5. We will first look at the special angles called the quadrantal angles.
90
180
270
0
The quadrantal anglesare those angles that lie on the axis of the Cartesian coordinate
system: 0° , 90
°, 180°, 270° and 360° .
6. We can count the quadrantal angles in terms of .
radians
2
π
radians
2
π
radians
2
2π
radians
2
3π
π= radians
360,0
90
180
270
radians57.1≈
radians14.3≈
radians71.4≈
radians2
radians,0
radians,0
π
π
radians28.6≈
9. Now let’s cut each quadrant in half,
which basically gives us 8 equal
sections.
0
4
π
4
2π
4
4π
4
6π
4
3π
4
5π
4
7π
4
8π
We can again count around the
circle, but this time we will count in
terms of
radians.
4
π
.
4
8
,
4
7
,
4
6
,
4
5
,
4
4
,
4
3
,
4
2
,
4
1 ππππππππ
and
4
π
2
π
=
π2=
2
3π
=
=π
Then reduce appropriately.
45=
90=
135=
180=
225=
270=
315=
360=
2
π
10. 45
45
The lengths of the legs of the 45 – 45 – 90
triangle are equal to each other because their
corresponding angles are equal.
If we let each leg have a length of 1, then we find
the hypotenuse to be using the Pythagorean
theorem.
2
1
1
2
You should memorize this triangle or at
least be able to construct it. These angles
will be used frequently.
Next we will look at two special triangles: the 45 – 45 – 90 triangle and the 30 – 60 – 90 triangle.
These triangles will allow us to easily find the trig functions of the special angles, 45 , 30 , and 60 .
(Knowing this derivation is not necessary,
but knowing the ratio of sides and angles is.)
12. For the 30 – 60 – 90 triangle, we will construct an equilateral triangle (a triangle with 3 equal angles of
each, which guarantees 3 equal sides).
60
If we let each side be a length of 2, then cutting
the triangle in half will give us a right triangle
with a base of 1 and a hypotenuse of 2. This
smaller triangle now has angles of 30, 60, and
90 .
We find the length of the other leg to be using
the Pythagorean theorem.
3
3
60
1
2
30
You should memorize this triangle or at least
be able to construct it. These angles, also,
will be used frequently.
(Knowing this derivation is not necessary,
but knowing the ratio of sides and angles is.)
13. Again, using the definition of the trigonometric functions as the ratios of the sides of a right triangle, we
can now list all the trig functions for a 30 angle and a 60 angle.
330cot
3
3
3
1
30tan
3
32
3
2
30sec
2
3
30cos
230csc
2
1
30sin
===
===
==
3
3
3
1
60cot360tan
260sec
2
1
60cos
3
32
3
2
60csc
2
3
60sin
===
==
===
60
30
1
3
2
14. All I
Sine II
III
Tangent
IV
Cosine
I. All
II. Students
III. Take
IV. Calculus
Positive Values for Trigonometric Functions
Note: Because they are reciprocals, the sign of cosecant matches the sign of sine,
the sign of secant matches the sign of cosine, and the sign of cotangent matches
the sign of tangent. Always.
28. y
x
Example 1: Find the six trig functions of 330 .
First draw the 330 degree angle.
3
60
1
2
30
AS
T C
330
30
(Answers on next page.)
Your reference angle is 30 .
29. y
x
Example 2: Find the six trig functions of .
3
60
1
2
30
3
4π
First determine the location of .
3
4π
3
π
3
2π
3
3π
3
3π
3
4π
3
π
(Answers on next page.)
Your reference angle is .
3
π
30. 3
60
1
2
30
AS
T C
Example 2: Find the six trig functions of . 3
4π
y
x
3
π
3
2π
3
4π
3
π
π
3
3
3
1
3
4
cot3
3
4
tan
2
3
4
sec
2
1
3
4
cos
3
32
3
2
3
4
csc
2
3
3
4
sin
===
−=−=
−=−=−=
ππ
ππ
ππ
Your reference angle is .
3
π
31. 0 radians
Example 3: Find the exact value of cos .
−
4
5π
We will first draw the angle to determine the quadrant.
−
4
5π
−
4
π
−
4
2π
−
4
3π
−
4
4π
AS
T C
45
45
1
1
2
4
π
Your reference angle is .
4
π
32. 0 radians
Problem 3: Find the sin .
AS
T C
−
6
5π
6
π
−
6
2π
−
6
3π
−6
4π
−
6
5π
−
is the reference angle.6
π
6
π
33. 0 radians
Problem 7: Find the exact value of cos .
We will first draw the angle to determine the quadrant.
AS
T C
Note that the reference angle is .
4
π
4
13π
4
π
4
2π
4
4π
4
6π
4
5π 4
7π
4
8π
4
9π4
10π
4
11π
4
3πWe see that the angle is located in the
3rd quadrant and the cos is negative in the
3rd quadrant.
4
13π
4
12π
4
13π
4
π
