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Sine & Cosine Functions (College Board AP® Precalculus): Revision Note

Written by: Roger B

Reviewed by: Mark Curtis

Updated on 3 July 2026

Sine & cosine function values

What are the sine and cosine functions?

The sine function is defined as $f (θ) = \sin θ$
- The domain of the sine function is all real numbers
  - I.e. any real number can be used as an input angle measure
- The range of the sine function is $[- 1, 1]$
  - I.e., the function's output value oscillates between -1 and 1 as the input values vary
The cosine function is defined as $f (θ) = \cos θ$
- The domain of the cosine function is all real numbers
  - I.e. any real number can be used as an input angle measure
- The range of the cosine function is $[- 1, 1]$
  - I.e., the function's output value oscillates between -1 and 1 as the input values vary

What is the connection between sine and cosine function values and a circle centered at the origin?

Given an angle $θ$ in standard position and a circle with radius $r$ centered at the origin
- the terminal ray for the angle will intersect the circle at a point $P$
- The coordinates of $P$ are $(r \cos θ, r \sin θ)$
  - The $x$ -coordinate is $r \cos θ$
  - The $y$ -coordinate is $r \sin θ$
- For the unit circle ( $r = 1$ ), these coordinates simplify to $(\cos θ, \sin θ)$

Diagram of a circle with radius r on an x-y plane, showing angle θ, point P at (r cos θ, r sin θ). The point (r, 0) on the x-axis is also marked. — **Coordinates of a point P on a circle centered at the origin**

How can I use triangles to find values of sine and cosine for special angles?

The two special right-angled triangles below can help you to find the exact values of sine and cosine for 30º $(= \frac{π}{6} radians)$ , 45º $(= \frac{π}{4} radians)$ and 60º $(= \frac{π}{3} radians)$
- Pythagoras' theorem and SOHCAHTOA are used to work out the values
- Remember that you can rationalize the denominator in a fraction like $\frac{1}{\sqrt{2}}$
  - I.e. $\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$

Triangle for 45°

Diagram of an isosceles right triangle with sides 1, hypotenuse √2, angle 45°. Shows sin and cos 45° as 1/√2, using Pythagoras.

So $\cos \frac{π}{4} = \sin \frac{π}{4} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$

Triangle for 30° and 60°

Diagram of a 30-60-90 triangle with sides 1, 2, and √3. Includes sine and cosine values for 30° and 60° angles, derived using Pythagoras' theorem.

So $\cos \frac{π}{6} = \frac{\sqrt{3}}{2}$ , $\sin \frac{π}{6} = \frac{1}{2}$ , $\cos \frac{π}{3} = \frac{1}{2}$ , $\sin \frac{π}{3} = \frac{\sqrt{3}}{2}$

Using the unit circle

How can I use the unit circle to find additional values for sine and cosine?

Given an angle $θ$ in standard position and a unit circle with radius $r$ centered at the origin
- we saw above that the terminal ray for the angle will intersect the circle at a point $P$ with coordinates $(\cos θ, \sin θ)$
  - The $x$ -coordinate is $\cos θ$
  - The $y$ -coordinate is $\sin θ$
Using the unit circle allows you to find additional values for sine and cosine
For example
- When $θ = 0$ (0°), the point $P$ is on the positive $x$ -axis with coordinates $(1, 0)$
  - so $\cos 0 = 1, \sin 0 = 0$
- When $θ = \frac{π}{2}$ (90°), the point $P$ is on the positive $y$ -axis with coordinates $(0, 1)$
  - so $\cos \frac{π}{2} = 0, \sin \frac{π}{2} = 1$
- When $θ = π$ (180°), the point $P$ is on the negative $x$ -axis with coordinates $(- 1, 0)$
  - so $\cos π = - 1, \sin π = 0$
- When $θ = \frac{3 π}{2}$ (270°), the point $P$ is on the negative $y$ -axis with coordinates $(0, - 1)$
  - so $\cos \frac{3 π}{2} = 0, \sin \frac{3 π}{2} = - 1$
Other values can be found using the symmetries of the unit circle
- and known values of sine and cosine for angles between 0 and $\frac{π}{2}$

Unit circle with centre O, angle θ at origin; points P(x, y), (1, 0), (-x, y), (x, -y), (-x, -y); axes x, y; right triangles inside. — **Symmetries of the unit circle**

For example, we saw above that $\cos \frac{π}{6} = \frac{\sqrt{3}}{2}$ and $\sin \frac{π}{6} = \frac{1}{2}$
- So the coordinates of point $P$ in the diagram would be $(\frac{\sqrt{3}}{2}, \frac{1}{2})$
- Using the other places in the diagram where an angle of size $θ$ appears, this allows you to determine that
  - $\cos (π - \frac{π}{6}) = \cos \frac{5 π}{6} = - \frac{\sqrt{3}}{2}$ , $\sin (π - \frac{π}{6}) = \sin \frac{5 π}{6} = \frac{1}{2}$
  - $\cos (π + \frac{π}{6}) = \cos \frac{7 π}{6} = - \frac{\sqrt{3}}{2}$ , $\sin (π + \frac{π}{6}) = \sin \frac{7 π}{6} = - \frac{1}{2}$
  - $\cos (2 π - \frac{π}{6}) = \cos \frac{11 π}{6} = \frac{\sqrt{3}}{2}$ , $\sin (2 π - \frac{π}{6}) = \sin \frac{11 π}{6} = - \frac{1}{2}$
- Remember that
  - A terminal ray along the negative $x$ -axis has an angle in standard position of $π$ radians
  - A terminal ray along the positive $x$ -axis has an angle in standard position of 0 or $2 π$ radians

What are the exact values of sine and cosine for all key angles?

The process above can also be applied to the results for sine and cosine of $\frac{π}{4}$ and $\frac{π}{3}$
- The results for all the key angles between 0 and 2π are shown in the following table

$θ$	$\cos θ$	$\sin θ$
$0$	$1$	$0$
$\frac{π}{6}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$
$\frac{π}{4}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{2}}{2}$
$\frac{π}{3}$	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$
$\frac{π}{2}$	$0$	$1$
$\frac{2 π}{3}$	$- \frac{1}{2}$	$\frac{\sqrt{3}}{2}$
$\frac{3 π}{4}$	$- \frac{\sqrt{2}}{2}$	$\frac{\sqrt{2}}{2}$
$\frac{5 π}{6}$	$- \frac{\sqrt{3}}{2}$	$\frac{1}{2}$
$π$	$- 1$	$0$
$\frac{7 π}{6}$	$- \frac{\sqrt{3}}{2}$	$- \frac{1}{2}$
$\frac{5 π}{4}$	$- \frac{\sqrt{2}}{2}$	$- \frac{\sqrt{2}}{2}$
$\frac{4 π}{3}$	$- \frac{1}{2}$	$- \frac{\sqrt{3}}{2}$
$\frac{3 π}{2}$	$0$	$- 1$
$\frac{5 π}{3}$	$\frac{1}{2}$	$- \frac{\sqrt{3}}{2}$
$\frac{7 π}{4}$	$\frac{\sqrt{2}}{2}$	$- \frac{\sqrt{2}}{2}$
$\frac{11 π}{6}$	$\frac{\sqrt{3}}{2}$	$- \frac{1}{2}$
$2 π$	$1$	$0$

These values can also all be seen on the following diagram of the unit circle

Unit circle diagram with angles in degrees and radians, showing cosine and sine values at key points, with coordinates for each angle marked. — **Exact values of angles which are multiples of π/6 and π/4**

Examiner Tips and Tricks

Make sure you know the exact values of sine and cosine for the key angles (i.e. multiples of $\frac{π}{6}$ and $\frac{π}{4}$ ).

These come up frequently, and the non-calculator section of the exam expects you to be able to work with them quickly

Understanding how these values come from the special triangles and the unit circle (rather than just memorizing the table) will help you reconstruct any value you forget under exam pressure.

Sine & cosine function graphs

What do the graphs of the sine and cosine functions look like?

As the input values (angle measures) increase
- the output values of both the sine and cosine functions oscillate between $- 1$ and $1$
  - taking every value in between
This oscillating behavior comes directly from the unit circle
- For sine, the output tracks the vertical displacement of points on the unit circle from the $x$ -axis as the angle increases
- For cosine, the output tracks the horizontal displacement of points on the unit circle from the $y$ -axis as the angle increases
Key features of both graphs
- The output values are always in the range $[- 1, 1]$
- Both functions are periodic with a period of $2 π$
- The graphs are smooth, continuous wave shapes
The domain of both functions is all real numbers
- so the graphs can be extended to negative values of the angle
  - this corresponds to an angle measured clockwise from the positive $x$ -axis on the unit circle
- or values of the angle greater than $2 π$
  - this corresponds to what happens if you 'keep going' counter-clockwise after making one complete revolution

Graph of the sine function, y = sin(θ), with marked points from -2π to 4π on the x-axis and range from -1 to 1 on the y-axis. — **Graph of y=sinθ**

Graph of the cosine function, y = cos(θ), with marked points from -2π to 4π on the x-axis and range from -1 to 1 on the y-axis. — **Graph of y=cosθ**

Note that the two graphs are horizontal translations of each other
- For example, shifting the graph of sine to the left by $\frac{π}{2}$ units gives the graph of cosine
- Or shifting the graph of cosine to the right by $\frac{π}{2}$ units gives the graph of sine

How can the graph of the sine function be understood from the unit circle?

The behavior of the sine graph can be understood by considering the unit circle
- Starting at $θ = 0$
  - the point on the unit circle is $(1, 0)$ , so $\sin 0 = 0$
- As $θ$ increases from $0$ to $\frac{π}{2}$
  - the $y$ -coordinate increases from $0$ to $1$
    - so sine increases to its maximum value of $1$
- From $\frac{π}{2}$ to $π$
  - the $y$ -coordinate decreases from $1$ back to $0$
    - so sine decreases back to $0$
- From $π$ to $\frac{3 π}{2}$
  - the $y$ -coordinate decreases from $0$ to $- 1$
    - so sine reaches its minimum value of $- 1$
- From $\frac{3 π}{2}$ to $2 π$
  - the $y$ -coordinate increases from $- 1$ back to $0$
    - completing one full cycle

Unit circle diagram connected to a sine wave on a graph, showing angles and sine values at key points from 0 to 2π on the x-axis. — **Construction of the sine graph from the unit circle**

How can the graph of the cosine function be understood from the unit circle?

The behavior of the cosine graph can also be understood by considering the unit circle
- Starting at $θ = 0$
  - the point on the unit circle is $(1, 0)$ , so $\cos 0 = 1$
    - the cosine function starts at its maximum
- As $θ$ increases from $0$ to $\frac{π}{2}$
  - the $x$ -coordinate decreases from $1$ to $0$
    - so cosine decreases back to 0
- From $\frac{π}{2}$ to $π$
  - the $x$ -coordinate decreases from $0$ to $- 1$
    - so cosine reaches its minimum value of $- 1$
- From $π$ to $\frac{3 π}{2}$
  - the $x$ -coordinate increases from $- 1$ back to $0$
    - so cosine increases back to 0
- From $\frac{3 π}{2}$ to $2 π$
  - the $x$ -coordinate increases from $0$ back to $1$
    - completing one full cycle

Unit circle diagram with x-coordinates mapped to y-coordinates on a cosine graph, showing angles 0 to 2π. Arrows indicate coordinate transitions. — **Construction of the cosine graph from the unit circle**

Worked Example

A circle centered at the origin has a radius of $10$ . An angle of measure $\frac{5 π}{4}$ radians is in standard position, and its terminal ray intersects the circle at point $Q$ .

(a) Find the exact coordinates of point $Q$ .

(b) Is the sine function increasing or decreasing at $θ = \frac{5 π}{4}$ ? Justify your answer by considering the unit circle.

Answer:

(a)

The coordinates of $Q$ are $(r \cos θ, r \sin θ)$

where $r = 10$ and $θ = \frac{5 π}{4}$
If you don't remember the values of sine and cosine for $\frac{5 π}{4}$ , you can work them out using the values for $\frac{π}{4}$ and the symmetries of the unit circle

$10 \cos \frac{5 π}{4} = 10 (- \frac{\sqrt{2}}{2}) = - 5 \sqrt{2}$

$10 \sin \frac{5 π}{4} = 10 (- \frac{\sqrt{2}}{2}) = - 5 \sqrt{2}$

So the coordinates are

$Q (- 5 \sqrt{2}, - 5 \sqrt{2})$

(b)

At $θ = \frac{5 π}{4}$ , the point on the unit circle is in the third quadrant III

As $θ$ increases from $π$ to $\frac{3 π}{2}$ , the point on the unit circle moves from $(- 1, 0)$ toward $(0, - 1)$
so the $y$ -coordinate is decreasing (becoming more negative).

Since the sine function tracks the $y$ -coordinate of the point on the unit circle as the angle increases in the counter-clockwise direction, and the point $Q$ is in the third quadrant, the sine function is decreasing at $θ = \frac{5 π}{4}$

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