P	original point	θ	20/29	21/29
Q	reflect P across the y-axis	π − θ	−20/29	21/29
R	reflect P through the origin	π + θ	−20/29	−21/29
S	reflect P across the x-axis	−θ	20/29	−21/29

Why those angles? Reflecting across the y-axis makes the ray sit $\theta$ radians short of the negative x-axis, which is $\pi - \theta$ . Reflecting through the origin points the ray in exactly the opposite direction — half a turn more — which is $\pi + \theta$ . Reflecting across the x-axis reverses the direction of rotation, which is $-\theta$ .

COMMON MISTAKE

"Q is the reflection across the y-axis, so change the sign of $y$ ."

Reflecting across the y-axis moves a point from the right side of the y-axis to the left side — so it is $x$ that changes sign, not $y$ . Q = (−20/29, $21/29$ ).

Quick check: the reflected point must land in the quadrant you expect. $Q$ should be in Quadrant II, where $x < 0$ and $y > 0$ . ✓

KEY RULE

Related angles have the same-sized sine and cosine — only the signs change, based on the quadrant.

Common slips

" $\cos \theta = 8/17$ , because lengths can't be negative."

$r$ is a distance, so it is always positive. But $x$ and $y$ are displacements — they carry a direction. $P$ is to the left of the y-axis, so its horizontal displacement is −8, and $\cos \theta = -8/17$ .

Dropping the sign would describe a point in Quadrant I, which belongs to a completely different angle.
" $\tan (\pi /2) = 0$ ."

At $\theta = \pi /2$ , $P = (0, 1)$ , so $\tan \theta = 1 \div 0$ — division by zero. The terminal ray is vertical, and a vertical line has no slope, so $\tan (\pi /2)$ is undefined.

Compare $\theta = \pi$ : $P = (-1, 0)$ , so $\tan \theta = 0 \div (-1) = 0$ . A horizontal ray has slope 0. Zero on top gives 0; zero on the bottom gives undefined.
"Q is the reflection across the y-axis, so change the sign of $y$ ."

Reflecting across the y-axis moves a point from the right side of the y-axis to the left side — so it is $x$ that changes sign, not $y$ . Q = (−20/29, $21/29$ ).

Quick check: the reflected point must land in the quadrant you expect. $Q$ should be in Quadrant ii, where $x < 0$ and $y > 0$ . ✓

Lock it in

Try the flashcards

5 cards · Unit circle

Start

Recap card

6 lines to re-read the night before.

01
If the terminal ray of $\theta$ meets a circle of radius $r$ at $P(x, y)$ , then $\sin \theta = y/r$ , $\cos \theta = x/r$ , and $\tan \theta = y/x$ .
02
$x$ and $y$ are signed displacements; $r$ is a distance and is always positive. The ratios do not depend on the size of the circle.
03
On the unit circle, $P = (\cos \theta , \sin \theta )$ and $\tan \theta = \sin \theta \div \cos \theta$ .
04
$-1 \le \sin \theta \le 1$ and $-1 \le \cos \theta \le 1$ . $\tan \theta$ is undefined when the terminal ray is vertical.
05
$\cos$ follows the sign of $x$ , $\sin$ follows the sign of $y$ , and $\tan$ is positive in Quadrants I and iii.
06
Reflecting a point across an axis or through the origin gives the angles $\pi - \theta$ , $\pi + \theta$ , and $-\theta$ , with the same-sized sine and cosine.

Practice 32 problems Next: Chapter 3.3 Sine and Cosine Function Values

Unlock every question

Free

Premium

Sign in to FiveWay

Sine, Cosine, and Tangent

Three Ratios from One Point

The Unit Circle

The quadrantal angles

Signs in Each Quadrant

Reflecting a Point on the Unit Circle

Recap card

Unlock Premium