Download Section 6.3 – Properties of Trigonometric Functions

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Section 6.3 – Properties of Trigonometric Functions
Objective 1:
Find the Domain & Range of the Trigonometric Functions
Recall the following Theorem from the previous section:
Let θ be an angle in standard position whose terminal side intersects the
circle x2 + y2 = r2 at the point (x, y). Then
sin(θ) =
csc(θ) =
y
r
r
y
cos(θ) =
sec(θ) =
x
r
r
x
tan(θ) =
cot(θ) =
y
x
x
y
Since r > 0, then the domain of the sine and cosine functions is all real
numbers. The tangent and secant functions are undefined when x is 0.
That occurs at θ =
π
2
,
3π
2
, and every odd multiple of
π
2
. The cotangent and
the cosecant functions are undefined when y = 0. This happens when θ =
0, π, and every multiple of π.
For the range, since x ≤ r, and y ≤ r, then both the sine and the cosine
functions will have a range of between – 1 and 1 inclusively and the secant
and cosecant functions will have a range of all values less than or equal to
– 1 or greater than or equal to 1. Since x is sometimes less than y,
sometimes equal to y , and sometimes larger than y, the range of the
tangent and cotangent functions id all real numbers.
Function
Domain
Range
sin(θ)
(– ∞, ∞)
[– 1, 1]
cos(θ)
(– ∞, ∞)
[– 1, 1]
tan(θ)
{θ | θ ≠
nπ
2
, n is an odd integer}
(– ∞, ∞)
sec(θ)
{θ | θ ≠
nπ
2
, n is an odd integer}
(– ∞, – 1] U [1, ∞)
csc(θ)
{θ| θ ≠ mπ, m is an integer}
(– ∞, – 1] U [1, ∞)
cot(θ)
{θ| θ ≠ mπ, m is an integer}
(– ∞, ∞)
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Objective 2:
Determine the Period of the Trigonometric Function
Periodic functions are functions that repeat at regular intervals. If we take
the sine and/or cosine of two angles that are coterminal, we will get the
same result. This means that the sine and cosine function are periodic
functions. This leads to the following definition:
Definition
A function f is periodic if there is a positive number p such that for any θ in
the domain of f, θ + p is also in the domain of f and f(θ + p) = f(θ). The
smallest such p value is called the (fundamental) period of f.
Since two coterminal angles differ by multiplies of 2π, the smallest such p
value would be 2π. Thus, the period of the sine and cosine function is 2π.
Since the secant and cosecant functions are reciprocals of the cosine and
sine function, then their periods are also 2π. The tangent and cotangent
functions are positive when x and y have the same sign, This occurs in the
first and third quadrants and thus tan(θ + π) = tan(θ) and cot(θ + π) =
cot(θ). Thus, the period of the tangent and cotangent function is π.
Function
sin(θ)
cos(θ)
tan(θ)
Period
2π
2π
π
Function
csc(θ)
sec(θ)
cot(θ)
Period
2π
2π
π
Find the exact value of the following:
Ex. 1a cos(
25 π
6
)
Ex. 1b csc(7π)
Ex. 1c tan(
4π
3
)
Solution:
25 π
6
) = cos(
π
6
+ 2(2π)) = cos(
π
6
)=
3
2
a)
cos(
b)
csc(7π) = csc(π + 3(2π)) = csc(π) which is undefined.
c)
tan(
Objective 3:
Since sin(θ) =
4π
3
) = tan(
π
3
+ π) = tan(
π
3
)=
3
Determine the Signs of the Trigonometric Functions
y
r
and csc(θ) =
r
,
y
then the sine and cosecant function are
positive in the first and second quadrant and negative in the third and fourth
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quadrant. Since cos(θ) =
x
r
and sec(θ) =
r
,
x
then the cosine and secant
are positive in the first and fourth quadrant and negative in the second and
third quadrant. Finally, tan(θ) =
y
x
and cot(θ) =
x
y
means they are positive
when x and y have the same signs (quadrant I and III) and negative when x
and y have opposite signs (quadrant II and IV). A phrase to remember this
is "All Suckers Take Calculus." All means all the trigonometric functions are
positive in quadrant I, Suckers means only the sine (and hence the
cosecant) is positive in quadrant II, Take means that only the tangent (and
hence the cotangent) is positive in quadrant III, and Calculus means only
the cosine (and hence the secant) is positive in quadrant IV.
sin(θ) > 0
csc(θ) > 0
all are
positive
suckers
all
tan(θ) > 0
cot(θ) > 0
cos(θ) > 0
sec(θ) > 0
take
calculus
Given the following information, determine the quadrant θ lies in:
Ex. 2a
sin(θ) < 0, cos(θ) > 0
Ex. 2b
tan(θ) > 0, sin(θ) < 0
Ex. 2c
csc(θ) > 0, sec(θ) < 0
Solution:
a) sin(θ) < 0 when θ is quadrant III and IV and cos(θ) > 0 when θ is
quadrant I and IV. The overlap occurs in quadrant IV so θ lies in
quadrant IV.
b) tan(θ) > 0 when θ is quadrant I and III and sin(θ) < 0 when θ is
quadrant III and IV. The overlap occurs in quadrant III so θ lies in
quadrant III.
c) csc(θ) > 0 when θ is quadrant I and II and sec(θ) < 0 when θ is
quadrant II and III. The overlap occurs in quadrant II so θ lies in
quadrant II.
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In the last section, we used symmetry to find the trigonometric values of the
angles in quadrants II. III. and IV. To make this process easier, we will
define what is called a reference angle.
Definition
A reference angle θR is the acute angle formed by the terminal side and
the x-axis. TrigFunction(θ) = TrigFunction(θR) or – TrigFunction(θR)
depending on the function and the quadrant.
Quadrant I
Quadrant II
θ
θ = θR
θR
θR = π – θ
Quadrant III
Quadrant IV
θ
θR
θ
θR
θR = θ – π
θR = 2π – θ
Find the exact value of the following:
Ex. 3a sin(210˚)
Ex. 3b sec(
7π
4
)
Ex. 3c tan(135˚)
Solution:
a)
210˚ is in Q III, so θR = 210˚ – 180˚ = 30˚. Since sine is negative
in Q III, then sin(210˚) = – sin(30˚) = –
1
.
2
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7π
4
b)
7π
π
=
4
4
π
sec( ) =
4
is Q IV, so θR = 2π –
Q IV, then sec(
7π
4
)=
. Since secant is positive in
2.
135˚ is in Q II, so θR = 180˚ – 135˚ = 45˚. Since tangent is
negative in Q II, then tan(135˚) = – tan(45˚) = – 1.
c)
Objective 3:
Finding Trigonometric Values using Trigonometric
Identities.
We have already established the reciprocal identities: csc(θ) =
sec(θ) =
1
,
cos(θ)
and cot(θ) =
1
.
tan(θ)
1
,
sin(θ)
We now want to establish the quotient
identities. Consider:
sin(θ)
= sin(θ)
cos(θ)
sin(θ)
.
tan(θ) =
cos(θ)
1
Also, cot(θ) =
=
tan(θ)
Since
÷ cos(θ) =
y
r
÷
sin(θ)
cos(θ)
=
cos(θ)
sin(θ)
1÷
x
r
=
y
r
•
r
x
=
y
x
, so cot(θ) =
= tan(θ), then
cos(θ)
sin(θ)
.
Use the information below to find the values of the four remaining
trigonometric functions:
Ex. 4
sin(θ) = –
1
3
and cos(θ) =
2 2
3
.
Solution:
tan(θ) =
1
3
2 2
3
−
sin(θ)
cos(θ)
=
1
tan(θ)
=1÷
=–
1
3
÷
2 2
3
=–
1
2 2
=–
2
4
(– 2 12 ) = – 2 2 .
1
1
= 1 ÷ (– ) = – 3
csc(θ) =
sin(θ)
3
2 2
3 2
1
3
=1÷(
=
=
sec(θ) =
)
3
4
cos(θ)
2 2
cot(θ) =
If we want to write [sin(θ)]2, we will write it as sin2(θ). Likewise, if we want to
write [tan(θ)]3, we will write tan3(θ).
Recall that for a circle of radius r centered on the origin, y2 + x2 = r2.
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Now, divide both sides by r2:
y2
x2
+
r2
y
r
r2
2
=
r2
r2
2
But, sin(θ) =
( ) + ( xr ) = 1
2
2
(sin(θ)) + (cos(θ)) = 1
y
r
and cos(θ) =
x
r
, so
sin2(θ) + cos2(θ) = 1
Our first Pythagorean Identity is sin2(θ) + cos2(θ) = 1. We will use this to
derive the other two:
(divide by sin2(θ))
sin2(θ) + cos2(θ) = 1
sin2 (θ)
+
(
=
1
sin2 (θ)
sin2 (θ)
2
2
cos(θ)
1
=
sin(θ)
sin(θ)
sin2 (θ)
1+
cos2 (θ)
) (
)
1 + cot2(θ) = csc2(θ)
or
sin2(θ) + cos2(θ) = 1
sin2 (θ)
+
2
cos (θ)
(
sin(θ)
cos(θ)
)
cos2 (θ)
2
cos (θ)
2
+1=
(
(simplify)
=
cot2(θ) + 1 = csc2(θ)
(divide by cos2(θ))
1
cos2 (θ)
2
1
cos(θ)
(simplify)
)
tan2(θ) + 1 = sec2(θ)
In summary, here are our fundamental identities:
Fundamental Identities
tan(θ) =
csc(θ) =
sin(θ)
cos(θ)
1
sin(θ)
cot(θ) =
sec(θ) =
sin2(θ) + cos2(θ) = 1
1
cos(θ)
tan2(θ) + 1 = sec2(θ)
cot(θ) =
cos(θ)
sin(θ)
1
tan(θ)
cot2(θ) + 1 = csc2(θ)
Find the exact value of each expression:
Ex. 5a
tan(40˚) – sin(40˚) sec(40˚)
Ex. 5b
(
cos(1.2)
sin(1.2)
)
2
– csc2(1.2)
Solution:
a)
tan(40˚) – sin(40˚) sec(40˚) = tan(40˚) – sin(40˚)
1
cos(40o )
30
= tan(40˚) –
b)
(
cos(1.2)
sin(1.2)
)
2
sin(40o )
cos(40o )
= tan(40˚) – tan(40˚) = 0
– csc2(1.2) = cot2(1.2) – csc2(1.2)
But, cot2(θ) + 1 = csc2(θ) implies that cot2(θ) – csc2(θ) + 1 = 0
which implies cot2(θ) – csc2(θ) = – 1. Hence,
cot2(1.2) – csc2(1.2) = – 1.
Objective 5:
Find the Exact Values of the Trigonometric Functions
Given One of the Functions and the Quadrant of the Angle.
Given the following information, find the remaining five trigonometric
functions:
Ex. 6
cos(θ) = 0.8 and sin(θ) < 0
Solution:
Since the cosine is positive and the sine is negative, the angle is in
Q IV. Let's use one of the trigonometric identities to find sin(θ):
sin2(θ) + cos2(θ) = 1
sin2(θ) + (0.8)2 = 1
sin2(θ) + 0.64 = 1
sin2(θ) = 0.36
sin(θ) = ± 0.6
Since, sin(θ) < 0, then sin(θ) = – 0.6
tan(θ) =
csc(θ) =
sec(θ) =
cot(θ) =
Ex. 7
− 0.6
sin(θ)
= 0.8 = – 0.75
cos(θ)
1
5
1
=
=– 3
− 0.6
sin(θ)
1
1
= 0.8 = 1.25
cos(θ)
1
4
1
=
=– 3
− 0.75
tan(θ)
tan(θ) =
1
3
and cos(θ) < 0
Solution:
Since the tangent is positive and the cosine is negative, the angle is
in Q III. Let's use one of the trigonometric identities to find sec(θ):
tan2(θ) + 1 = sec2(θ)
( 31 )
2
+ 1 = sec2(θ)
31
1
9
+ 1 = sec2(θ)
sec2(θ) =
sec(θ) = ±
cos(θ) =
cot(θ) =
1
sec(θ)
1
tan(θ)
Since tan(θ) =
=
10
9
10
3
1
10
3
−
=
1
1
3
Since, sec(θ) < 0, then sec(θ) = –
3
=–
10
1
sin(θ)
sin(θ)
cos(θ)
, then
Objective 6:
1
=
−
3 10
10
=–
=3
1
•
3
10
sin(θ) = tan(θ)•cos(θ) =
csc(θ) =
10
3
10
10
=–
10
(–
3 10
10
=–
)=–
10
10
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Even-Odd Properties of Trigonometric Functions.
Let θ be an angle in standard position with the terminal side passing
through the point (xuc, yuc) on the unit circle. Then the angle – θ will have
the terminal side pass through the point (xuc, – yuc) on the unit circle.
Since cos(θ) = xuc = cos(– θ) on the unit circle, then the cosine function is
even. But – sin(θ) = – yuc = sin(– θ), which means that the sine function is
odd. Similarly, the secant function is even (reciprocal of the cosine function)
and the cosecant function is odd (reciprocal of the sine function). Now, let's
examine the tangent function:
tan(– θ) =
sin(−θ)
cos(−θ)
=
− sin(θ)
cos(θ)
= – tan(θ)
So, the tangent function is odd and likewise, the cotangent function is also
odd.
Even-Odd Properties of Trigonometric Functions:
sin(– θ) = – sin(θ)
cos(– θ) = cos(θ
θ)
tan(– θ) = – tan(θ)
csc(– θ) = – csc(θ)
sec(– θ) = sec(θ
θ)
cot(– θ) = – cot(θ)
Only the cosine and the secant functions are even; all the other ones are
odd.
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Find the exact value of the following:
Ex. 8a
cos(– 30˚)
Ex. 8b
sin(–
Ex. 8c
cot(– 240˚)
Ex. 8d
csc(–
π
)
2
5π
3
)
Solution:
3
2
a)
cos(– 30˚) = cos(30˚) =
b)
sin(–
c)
cot(– 240˚) = – cot(240˚)
(odd function)
240˚ is in quadrant III, so the reference angle is 240˚ – 180˚
= 60˚. Also, the cotangent is positive in Q III. So, cot(240˚)
= cot (60˚)
π
2
π
) = – sin( 2 ) = – 1
(even function)
(odd function)
Thus, – cot(240˚) = – cot(60˚) = –
d)
csc(–
5π
3
5π
3
) = – csc(
5π
3
)
1
o
tan(60 )
= –
1
3
3
3
=–
(odd function)
is in quadrant IV, so the reference angle is 2π –
5π
3
=
Since the cosecant is negative in Q IV, then
5π
)=
3
5π
csc( 3 )
csc(
–
=
2 3
3
π
– csc( 3 ). Hence,
π
π
= – [– csc( 3 )] = csc( 3 ) =
1
sin(
π
3
)
=
1
3
2
=
2
3
π
3
.