Download Section 7.5 – Trigonometric Functions: Unit Circle Approach

39
Section 7.5 – Trigonometric Functions: Unit Circle Approach
The unit circle is a circle of radius 1 centered at the origin. If we have an
angle in standard position superimposed on the unit circle, the terminal side
will intersect the unit circle at a particular point. The point of intersection will
depend on the value of the angle. This implies that the x-coordinate and the
y-coordinate of the point will depend on the value of the angle.
(–
(–
(–
2
2
3
2
,
)
(
–
(–
,–
2
2
π
1
2
,–
(–
2
2
1
,
2
)
π
2
2π
3
120˚
135˚
150˚
90˚
( 21 ,
π
3
60˚
)
–
3
2
(
45˚
30˚
4π
3
330˚
315˚
7π
270˚ 300˚ 4
3π
2
)
(0, – 1)
5π
3
)
2
2
11π
6
(
( 21 , –
2
2
,
(
π
6
0˚
210˚
225˚
5π
240˚
4
3
2
π
4
180˚
7π
6
)
(0, 1)
3π
4
5π
6
(– 1, 0)
3
2
3
2
)
2
2
,
1
2
1
,
2
3
2
,
)
1
2
)
(1, 0)
0
(
2
2
3
2
3
2
,–
)
,–
2
2
1
2
)
)
Thus, for an angle that measures 30˚ in standard position, the terminal side
will pass through the point
(
3
2
,
1
2
) on the unit circle. For an angle that
measures 270˚ in standard position, the terminal side will pass through the
point (0, – 1) on the unit circle. Since the x- and y-coordinates of the point
on the unit circle intersected by the terminal side depend on the measure of
the angle, we can define each coordinate as a function of the angle.
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Definition
If θ is an angle in standard position and if the terminal side of θ intersects
the unit circle at (a, b), then
sin(θ) = b
(sine of θ)
cos(θ) = a
(cosine of θ)
There are four other trigonometric functions we can define in reference to
our unit circle. We can first talk about the slope of the terminal side of θ in
standard position. Two points on that line are (0, 0) and (a, b), so the slope
b−0
b
m=
= . This value again depends on the value of θ, so we will
a−0
€
a
define the tangent function (tan(θ)) to be equal to the slope of the terminal
side. The other three trigonometric functions, the cosecant (csc(θ)), the
secant
€ (sec(θ)), and the cotangent (cot(θ)),are the reciprocals of the first
three trigonometric functions sine, cosine, and tangent.
Definition
If θ is an angle in standard position and if the terminal side of θ intersects
the unit circle at (a, b), then
b
sin(θ) = b
cos(θ) = a
tan(θ) =
csc(θ) =
1
b
sec(θ) =
Note that csc(θ) =
€ 1:
Objective
1
,
sin(θ)
sec(θ) =
1
a
1
,
cos(θ)
cot(θ) =
a
a
b
1
and cot(θ) =
.
€tan(θ)
€ Values of the Trigonometric
€
Find the Exact
Functions
Using a Point on the Unit Circle.
€
€
€
Let P = (x, y) be the point on the unit circle that corresponds to t. Find
the values of the six trigonometric functions of t:
( 23 , –
Ex. 1a
Solution:
a)
€
5
3
)
Ex. 1b
5
3
sin(t) = b = –
€
tan(t) =
€
€
€
b
a
=
−
5
3
2
3
€
(0, 1)
cos(t) = a =
=–
5
3
÷
2
3
=–
5
3
•
€
€
€
€
€
2
3
3
2
=–
5
2
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csc(t) =
sec(t) =
1
b
=
1
a
=
1
−
1
2
3
5
3
=
=–
3
5
=–
3 5
5
3
2
€
€ 2
€ xuc
3
2
2
cot(t) =
= 3 =
• –
=–
=–
y uc
5
3
5
5
−
€
€
3
€
sin(t) = b = 1
cos(t) = a = 0
1
b
tan(t) =
= which is undefined
0
a
€
€
€
€
1
1
€
csc(t) = = = 1
€
b)
(
sec(t) =
€
cot(t) =
€
€
Objective 2:
€
b
1
a
a
b
=
=
1
1
0
0
1
)
2 5
5
which is undefined
=0
Use a Circle of Radius r to Evaluate the Trigonometric
Functions.
Now, consider an angle that is in standard position with a terminal side that
intersects a circle of radius of r at the point (a, b). By the Pythagorean
Theorem, a2 + b2 = r2. Using similar triangles, it can be shown that:
b
a
b
sin(θ) =
cos(θ) =
tan(θ) =
csc(θ) =
r
r
b
sec(θ) =
r
r
a
cot(θ) =
a
a
b
€Theorem
€
€
Let θ be an angle in standard position whose terminal side intersects the
€
€
circle x2 + y2 = r2 at the point €
(a, b). Then
b
a
b
sin(θ) =
cos(θ) =
tan(θ) =
csc(θ) =
r
r
b
sec(θ) =
r
r
a
cot(θ) =
a
a
b
€Find the six trigonometric values
€
€ the following
of an angle θ with
conditions:
€
€
Ex. 2 (– 5, 12) is a point on€the terminal side.
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Solution:
First, we need to find the radius of the circle:
r2 = a 2 + b 2
r2 = (– 5)2 + (12)2 = 25 + 144 = 169
r = 169 = 13 (ignore the – answer)
Thus, a = – 5, b = 12, and r = 13:
sin(θ) =
csc(θ)€=
12
13
13
12
cos(θ) =
sec(θ) =
−5
13
13
−5
5
13
13
–
5
=–
tan(θ) =
=
cot(θ) =
12
=–
−5
−5
=–
12
12
5
5
12
€Objective 3: Domain and
€ Range
€ of Trigonometric Functions.
€
€
€
€
€
Let's re-examine the following
theorem:
€
€
Theorem
Let θ be an angle in standard position whose terminal side intersects the
circle x2 + y2 = r2 at the point (a, b). Then
b
a
b
sin(θ) =
cos(θ) =
tan(θ) =
csc(θ) =
r
r
b
sec(θ) =
r
r
a
cot(θ) =
a
a
b
of the sine and cosine functions
is all real
€Since r > 0, then the domain €
€
numbers. The tangent and secant functions are undefined when a is 0.
€
€
€π
π 3π
That occurs at θ = ,
, and every odd multiple of
or 90˚. The
2
2
2
cotangent and the cosecant functions are undefined when b = 0. This
happens when θ = 0, π, and every multiple of π or 180˚.
€ €since a ≤ r, and b ≤ r, then both
€ the sine and the cosine
For the range,
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 a is sometimes less than b,
sometimes equal to b , and sometimes larger than b, the range of the
tangent and cotangent functions is all real numbers.
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Function
Domain
Range
sin(θ)
(– ∞, ∞)
[– 1, 1]
cos(θ)
(– ∞, ∞)
[– 1, 1]
tan(θ)
sec(θ)
csc(θ)
cot(θ)
{θ| θ ≠
nπ
2
, n is an odd integer}
nπ
2
, n is an odd integer}
{θ| θ ≠ n(90˚), n is an odd integer}
{θ| θ ≠
€{θ| θ ≠ n(90˚), n is an odd integer}
{θ| θ ≠ mπ, m is an integer}
{θ| θ ≠ n(180˚), n is an integer}
€
{θ| θ ≠ mπ, m is an integer}
{θ| θ ≠ n(180˚), n is an integer}
Objective 4:
(– ∞, ∞)
(– ∞, – 1] U [1, ∞)
(– ∞, – 1] U [1, ∞)
(– ∞, ∞)
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.
Periodic Properties:
sin(θ + 2π) = sin(θ)
csc(θ + 2π) = csc(θ)
cos(θ + 2π) = cos(θ)
sec(θ + 2π) = sec(θ)
tan(θ + π) = tan(θ)
cot(θ + π) = cot(θ)
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Function
sin(θ)
cos(θ)
tan(θ)
Period
2π
2π
π
Function
csc(θ)
sec(θ)
cot(θ)
Find the exact value of the following:
25π
Ex. 3a sin(
)
Ex. 3b cot(7π)
Ex. 3c tan(
6
Solution:
a)
sin(
€ b)
c)
25π
6
) = sin(
Period
2π
2π
π
π
6
+ 2(2π)) = sin(
3
3
π
6
)=
4π
3
)
1
2
cot(7π) = cot(π + 6(π)) = cot(π) which is undefined.
€
7π
π
π
sec(
) = sec( + 2π) = sec( ) = 2
3
€
Objective 6:
€
€
€
Even-Odd Properties of Trigonometric Functions.
€ an angle €
€ with the terminal side passing
Let θ be
in standard position
through the point (a, b). Then the angle – θ will have the terminal side pass
through the point (a, – b).
(a, b)
r
θ
r
(a, – b)
Since cos(θ) =
=–
€
b
a
a
r
–θ
= cos(– θ), then the cosine function is even. But – sin(θ)
= 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.
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Even-Odd Properties of Trigonometric Functions:
sin(– θ) = – sin(θ)
cos(– θ) = cos(θ)
tan(– θ) = – tan(θ)
csc(– θ) = – csc(θ)
sec(– θ) = sec(θ)
cot(– θ) = – cot(θ)
Only the cosine and the secant functions are even; all the other ones are
odd.
Find the exact value of the following:
Ex. 8a
cos(– 30˚)
Ex. 8b
sin(–
Ex. 8c
csc(–
cot(– 240˚)
Ex. 8d
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
) = – sin(
π
2
)=–1
(even function)
(odd function)
Thus, – cot(240˚) = – cot(60˚) = –
d)
csc(–
5π
3
5π
3
) = – csc(
5π
3
)
1
= –
tan(60o )
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
π
csc(
) = – csc( ). Hence,
3
– csc(
=
2 3
3
5π
3
3
) = – [– csc(
π
3
)] = csc(
π
3
)=
1
sin(
π
3
)
=
1
3
2
=
2
3
π
3
.