AP®MathsCollege BoardPrecalculusRevision NotesTrigonometric & Polar FunctionsTrigonometric FunctionsSine, Cosine & Tangent in Terms of Standard Position

Sine, Cosine & Tangent in Terms of Standard Position (College Board AP® Precalculus): Revision Note

Written by: Roger B

Reviewed by: Mark Curtis

Updated on 15 April 2026

Sine, cosine & tangent in terms of standard position

How are sine, cosine, and tangent connected to right triangle trigonometry?

From previous courses, you may have learned sine, cosine, and tangent as ratios of side lengths in a right triangle (SOHCAHTOA)
- $\sin θ = \frac{opposite}{hypotenuse}$ , $\cos θ = \frac{adjacent}{hypotenuse}$ , $\tan θ = \frac{opposite}{adjacent}$
In this course, these definitions are extended
- using angles in standard position
- and circles centered at the origin
This more general approach allows sine, cosine, and tangent to be defined for any angle
- not just acute angles in a right triangle
However the two approaches are fully consistent
- The right triangle definitions are a special case of the standard position definitions

What is the sine of an angle in standard position?

Given an angle $θ$ in standard position and a circle centered at the origin
- the terminal ray of the angle
- intersects the circle at a point $P$
The sine of the angle is the ratio of
- the vertical displacement of $P$ from the $x$ -axis
- to the distance between the origin and $P$ :

$\sin θ = \frac{vertical displacement of P from the x -axis}{distance from the origin to P}$

For a circle of radius $r$ , this means
- $\sin θ = \frac{y}{r}$
  - where $y$ is the $y$ -coordinate of point $P$
  - and $r$ is the radius of the circle
For a unit circle (where $r = 1$ ), this simplifies to:
- $\sin θ = y$
  - i.e. the sine of the angle is simply the $y$ -coordinate of the point where the terminal ray meets the unit circle

What is the cosine of an angle in standard position?

For the same angle $θ$ in standard position, the cosine of the angle is the ratio of
- the horizontal displacement of $P$ from the $y$ -axis
- to the distance between the origin and $P$ :

$\cos θ = \frac{horizontal displacement of P from the y -axis}{distance from the origin to P}$

For a circle of radius $r$
- $\cos θ = \frac{x}{r}$
  - where $x$ is the $x$ -coordinate of point $P$
  - and $r$ is the radius of the circle

For a unit circle (where $r = 1$ )
- $\cos θ = x$
  - i.e. the cosine of the angle is simply the $x$ -coordinate of the point where the terminal ray meets the unit circle

Diagram of a unit circle with trigonometric functions. Includes labels for sine, cosine, tangent, and triangle side definitions. Circle centre is (0,0). — **Values of sine, cosine and tangent on a unit circle**

What is the tangent of an angle in standard position?

The tangent of an angle $θ$ in standard position is the slope of the terminal ray
- when the slope exists
Because the slope of a line through the origin is the ratio of the $y$ -coordinate to the $x$ -coordinate for any point on the line
- $\tan θ = \frac{y}{x}$
  - where $(x, y)$ is any point on the terminal ray (other than the origin)

This is equivalent to:
- $\tan θ = \frac{\sin θ}{\cos θ}$
  - provided that $\cos θ \neq 0$

The tangent is undefined when the terminal ray is vertical (i.e. when $\cos θ = 0$ )
- This occurs at $θ = \frac{π}{2} + k π$ for integer values of $k$
  - I.e. when $θ$ is equal to $. . ., - \frac{5 π}{2}, - \frac{3 π}{2}, - \frac{π}{2}, \frac{π}{2}, \frac{3 π}{2}, \frac{5 π}{2}, . . .$

How do the signs of sine, cosine, and tangent depend on the quadrant?

The signs of the $x$ - and $y$ -coordinates of point $P$ change depending on which quadrant the terminal ray lies in
- This means the signs of sine, cosine, and tangent also vary by quadrant

Quadrant	$x$	$y$	$\sin θ$	$\cos θ$	$\tan θ$
I	$+$	$+$	$+$	$+$	$+$
II	$-$	$+$	$+$	$-$	$-$
III	$-$	$-$	$-$	$-$	$+$
IV	$+$	$-$	$-$	$+$	$-$

A common mnemonic for remembering which trig ratios are positive in each quadrant is "All Students Take Calculus"
- Quadrant I: All are positive
- Quadrant II: Sine is positive
- Quadrant III: Tangent is positive
- Quadrant IV: Cosine is positive

How does this relate to right triangle trigonometry?

When the terminal ray is in the first quadrant a right triangle can be formed by drawing a vertical line from $P$ down to the $x$ -axis
- The horizontal leg has length $x$
- the vertical leg has length $y$
- and the hypotenuse has length $r$
The standard position definitions produce the same ratios as SOH CAH TOA
The advantage of the standard position approach is that it works for any angle
- including angles in the second, third, and fourth quadrants
- as well as angles greater than $360 °$ or negative angles
The signs of the coordinates automatically account for the signs of sine, cosine and tangent

Examiner Tips and Tricks

Pay close attention to whether a question uses the unit circle (radius $1$ ) or a circle with a different radius.

On the unit circle, $\sin θ$ is the $y$ -coordinate and $\cos θ$ is the $x$ -coordinate directly.
For any other circle of radius $r$ , you need to divide by $r$ — that is
- I.e. $\sin θ = \frac{y}{r}$ and $\cos θ = \frac{x}{r}$

Confusing these is a common source of errors.

Worked Example

Diagram showing a circle of radius 7 centered at the origin O, intersected by x and y axes. Angle θ marks the angle of the ray through P on the circumference in the third quadrant. Point R is a reflection of P over the y-axis.

The figure shows a circle centered at the origin with radius 7, and an angle of measure $θ$ radians in standard position. The terminal ray of the angle intersects the circle at point $P$ , and point $R$ also lies on the circle. The coordinates of $P$ are $(x, y)$ , and the coordinates of $R$ are $(- x, y)$ . Which of the following is true about the cosine of $θ$ ?

(A) $\cos θ = \frac{- x}{7}$ , because it is the ratio of the horizontal displacement of $R$ from the $y$ -axis to the distance between the origin and $R$ .

(B) $\cos θ = \frac{x}{7}$ , because it is the ratio of the horizontal displacement of $P$ from the $y$ -axis to the distance between the origin and $P$ .

(C) $\cos θ = \frac{y}{7}$ , because it is the ratio of the vertical displacement of $P$ from the $y$ -axis to the distance between the origin and $P$ .

(D) $\cos θ = \frac{y}{7}$ , because it is the ratio of the vertical displacement of $R$ from the $x$ -axis to the distance between the origin and $R$ .

Answer

The cosine of an angle is the ratio of

the horizontal displacement of the point where the terminal ray intersects the circle (point $P$ )
to the distance from the origin to that point (the radius, $7$ )

Since $P$ has coordinates $(x, y)$ , the horizontal displacement of $P$ from the $y$ -axis is $x$

Therefore $\cos θ = \frac{x}{7}$

Note that because $P$ is in the third quadrant, $x$ is negative, so the cosine is negative

This is consistent with cosine being negative in the third quadrant.

(B) $\cos θ = \frac{x}{7}$ , because it is the ratio of the horizontal displacement
of $P$ from the $y$ -axis to the distance between the origin and $P$

Examiner Tips and Tricks

Make sure you are clear about the distinction between distance and displacement in the definitions.

Distance is always positive
Displacement can be positive or negative
- A point with a negative $x$ -coordinate has a negative horizontal displacement from the $y$ -axis
- A point with a negative $y$ -coordinate has a negative vertical displacement from the $x$ -axis

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Sine, Cosine & Tangent in Terms of Standard Position (College Board AP® Precalculus): Revision Note

Sine, cosine & tangent in terms of standard position

How are sine, cosine, and tangent connected to right triangle trigonometry?

What is the sine of an angle in standard position?

What is the cosine of an angle in standard position?

What is the tangent of an angle in standard position?

How do the signs of sine, cosine, and tangent depend on the quadrant?

How does this relate to right triangle trigonometry?

Examiner Tips and Tricks

Worked Example

Examiner Tips and Tricks

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Roger B

Reviewer: Mark Curtis

Quadrant	$x$	$y$	$\sin θ$	$\cos θ$	$\tan θ$
I	$+$	$+$	$+$	$+$	$+$
II	$-$	$+$	$+$	$-$	$-$
III	$-$	$-$	$-$	$-$	$+$
IV	$+$	$-$	$-$	$+$	$-$

Quadrant	$x$	$y$	$\sin θ$	$\cos θ$	$\tan θ$
I	$+$	$+$	$+$	$+$	$+$
II	$-$	$+$	$+$	$-$	$-$
III	$-$	$-$	$-$	$-$	$+$
IV	$+$	$-$	$-$	$+$	$-$

Quadrant	$x$	$y$	$\sin θ$	$\cos θ$	$\tan θ$
I	$+$	$+$	$+$	$+$	$+$
II	$-$	$+$	$+$	$-$	$-$
III	$-$	$-$	$-$	$-$	$+$
IV	$+$	$-$	$-$	$+$	$-$