Sine, Cosine and Tangent in Four Quadrants

Sine, Cosine and Tangent

The three main functions in trigonometry are Sine, Cosine and Tangent.

triangle showing Opposite, Adjacent and Hypotenuse

They are easy to calculate:

Divide the length of one side of a
right angled triangle by another side

... but we must know which sides!

For an angle θ, the functions are calculated this way:

Sine Function: 
sin(θ) = Opposite / Hypotenuse
Cosine Function: 
cos(θ) = Adjacent / Hypotenuse
Tangent Function: 
tan(θ) = Opposite / Adjacent

Example: What is the sine of 35°?

triangle 2.8 4.0 4.9

Using this triangle (lengths are only to one decimal place):

sin(35°) = Opposite / Hypotenuse = 2.8/4.9 = 0.57...

Cartesian Coordinates

Using Cartesian Coordinates we mark a point on a graph by how far along and how far up it is:

graph with point (12,5)
The point (12,5) is 12 units along, and 5 units up.



Four Quadrants

When we include negative values, the x and y axes divide the space up into 4 pieces:

Quadrants I, II, III and IV

(They are numbered in a counter-clockwise direction)

  • In Quadrant I both x and y are positive,
  • in Quadrant II x is negative (y is still positive),
  • in Quadrant III both x and y are negative, and
  • in Quadrant IV x is positive again, and y is negative.

Like this:

cartesian coordinates

Quadrant X
I Positive Positive (3,2)
II Negative Positive  
III Negative Negative (−2,−1)
IV Positive Negative  

Example: The point "C" (−2,−1) is 2 units along in the negative direction, and 1 unit down (i.e. negative direction).

Both x and y are negative, so that point is in "Quadrant III"

Sine, Cosine and Tangent in the Four Quadrants

Now let us look at what happens when we place a 30° triangle in each of the 4 Quadrants.

In Quadrant I everything is normal, and Sine, Cosine and Tangent are all positive:

Example: The sine, cosine and tangent of 30°

triangle 30 quadrant I

sin(30°) = 1 / 2 = 0.5
cos(30°) = 1.732 / 2 = 0.866
tan(30°) = 1 / 1.732 = 0.577


But in Quadrant II, the x direction is negative, and both cosine and tangent become negative:

Example: The sine, cosine and tangent of 150°

triangle 30 quadrant II

sin(150°) = 1 / 2 = 0.5
cos(150°) = −1.732 / 2 = −0.866
tan(150°) = 1 / −1.732 = −0.577


In Quadrant III, sine and cosine are negative:

Example: The sine, cosine and tangent of 210°

triangle 30 quadrant III

sin(210°) = −1 / 2 = −0.5
cos(210°) = −1.732 / 2 = −0.866
tan(210°) = −1 / −1.732 = 0.577

Note: Tangent is positive because dividing a negative by a negative gives a positive.


In Quadrant IV, sine and tangent are negative:

Example: The sine, cosine and tangent of 330°

triangle 30 quadrant IV

sin(330°) = −1 / 2 = −0.5
cos(330°) = 1.732 / 2 = 0.866
tan(330°) = −1 / 1.732 = −0.577

There is a pattern! Look at when Sine Cosine and Tangent are positive ...

  • All three of them are positive in Quadrant I
  • Sine only is positive in Quadrant II
  • Tangent only is positive in Quadrant III
  • Cosine only is positive in Quadrant IV

This can be shown even easier by:

trig ASTC is All,Sine,Tangent,Cosine

Some people like to remember the four letters ASTC by one of these:

  • All Students Take Chemistry
  • All Students Take Calculus
  • All Silly Tom Cats
  • All Stations To Central
  • Add Sugar To Coffee

You can remember one of these, or maybe you could make up
your own. Or just remember ASTC.

trig graph 4 quadrants
This graph shows "ASTC" also.

Two Values

Have a look at this graph of the Sine Function::

sine crosses 0.5 at 30,150,390, etc
There are two angles (within the first 360°) that have the same value!

And this is also true for Cosine and Tangent.

The trouble is: Your calculator will only give you one of those values ...

... but you can use these rules to find the other value:

First value Second value
Sine θ 180º − θ
Cosine θ 360º − θ
Tangent θ θ − 180º

And if any angle is less than 0º, then add 360º.

We can now solve equations for angles between 0º and 360º (using Inverse Sine Cosine and Tangent)

Example: Solve sin θ = 0.5

We get the first solution from the calculator = sin-1(0.5) = 30º (it is in Quadrant I)

The other solution is 180º − 30º = 150º (Quadrant II)

Example: Solve tan θ = −1.3

We get the first solution from the calculator = tan-1(−1.3) = −52.4º

This is less than 0º, so we add 360º: −52.4º + 360º = 307.6º (Quadrant IV)

The other solution is 307.6º − 180º  = 127.6º (Quadrant II)

Example: Solve cos θ = −0.85

We get the first solution from the calculator = cos-1(−0.85) = 148.2º (Quadrant II)

The other solution is 360º − 148.2º = 211.8º (Quadrant III)