Circular Trigonometric
Functions
Y
• circle…center at (0,0)
r
θ
•radius r…vector with
length/direction
X
• angle θ… determines
direction
Y-axis
90º
Quadrant II
Quadrant I
r
180º
Terminal
side r
θ
0º X-axis
Initial side
Quadrant III
360º
Quadrant IV
270º
Y-axis
-270º
Quadrant II
-180º
Terminal
side r
Quadrant I
Initial side
θ
Quadrant III
-360º X-axis
0º
Quadrant IV
-90º
• angle θ…measured from positive x-axis,
or initial side, to terminal side
counterclockwise: positive direction
clockwise: negative direction
• four quadrants…numbered I, II, III, IV
counterclockwise
• six trigonometric functions for angle θ
whose terminal side passes thru point
(x, y) on circle of radius r
sin θ = y / r
csc θ = r / y
cos θ = x / r
sec θ = r / x
tan θ = y / x
cot θ = x / y
These apply to any angle in any quadrant.
For any angle in any quadrant
x2 + y2 = r2 …
So, r is positive by Pythagorean theorem.
(x,y)
r
θ
x
y
Y
r
θ
x
(x,y)
y
X
NOTE:
right-triangle
definitions are
special case
of circular
functions
when θ is in
quadrant I
*Reciprocal Identities
sin θ = y / r and csc θ = r / y
cos θ = x / r and sec θ = r / x
tan θ = y / x and cot θ = x / y
*Ratio Identities
sinθ
tanθ =
cosθ
cosθ
cot θ =
sinθ
*Both sets of identities are useful
to determine trigonometric
functions of any angle.
Positive trig values in each quadrant:
Y
Students
All
sin positive
(csc)
all six positive
(-, +)
(+, +)
II I
Take
tan positive
(cot)
III IV
(-, -) (+, -)
X
Classes
cos positive
(sec)
Y
REMEMBER:
In the ordered pair (x, y),
x represents cosine and
y represents sine.
(-, +)
(+, +)
II I
III IV
(-, -) (+, -)
X
#1 Draw each angle whose terminal side
passes through the given point, and find
all trigonometric functions of each angle.
θ1: (4, 3)
θ2: (- 4, 3)
θ3: (- 4, -3)
θ4: (4, -3)
SOLUTION
I
(4,3)
θ1
SOLUTION
x=
y=
r=
sin θ =
cos θ =
tan θ =
csc θ =
sec θ =
cot θ =
II
(-4,3)
θ2
x=
y=
r=
sin θ =
cos θ =
tan θ =
csc θ =
sec θ =
cot θ =
SOLUTION
x=
y=
r=
sin θ =
cos θ =
tan θ =
csc θ =
sec θ =
cot θ =
θ3
(-4,-3)
III
SOLUTION
x=
y=
r=
θ4
(4,-3)
IV
SOLUTION
sin θ =
cos θ =
tan θ =
csc θ =
sec θ =
cot θ =
Y
II
ref θ2
ref θ3
III
I
θ1
ref θ4
IV
Perpendicular
line from
point on circle
always drawn
X
to the x-axis
forming a
reference triangle
Y
II
ref θ2
ref θ3
III
I
θ1
ref θ4
IV
Value of trig
function
of angle in
any quadrant
is equal to trig
X
function of its
reference angle,
or it differs
only in sign.
#2 Given: tan θ = -1 and cos θ is positive:
• Draw θ. Show the values for x, y, and r.
SOLUTION
Given: tan θ = -1 and cos θ is positive:
• Find the six trigonometric functions of θ.
SOLUTION
# 1 Find the value of sin 110º.
(First determine the reference angle.)
SOLUTION
#2 Find the value of tan 315º.
(First determine the reference angle.)
SOLUTION
#3 Find the value of cos 230º.
(First determine the reference angle.)
SOLUTION
#1 Draw the angle whose terminal side
passes through the given point 1, 3 .

SOLUTION

Find all trigonometric functions for angle
whose terminal side passes thru 1, 3 .

SOLUTION

#2 Draw angle: sin θ = 0.6, cos θ is negative.
SOLUTION
Find all six trigonometric functions:
sin θ = 0.6, cos θ is negative.
SOLUTION
#3 Find remaining trigonometric functions:
sin θ = - 0.7071, tan θ = 1.000
SOLUTION
Find remaining trigonometric functions:
sin θ = - 0.7071, tan θ = 1.000
SOLUTION
#1 Express as a function of a reference
angle and find the value: cot 306º .
SOLUTION
#2 Express as a function of a reference
angle and find the value: sec (-153º) .
SOLUTION
#3 Find each value on your calculator.
(Key in exact angle measure.)
sin 260.5º
tan 150º 10’
SOLUTION
cot (-240º)
csc 450º
SOLUTION
cos 5.41
sec (7/4)
SOLUTION
π/2 = 1.57
0
2π = 6.28
π = 3.14
3π/2 = 4.71
# 1 The refraction of a certain prism is
sin 100°
n=
sin 47°
Calculate the value of n.
SOLUTION
#2 A force vector F has components
Fx = - 4.5 lb and Fy = 8.5 lb.
Find sin θ and cos θ.
Fy = 8.5 lb
θ
Fx=-4.5 lb
SOLUTION
Fy = 8.5 lb
θ
Fx=-4.5 lb
SOLUTION