MORE ON TRIANGLES
Trigonometric Functions of Any Angle
We know how to find the trig function values of an acute angle from a right triangle. Let the point
P = ( x , y ) be a point “ r ” units from the origin on the terminal side of angle θ . A right triangle can be
formed by drawing a perpendicular line from P to the x-axis.
Below are examples of the right triangles formed from the angle θ . Each example has θ in a different one
of the four quadrants in the xy -plane.
y
y
y
y
P =( x , y )
P =( x , y )
θ
θ
θ
x
x
θ
x
x
P =( x , y )
P =( x , y )
θ lies in
Quadrant I
θ lies in
Quadrant II
θ lies in
Quadrant III
θ lies in
Quadrant
We know how to use right triangles to evaluate the trigonometric functions of an acute angle. However, in
the figures above the angle θ is not always an acute angle. We will often evaluate trigonometric functions
of positive angles greater than 90o and all negative angles by making use of a positive acute angle. This
positive acute angle is called a REFERENCE ANGLE .
Definition of Reference Angle – Let θ be a non-acute angle in standard position that lies in a quadrant.
Its reference angle is the positive acute angle α formed by the terminal side of the angle θ and the x -axis.
Below are examples of the right triangles formed by the angle θ with the reference angle, α , labeled.
y
y
y
y
P =( x , y )
P =( x , y )
α
α =θ
x
θ
θ
x
x
θ
α
α
P =( x , y )
P =( x , y )
θ =α
Quadrant I
θ lies in
Quadrant II
θ lies in
Quadrant III
θ lies in
Quadrant IV
x
Examples: Find the reference angle α for each of the following angles θ .
(i) θ = 60o then α = __________
(ii) θ = 135o then α = __________
(iii) θ = 210o then α = __________
(iv) θ = −45o then α = __________
Once again we have our examples of the right triangles formed by the angle θ with the reference angle, α ,
labeled.
P =( x , y )
P =( x , y )
θ
r
α =θ
x
y
θ
r
y
α
θ
α
x
x
x
α
r
y
r
y
P =( x , y )
P =( x , y )
Quadrant I
Quadrant II
Quadrant III
Quadrant IV
Notice that for each right triangle, the length of the side opposite α is y (the y -coordinate of the point P ).
The length of the side adjacent to α is x (the x -coordinate of the point P ). Using Pythagorean’s
Theorem, we can find the length of the hypotenuse, r . So the trigonometric functions evaluated at α are
defined as:
sin α =
opposite
hypotenuse
csc α =
=
hypotenuse
opposite
y
r
=
cos α =
r
y
sec α =
adjacent
x
=
hypotenuse r
hypotenuse
adjacent
=
r
x
tan α =
opposite y
=
adjacent x
cot α =
adjacent
opposite
=
x
y
Using Reference angles to evaluate trigonometric Functions
The values of the trigonometric functions of a given angle, θ , are the same as the values of the trigonometric
functions of the reference angle, α , except for possibly the sign. A function value of the acute reference angle α is
always positive. However, the same function value for θ may be positive or negative.
Example: Let the point P = ( x , y ) = ( 3, − 4 ) be on the terminal side of the angle θ .
Find sin θ and cosθ .
θ
α
Using Pythagorean’s Theorem and the facts that we know from right triangles:
x=3
90o
2
2
2
x +y =r
α
32 + 42 = r 2
9 + 16 = r
25 = r
P = (3 , − 4)
y=4
r =5
2
2
5=r
Recall that when using the reference angle α the x , y , and r values are all positive. So,
sin α =
y 4
=
r 5
cos α =
and
x 3
= .
r 5
"− y " − 4
.
=
r
5
"+ x " 3
However, also since θ lies in quadrant IV, the x -coordinate is positive. So, cos θ = cos α =
= .
r
5
Since θ lies in quadrant IV, the y -coordinate is negative. So, sin θ = − sin α =
Example 2: Find the sin θ and cosθ for θ = 150o .
When the reference angle, α , is a common angle (30o , 45o , 60o )
we do not need to use Pythagorean’s Theorem to help us establish a right triangle.
Instead the right triangle defined by the reference angle will be one of our basic triangles.
θ
If θ = 150o then the reference will be α = 30o .
So the right triangle is a 30 − 60 − 90 triangle:
o
o
α
o
60o
y=
1
r =1
2
α = 30o
90o
3
x=
2
Recall that when using the reference angle α the x , y , and r values are all positive. So,
sin α =
y 1/ 2 1
=
=
r
1
2
and
Since θ lies in quadrant II, the y -coordinate is positive. So,
sin θ = sin α =
"+ y " 1/ 2 1
=
= .
r
1
2
cos α =
x
=
r
However, also since θ lies in quadrant II, the x -coordinate is negative. So,
cos θ = − cos α =
"− x " − 3 / 2
3
.
=
=−
r
1
2
3/2
3
.
=
1
2
Reference for information:
Algebra & Trigonometry 2nd Ed.
by Robert Blitzer