2.4 Trigonometric Functions of General Angles
So far we have been talking about trig functions of acute angles.
We employ the rectangular coordinate system for trig functions of all angles (acute
or not). We place the angle in standard position (vertex at the origin, initial side
on positive x-axis). We’ll let the point (a,b) be on the terminal side of the angle.
y- axis
3
Opposite side
(-2,2)
-4
-3
2
H
yp
ot
en
us 1
e
Adjacent side
-2
θ
x- axis
0
-1
0
-1
1
origin
(0,0)
2
3
4
-2
-3
-4
We can form a right triangle from any point
(a,b) and the origin (0,0). In the example
above, (a,b) = (-2,2).
The adjacent side = a = -2
Using the actual value of the x-coordinate, a,
instead of the distance identifies the
quadrant of this angle.
The opposite side = b = 2
The hypotenuse, c, is the distance from
(a,b) to the origin
c=
2
2
2
2
(a − 0) + (b − 0) = a + b =
(−2)2 + (2)2 = 8 = 2 2
Trig
Fctn
Definition
Exact Value
sin(θ)
b/c=opp/hyp
b/c=
cos(θ)
a/c=adj/hyp
a/c= − 2 = − 2
2
2
=
2
2 2
2 2
2
tan(θ)
b/a=opp/adj
b/a= 1/-1 = -1
csc(θ)
c/b=hype/opp
c/b= 2 2 = 2
sec(θ)
c/a=hyp/adj
c/a=
cot(θ)
a/b=adj/opp
a/b= -1/1 = -1
2
2 2
=− 2
−2
Finding the Exact Value of the Six Trig Functions of Quadrantal Angles
3
α= π/2rad = 90°
2
Example 2 on p.133
a)
θ = 0rad = 0°
b) θ = π/2rad = 90°
c)
d) θ = 3π/2rad = 270°
θ = πrad = 180°
1
β= πrad = 180°
-4
a)
y
-3
-2
-1
β0
φ
α
δ= 0rad = 0°
0
1
2
3
δ
0° is on the positive x-axis, so choose a
point on the positive x-axis. (a,b)=(1,0)
-1
c = square root of (a2 + b2)=square root
of (12 + 02) = 1.
Φ= 3π/2rad = 270°
-2
-3
-4
Trig
Funct.
Def.
Exact Value for
θ = 0rad = 0°
[use (a,b)=(1,0)]
Exact Value for
θ = π/2rad = 90°
[use (a,b)=(0,1)]
Exact Value for
θ = πrad = 180°
[use (a,b)=(-1,0)]
Exact Value for
θ = 3π/2rad = 270°
[use (a,b)=(0,-1)]
sin(θ)
b/c=
opp/hyp
b/c=0/1 = 0
b/c=1/1 = 1
b/c=0/1 = 0
b/c=-1/1 = 1
cos(θ)
a/c=
adj/hyp
a/c= 1/1 = 1
a/c= 0/1 = 0
a/c= -1/1 = -1
a/c= 0/1 = 0
tan(θ)
b/a=
opp/adj
b/a= 0/1 = 0
b/a= 1/0
undefined
b/a= 0/-1 = 0
b/a= -1/0
undefined
csc(θ)
c/b=
hyp/opp
c/b=1/0
undefined
c/b=1/1=1
c/b=1/0
undefined
c/b=1/-1= -1
sec(θ)
c/a=
hyp/adj
c/a=1/1=1
c/a=1/0
undefined
c/a=1/-1= -1
c/a=1/0 undefined
cot(θ)
a/b=
adj/opp
a/b= 1/0
undefined
a/b= 0/1 = 0
a/b= =1/0
undefined
a/b= 0/1 = 1
4
Coterminal Angles
If θ is an angle measured in degrees, then θ ±360°(k) , where k is any integer, is an
angle coterminal with θ.
If θ is an angle measured in radians, then θ ± 2πk , where k is any integer, is an
angle coterminal with θ.
Since coterminal angles are basically equal, all their trigonometry functions are
equal.
Example 3 on p.134
Find the exact value of each of the following:
a)
sin 390°
Reduce 390° to an angle less than 360° by subtracting the highest multiple
of 360° that is less than 390°.
That would be 360x1 = 360
sin(390°) = sin(390° - 360°) = sin(30°) = ½
c)
tan 9π/4
Reduce 9π/4 to an angle less than 2π by subtracting the highest multiple of
2π that is less than 9π/4.
2π = 8π/4
tan 9π/4 = tan (8π/4 + π/4) = tan π/4 = 1
d)
sec(-7π/4)
Change -7π/4 to an angle between 0 and 2π by adding a multiple of 2π.
-7π/4 + 8π/4 = π/4
sec(-7π/4) = sec(-7π/4 + 8π/4) = sec(π/4) = square root of 2.
Signs of Trigonometric Functions
Quadrant
sin, csc
cos, sec
tan, cot
I
+
+
+
II
+
-
-
III
-
-
+
IV
-
+
-
3
II
(-,+)
α= π/2rad = 90°
Sin, csc >0
Cos, sec, tan, cot <0
1
β= πrad = 180°
-4
-3
All trig functions >0
0
-2
-1
δ= 0rad = 0°
0
-1
III
(-,-)
I
(+,+)
2
-2
Φ= 3π/2rad = 270°
Tan, cot > 0
sin, cos, sec, csc <0
-3
1
2
3
4
II
(+,-)
Cos,sec > 0
sin, csc, tan,cot <0
-4
Homework
p.141 #3,9,11,15,19,27,33,35,37,47
Reference Angle
Let θ denote a nonacute angle that lies in a quadrant. A
reference angle for θ is the acute angle formed by the
terminal side of θ and either the positive x-axis (0°) or the
negative x-axis (180°).
yaxis
3
Reference angle for θ
-4
-3
-2
y-3 axis
2
2
1
1
θ
0
-1
0
-1
2
2
1
2
xaxis
3
4
-3
-2
-1
20
-1
2
-2
-2
-3
-3
-4
-4
1
2
In this example, the trig functions for the
reference angle are the same as for θ except
sin, csc, cos, & sec are negative for θ.
Example 6 on p. 138
a) Use the reference angle for 135° to find exact
values of the trig functions.
The closest x-axis to this angle is the negative
axis which is at 180°. The reference angle is
180°-135°=45°
Reference
135° is in Quadrant II, so
angle for θ
cos(135°) = -cos (45°)= - sqrt(2)/2
-4
-3
y-3 axis
2
1
θ
0
-2
-1
0
sec (135°) = -sec(45°)= -sqrt(2)
-1
sin (135°) = sin(45°) = sqrt(2)/2
-2
csc (135°) = csc (45°) = sqrt(2)
-3
tan(135°) = -tan(45°) = -1
-4
cot (135°) = -cot(45°) = -1
xaxis
3
4
0
-4
Reference angle for θ
In this example, the trig functions for the
reference angle are the same as for θ except
cos, sec, tan, & cot are negative for θ.
θ
You do #53 on p.141
1
2
xaxis
3
4
Finding the Exact Value of Trigonometric Functions
Example 7 on p.139
Given that cos (θ) = -2/3, and π/2 < θ < π. Find the exact value
of the remaining trig functions.
Since π/2 < θ < π, θ is in Quadrant II. So sin & csc are positive
and all the other trig functions are negative.
Using a right triangle, we can figure out the hyp, adj, and opp
sides. Cos(θ) = -2/3 = adj/hyp. So adj = -2, hyp =3
opp =
32 − (−2) 2 = 9 − 4 = 5
hy
p
Trig
Definition
Exact Value
=3
Fctn
opp
θ
adj=-2
sin(θ)
opp/hyp
5
3
cos(θ)
adj/hyp
−2
3
tan(θ)
opp/adj
5
− 2
csc(θ)
hyp/opp
3
3 5
=
5
5
sec(θ)
hyp/adj
cot(θ)
adj/opp
3
−2
−2 −2 5
=
5
5
Finding Exact Values of Trig functions
Example 8
If tan(θ) =-4 =-4 and sin(θ) < 0, find the exact value of each of the
remaining trig functions.
In which quadrant is tan(θ) and sin(θ) < 0 ?
IV
In Quadrant IV, cos and sec are positive but all other trig functions are
negative.
To find trig functions, tan(θ) = opp/adj = -4/1. Since θ is in Quadrant
IV, we’ll let the opp side be -4, and adj side = 1
hyp2 = (-4)2 + (1)2 = 17
1
hyp =
θ
-4
17
Trig
Fctn
Definition
Exact Value
sin(θ)
opp/hyp
− 4 − 4 17
=
17
17
cos(θ)
adj/hyp
1
17
=
17
17
tan(θ)
opp/adj
csc(θ)
hyp/opp
17
−4
sec(θ)
hyp/adj
17
= 17
1
cot(θ)
adj/opp
Now do #89
− 4 / 1 = −4
1
−4