Section 5-3
Trigonometric Functions on the
Unit Circle
Objective: Students will be able to:
1. Find the values of the six trigonometric functions
using the unit circle.
2. Find the values of the six trigonometric functions of
an angle in standard position given a point on its
terminal side.
unit circle is a circle of radius 1 and whose center is at the origin of a coordinate
The ___________
plane.
The unit circle is symmetric with respect to the x-axis, y-axis, and origin.
1
1
y
𝜽
𝜽
x
-1
1
-1
** If we use are trig functions we learned in section 5.2 we can find the value
of different degrees on the unit circle.
We can find values for 𝐬𝐢𝐧 𝜽 and 𝐜𝐨𝐬 𝜽 using the definitions used in section 5.2.
sin 𝜃 =
𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
cos 𝜃 =
𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑦
𝑥
=
=
1
1
Since there is exactly one point P(x,y) for an angle 𝜃, the relations cos 𝜃 = x and sin 𝜃 = y
are functions of 𝜃.
Because both of these functions are defined using the unit circle, they are often called
circular functions
________________.
The four other trig functions can also be defined using the unit circle:
tan 𝜃 =
𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
sec 𝜃 =
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
=
=
𝑦
𝑥
1
𝑥
csc 𝜃 =
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
cot 𝜃 =
𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
=
=
1
𝑦
𝑥
𝑦
Example 1:
a.
Use the unit circle to find each value.
tan 180°
𝑦
tan 180° =
𝑥
0
tan 180° =
−1
tan 180° = 0
b.
1
csc (-90°) =
𝑦
1
csc (-90°) =
−1
csc (-90°) = -1
sin 1800 = 0
cos 1800 = -1
sec 1800 = -1
cot 1800 = undefined
csc 1800 = undefined
csc (-90°)
sin −900 = -1
cot −900 = 0
cos −900 = 0
tan −900 = undefined
sec −900 = undefined
sin , csc
cos, tan, sec, cot
1
2 2
2 2
3 1
2 2
3
2
-2
90
-
-
135
sin, cos, tan, sec, csc, cot
none
0 1
1 3
2 2
60
120
150
30
0
360
210
-
3
2
1
-2
2
- 2
225
-
2
2
1
3 1
2 2
45
180
-1 0
2 2
2 2
-2 -
330
315
240
3
2
tan, cot
sin, cos, sec, csc
1 0
300
270
0 -1
1
2
-
3
2
2
2
-
3
2
2
2
1
-2
cos, sec
sin, tan, csc, cot
The radius of a circle is defined as a positive value. Therefore the signs of the six trig
x and ___
y in each
functions are determined by the signs of the coordinates of ___
quadrant.
Example 2:
Use the unit circle to find the values of the six trigonometric functions
for a 300° angle.
** The reference angle would be 60°.
** The terminal side of the angle intersects
the unit circle at a point.
Quad. II
x is neg. & y is pos.
** We need to find the coordinates of this point
from our special triangles from Geometry.
Special Triangles
30° - 60° - 90°
45°
x
2x
45°
y
Quad. IV
x is pos. & y is neg.
sin 60° =
𝑜𝑝𝑝.
ℎ𝑦𝑝𝑜𝑡.
cos 60° =
𝑎𝑑𝑗.
ℎ𝑦𝑝𝑜𝑡.
y 2
y
30°
x 3
300°
Quad. III
x & y are neg.
45° - 45° - 90°
60°
Quad. I
x & y are pos.
The pt. is
1
2
=
=
,−
𝑥 3
2𝑥
𝑥
2𝑥
3
2
=
=
1
2
3
2
** Now that we know the point on the terminal side that intersects the unit circle, we can
1
3
,
−
2
2
now find the 6 trig. functions.
sin 300° = y = - 3
2
cos 300° = x =
1
2
3
𝑦
−
tan 300° =
= 12 = - 3
𝑥
2
1
csc 300° =
𝑦
=
1
−
3
2
1
1
sec 300° =
= 1
𝑥
2
= -
2
3
= -
2 3
3
= 2
1
𝑥
= 23
cot 300° =
𝑦
− 2
= -
1
3
= -
3
3
The sine and cosine functions of an angle in standard position may also be determined
terminal side
using the ordered pair of any point on its ________________
and the distance between
point and the_______.
origin
that _______
P(x, y)
r
y
x
All six trig functions can be determined using x, y, and r. (𝑟 =
𝑥2 + 𝑦2 )
Trig Functions of an Angle in Standard Position
𝑦
sin 𝜃 =
𝑟
𝑟
csc 𝜃 =
𝑦
𝑥
cos 𝜃 =
𝑟
𝑟
sec 𝜃 =
𝑥
𝑦
tan 𝜃 =
𝑥
cot 𝜃 =
𝑥
𝑦
Example 3:
Find the values of the six trigonometric functions for angle 
in standard position if a point with coordinates (-6, 8) lies on
its terminal side.
r = (−6)2 + 82
** 1st find r :
r = 100
r = 10
sin 𝜃 =
𝑦
𝑟
cos 𝜃 =
𝑥
𝑟
tan 𝜃 =
csc 𝜃 =
𝑦
𝑥
𝑟
𝑦
4
5
=
8
10
=
=
−6
10
= -
=
8
−6
= -
=
10
8
=
5
4
3
5
4
3
sec 𝜃 =
𝑟
𝑥
=
10
−6
= -
5
3
cot 𝜃 =
𝑥
𝑦
=
−6
8
= -
3
4
If you know the value of one of the trig functions and the quadrant in which the terminal
side lies in, you can find the other five trig functions.
Example 4:
Suppose  is an angle in standard position whose terminal side lies in Quadrant II.
If csc  =
8
, find the values of the remaining five trigonometric functions of .
3
** Since the point lies in Quad. II, x is neg. & y is pos.
csc 𝜃 =
Find x :
𝑟
𝑦
, so r = 8 and y = 3
sin 𝜃 =
𝑥 2+ 𝑦 2 = 𝑟2
𝑥 2 + 9 = 64
𝑥 2 = 55
x = ± 55
𝑦
𝑟
=
3
8
cos 𝜃 =
𝑥
𝑟
=
− 55
8
tan 𝜃 =
𝑦
𝑥
=
3
− 55
= -
sec 𝜃 =
𝑟
𝑥
=
8
− 55
= -
cot 𝜃 =
𝑥
𝑦
=
− 55
3
x has to be - 55
3 55
55
8 55
55