Sine, Cosine and Tangent in Four Quadrants
Sine, Cosine and Tangent
The three main functions in trigonometry are Sine, Cosine and Tangent.
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:
|sin(θ) = Opposite / Hypotenuse|
|cos(θ) = Adjacent / Hypotenuse|
|tan(θ) = Opposite / Adjacent|
Example: What is the sine of 35°?
Using this triangle (lengths are only to one decimal place):
sin(35°) = Opposite / Hypotenuse = 2.8/4.9 = 0.57...
Using Cartesian Coordinates we mark a point on a graph by how far along and how far up it is:
The point (12,5) is 12 units along, and 5 units up.
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.
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°
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°
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°
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°
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:
This graph shows "ASTC" also.
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.
Have a look at this graph of the Sine Function:
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)