6.3 The Trigonometric Functions
For any point you plot on a Cartesian graph, you are able to draw a line segment to the
origin from that point making an angle θ with the positive x-axis. From that point you
can also draw a vertical line segment to the x-axis forming a right triangle. With this
right triangle we are able to define six trigonometric ratios.
Example
Plot the point (4, 3). Draw a right triangle to the x-axis and origin from this point.
Determine the value of sin θ.
. (4,3)
3
θ
4
You can apply the Pythagorean Theorem to determine the length of the hypotenuse of the
right triangle.
5
opp 3
a2 + b2 = c2
3
sin θ =

2
2
2
hyp 5
4 +3 =c
θ
16 + 9 = c2
4
25 = c2
5=c
We can also determine the other five trig function values given one of the trig values and
the location of θ.
Example
If tan θ = -4/3 and θ is in QII, determine the values of the other five trig functions.
First, you want to sketch what you do know.
.
(-3,4)
4
By applying the Pythagorean
Theorem you can determine the
length of the hypotenuse to be 5.
5
θ
-3
sin θ = 4/5
cos θ = -3/5
cot θ = -3/4
sec θ = -5/3
csc θ = 5/4
Note: This triangle corresponds to the creation of a right triangle with the x-axis, origin,
and the point (-3,4)
Try the following:
If cos θ = -5/13 and θ is in QIII, determine the values of the other five trig functions.
Answers: -12/13; 12/5; 5/12; -13/5; -13/12
At times you will know the angle in question and have no knowledge of the lengths of the
legs of the corresponding triangle. Let’s say we want to know the value of sin 150˚.
First, we need to know which quadrant the angle is located in. The angle measure of
150˚ is located in quadrant II. We can make a right triangle with the x-axis from some
place on the terminal side.
150˚
150˚
reference
angle
The reference angle is the acute angle measure at the base of the right triangle. In this
case the angle measure and its reference angle are supplementary. Hence, the measure of
the reference angle is 30˚. Referring back to the ratios of the 30-60-90 triangle, we know
this much:
1
2
- 3
Keep in mind the negative side of the x-axis is being used, and the point corresponding to
this right triangle is (- 3 ,1). Using the reference angle we can evaluate sin 150˚, which
is the same value as sin 30˚.
sin 30˚ = ½
therefore, sin 150˚ = ½ .
Think about what would happen when the angle is located in quadrant III or IV. For
instance, 225˚ is an angle located in QIII, with a reference angle 45˚. In addition, 300˚ is
an angle located in QIV with a reference angle of 60˚.
225˚
300˚
60˚
45˚
225˚ - 180˚ = 45˚
360˚ - 300˚ = 60˚
Example
Find the six trig values for 225˚ and 300˚.
sin 225˚ = 
2
2
csc 225˚ =  2
cos 225˚ = 
2
2
sec 225˚ =  2
tan 225˚ = 1
sin 300˚ = 
cot 225˚ = 1
3
2
csc 300˚ = 
cos 300˚ = ½
sec 300˚ = 2
tan 300˚ =  3
cot 300˚ = 
2 3
3
3
3
Try the following:
Find the six trig value for 210˚.
Answers:
Examine the signs of the example above. We can use the following to assist us in
knowing the sign of the six trig functions without sketching a triangle:
Sin (+)
All (+)
Acronym: “All Students Take Calculus”
Tan (+)
Cos (+)
The reciprocals of sine, cosine, and tangent are also positive in these quadrants.
Example
Find csc θ and cos θ if tan θ = 2/5 and sin θ < 0.
We know the value of the tangent function is positive in QI and QIII.
The values for sin θ < 0 are located in QIII and QIV. The common quadrant is QIII.
-2
csc θ = - 29 /2
-5
θ
29
cos θ = - 29 /5
Degenerate triangles can also be evaluated. If you want to evaluate sin 90˚ or cos 270˚,
you can imagine drawing a triangle applying the definition of the trig function.
sin θ =
opp
hyp
hyp
opp
θ
or sin θ =
y
r
adj
No x is used to evaluate, yet we need a value for the triangle. Let’s use zero for the
adjacent side. Pick a number for y and find r, which happens to be the same when
applying the Pythagorean Theorem.
(0,2)
.
2
90˚
2
02 + 22 = r2
0 + 4 = r2
2=r
sin 90˚ = y/r = 2/2 = 1
-1
02 + (-1)2 = r2
0 + 1 = r2
1=r
cos 270˚ = x/r = 0/1 = 0
θ
0
Same holds true for cos 270˚.
1
270˚
. (0,-1)
θ
0
If we consider drawing a circle about the origin with radius 1, we are able to draw a right
triangle from any point on the circle with the x-axis. Let’s consider the basic angles you
have been exposed to in QI.
1 3
 ,

 2 2  2 2 
,


2
2 

 3 1
, 

 2 2
60˚ : ( ½ , sqrt3/2)
45˚ : (sqrt2/2, sqrt2/2)
30˚ : (sqrt3/2, ½ )
(x, y) = (cos θ, sin θ)
i.e. cos 60˚ = ½
The Unit Circle is a circle with radius 1.
You are basically taking the 30-60-90 and 45-45-90 triangles and creating similar
triangles to them where the hypotenuse is a length of 1.
These points can be reflected into the other quadrants to assist in evaluating trig
functions. Refer to the table of values provided by Mr. Burger.