Download Section 6.2 – Trigonometric Functions: Unit Circle Approach

12
Section 6.2 – 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
,
,
1
2
3
2
2
2
)
(
–
(–
,–
2
2
)
π
1
2
)
π
2
2π
3
120˚
135˚
150˚
90˚
,–
2
2
)
π
3
60˚
(–
1
,
2
–
3
2
(
π
4
45˚
4π
3
2
2
2
2
,
(
3
2
,
)
1
2
)
30˚
330˚
315˚
7π
270˚ 300˚ 4
3π
2
)
)
π
6
0˚
210˚
225˚
5π
240˚
4
3
2
( 21 ,
180˚
7π
6
)
(0, 1)
3π
4
5π
6
(– 1, 0)
3
2
3
2
1
,
2
(0, – 1)
5π
3
11π
6
(
( 21 , –
(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.
13
Definition
If θ is an angle in standard position and if the terminal side of θ intersects
the unit circle at (xuc, yuc), then
sin(θ) = yuc
(sine of θ)
(cosine of θ)
cos(θ) = xuc
The sine and cosine functions belong to a class of functions that we call the
trigonometric functions.
Find the exact values of the sine and cosine functions for the given
angles:
Ex. 1a
π
Ex. 1b
330˚
Ex. 1c
3π
4
Ex. 1d
240˚
Solution:
a)
Since the terminal side of θ passes through (– 1, 0), then
sin(θ) = 0 and cos(θ) = – 1.
b)
Since the terminal side of θ passes through (
sin(θ) = –
c)
1
2
and cos(θ) =
2
2
and cos(θ) = –
2
2
3
2
and cos(θ) = –
1
),
2
then
2
2
,
2
2
),
–
3
2
), then
.
Since the terminal side of θ passes through (–
sin(θ) = –
,–
.
Since the terminal side of θ passes through (–
then sin(θ) =
d)
3
2
3
2
1
,
2
1
.
2
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 (xuc, yuc), so the
slope m =
y uc −0
xuc −0
=
y uc
xuc
. This value again depends on the value of θ, so we
will 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.
14
Definition
If θ is an angle in standard position and if the terminal side of θ intersects
the unit circle at (xuc, yuc), then
sin(θ) = yuc
csc(θ) =
1
y uc
tan(θ) =
1
xuc
cot(θ) =
sec(θ) =
Note that csc(θ) =
Objective 1:
cos(θ) = xuc
1
,
sin(θ)
sec(θ) =
1
,
cos(θ)
y uc
xuc
xuc
y uc
1
.
tan(θ)
and cot(θ) =
Find the Exact Values of the Trigonometric 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:
Ex. 2a
( 23 , –
5
3
)
Ex. 2b
(0, 1)
Solution:
a)
b)
5
3
sin(t) = yuc = –
−
tan(t) =
y uc
xuc
=
csc(t) =
1
y uc
=
sec(t) =
1
xuc
=
cot(t) =
xuc
y uc
1
−
tan(t) =
csc(t) =
2
3
−
sec(t) =
cot(t) =
=
=
=
=
2
3
=
=
5
3
1
sin(t) = yuc = 1
y uc
xuc
1
y uc
1
xuc
xuc
y uc
5
3
2
3
5
3
cos(t) = xuc =
=–
=–
5
3
3
2
3
÷
5
3
3
•2 =–
5
2
3 5
5
=–
5
=–
2
3
3
2
=
2
3
(
• –
3
)=–
5
2
5
=–
cos(t) = xuc = 0
1
which is undefined
0
1
=1
1
1
which is undefined
0
0
=0
1
2 5
5
15
Objective 2:
Find the Exact Value of the Six Trigonometric Functions
of Quadrantal Angles.
Find the exact values of the six trigonometric functions of:
Ex. 3a
θ = 0 = 0˚
Ex. 3b
θ=
Ex. 3c
θ = π = 180˚
Ex. 3d
θ=
π
= 90˚
2
3π
= 270˚
2
Solution:
a)
The terminal side of θ intersects (1, 0) on the unit circle. Thus,
sin(0) = yuc = 0
csc(0) =
cot(0) =
b)
1
y uc
xuc
y uc
=
=
1
0
1
0
which is undefined
sec(0) =
y uc
xuc
1
xuc
0
1
1
1
=
=
=0
=1
which is undefined
π
1
cos( 2 ) = xuc = 0
=
=
1
0
1
0
π
which is undefined
csc( 2 ) =
which is undefined
cot( 2 ) =
π
1
y uc
xuc
y uc
=
=
1
=1
1
0
=0
1
The terminal side of θ intersects (– 1, 0) on the unit circle. Thus,
sin(π) = yuc = 0
csc(π) =
cot(π) =
d)
tan(0) =
The terminal side of θ intersects (0, 1) on the unit circle. Thus,
π
sin( 2 ) = yuc=
π
y
tan( 2 ) = x uc
uc
π
1
sec( 2 ) = x
uc
c)
cos(0) = xuc = 1
1
y uc
xuc
y uc
=
=
cos(π) = xuc = – 1
1
0
1
0
which is undefined
tan(π) =
sec(π) =
y uc
xuc
1
xuc
=
=
0
−1
1
−1
=0
=–1
which is undefined
The terminal side of θ intersects (0, – 1) on the unit circle. Thus,
3π
sin( 2 ) = yuc = –
3π
y
tan( 2 ) = x uc =
uc
3π
1
sec( 2 ) = x =
uc
1
1
0
1
0
cos(
3π
2
) = xuc = 0
which is undefined
csc(
which is undefined
cot(
We can summarize our results:
3π
1
)
=
2
y uc
3π
xuc
)
=
2
y uc
=
=
1
−1
0
1
=–1
=0
16
θ
θ
sin(θ
θ)
cos(θ
θ)
tan(θ
θ)
csc(θ
θ)
sec(θ
θ)
cot(θ
θ)
0
0˚
0
1
0
undefined
1
undefined
π
2
90˚
1
0
undefined
1
undefined
0
π
180˚
0
–1
0
undefined
–1
undefined
3π
2
270˚
–1
0
undefined
–1
undefined
0
Definition:
Two angles are coterminal if they have the same initial and terminal sides.
Properties of Coterminal Angles
1)
Two coterminal angles with differ by integer multiplies of 2π (360˚)
2)
The Trig. value of two coterminal angles are equal.
Find the following:
Ex. 4a
sin(– 450˚)
Solution:
A coterminal angle to – 450˚ is
– 450˚ + 2•360˚ = 270˚.
So, sin(– 450˚) = sin(270˚) = – 1
Ex. 4b
tan(9π)
Solution:
A coterminal angle to 9π
is 9π – 4•2π = π
So, tan(9π) = tan(π) = 0
Objectives 3 & 4: Find the Exact Values of the Trigonometric Functions of
for Special Angles
The first special right triangle we want to examine is a 45˚-45˚-90˚ right
triangle. Since two of the angles are the same, that means two of the sides
are equal, so we have an isosceles triangle. Also, the two equal sides of
the isosceles triangle are the legs of the right triangle since the hypotenuse
is always opposite of the right angle. The hypotenuse of the triangle will be
equal to 1 since we are on the unit circle. If we let a be the length of one of
the equal sides, then we can the Pythagorean Theorem to find a.
12 = a2 + a2 = 2a2
1
2
=a
a=
2
1
2
(take the square root)
=
2
2
(ignore – answer)
1
45˚
a
45˚
a
17
Thus, the terminal side of a 45˚ angle in standard position intersects the
unit circle at
2
2
(
2
2
,
).
The second special right triangle we want to examine is a 30˚-60˚-90˚ right
triangle. If we take an equilateral triangle and cut it in half along the altitude,
we get two 30˚-60˚-90˚ triangles. The hypotenuse of the triangle will be
equal to 1 since we are on the unit circle. The shortest side is half of the
hypotenuse so it is 1/2. If we let a be the length of the longer leg, then can
use the Pythagorean Theorem to find the its length.
1
(1)2 = a2 + ( 2 )2
1 = a2 +
a2 =
1
4
(take the square root)
3
4
30˚
1
3
4
a=
(simplify)
3
2
=
60˚
(ignore the – answer)
1
2
Thus, we get two triangles:
1
60˚
30˚
a
1
2
30˚
1
3
2
60˚
3
2
1
2
Thus, the terminal side of a 30˚ angle in standard position intersects the
(
3
2
) and the terminal side of a 60˚ angle in standard
3
1
position intersects the unit circle at ( ,
).
2
2
unit circle at
,
1
2
Find the exact values of the six trigonometric functions of:
Ex. 5a
θ=
Ex. 5c
θ=
π
6
π
3
= 30˚
Ex. 5b
θ=
π
4
= 45˚
= 60˚
Solution:
a) The terminal side of θ intersects
(
3
2
,
1
2
) on the unit circle. Thus,
18
sin(
π
6
tan(
π
6
csc(
π
6
cot(
π
6
) = yuc =
1
2
)=
y uc
xuc
=
)=
1
y uc
=
)=
xuc
y uc
1
2
3
2
1
1
2
=
3
2
÷
=2
1
2
3
2
1
2
=
3
2
=
π
4
tan(
π
4
)=
sec(
π
4
)=
y uc
xuc
2
2
2
2
=
1
2
2
=
2
2
•
2
3
=
1
3
)=
1
xuc
=
1
π
6
1
2
=
÷
=1
=
1
2
sec(
2
2
) = yuc =
π
6
=
b) The terminal side of θ intersects
sin(
cos(
2
•
2
1
2
2
,
2
2
cos(
π
4
) = xuc =
csc(
π
4
)=
1
y uc
=
cot(
π
4
)=
xuc
y uc
=
(
( 21 ,
3
2
cos(
π
3
tan(
π
3
csc(
cot(
) = yuc =
)=
y uc
xuc
=
π
3
)=
1
y uc
=
π
3
)=
xuc
y uc
=
3
2
1
2
1
3
2
1
2
3
2
=
2
3
=
=
3
2
1
2
÷
÷
=
1
2
=
3
2
2 3
3
3
2
We can summarize our results:
=
3
2
π
3
•
2
3
2 3
3
) on the unit circle. So,
2
2
1
=
2
2
2
2
2
2
2
2
=
2
= 1
) on the unit circle. Thus,
2
1
sec(
1
2
=
3
) = xuc =
•
2
3
=
3
2
=
3
3
=
3
2
c) The terminal side of θ intersects
sin(
3
2
) = xuc =
=
=
3
π
3
)=
1
3
1
2
=
1
xuc
=
3
3
1
1
2
=2
19
θ
θ
sin(θ
θ)
cos(θ
θ)
tan(θ
θ)
csc(θ
θ)
sec(θ
θ)
π
6
30˚
1
2
3
2
3
3
2
2 3
3
π
4
45˚
2
2
2
2
1
π
3
60˚
3
2
1
2
2
2 3
3
3
2
2
cot(θ
θ)
3
1
3
3
Find the exact value of the following expression:
Ex. 6a
cos(30˚) sin(270˚)
Ex. 6c
(
Ex. 6b
cot(
Ex. 6d
sin(
2
tan(60˚)
)
+ csc(30˚)
π
) – cos(π)
4
π
π
) – sin( )
6
2
Solution:
a)
cos(30˚) sin(270˚) =
b)
cot(
c)
(
d)
sin(
Objective 5:
π
4
•(– 1) = –
3
2
) – cos(π) = 1 – (– 1) = 2
2
tan(60˚)
π
6
3
2
)
+ csc(30˚) = ( 3 )2 + 2 = 5
) – sin(
π
2
)=
1
2
–1=–
1
2
Find the Exact Value of the Trigonometric Functions for
Certain Integer Multiplies of the Special Angles.
Since we know the values of the trigonometric functions for 30˚, 45˚, and
60˚, we can use symmetry to find the values of the trigonometric functions
for 120˚, 210˚, 330˚, 135˚, 225˚, 315˚, 150˚, 240˚ and 300˚.
20
3
2
(–
(
3
2
–
(–
,
,–
2
2
1
2
)
1
2
,
5π
6
7π
6
)
2
2
)
210˚
π
4
225˚
(–
,–
2
2
)
330˚
3π
4
135˚
2
2
30˚
150˚
5π
4
π
6
11π
6
(
3
2
,
1
2
)
(
3
2
,–
1
2
(
2
2
,
(
2
2
,–
2
2
)
45˚
315˚
7π
4
2
2
)
)
21
3
2
1
,
2
(–
)
π
3
2π
3
4π
3
3
2
–
)
300˚
240˚
1
,
2
3
2
60˚
120˚
(–
( 21 ,
5π
3
( 21 , –
)
3
2
)
Find the exact values of the following:
Ex. 7a
cos(135˚)
Ex. 7c
tan(
11π
6
)
Ex. 7d
Ex. 7e
cot(210˚)
Solution:
Ex. 7f
a)
c)
tan(
d)
csc(
7π
4
e)
f)
)=
)=
−
y uc
xuc
1
y uc
cot(210˚) =
sec(150˚) =
=
1
2
3
2
=–
2
2
−
3
2
1
2
−
xuc
y uc
=
1
xuc
=
sec(150˚)
b)
1
=
sin(
2
2
cos(135˚) = xuc = –
11π
6
4π
)
3
7π
csc( )
4
Ex. 7b
−
1
−
3
2
3
3
=–
2
=
3
=–
2 3
3
sin(
4π
3
) = yuc = –
3
2
22
Objective 6:
Approximate the Value of a Trigonometric Function.
On a scientific calculator, SIN key is for the sine ratio, COS key is for the
cosine ratio, and TAN key is for the tangent ratio. To evaluate a
trigonometric ratio for a specific angle, you first want to be in the correct
mode. If the angle is in degree, then the calculator needs to be in degree
mode while if the angle is in radians, the calculator needs to be in radian
mode. Next, type in the angle and then hit the appropriate trigonometric
ratio. On some calculators, you might need to enter the trigonometric ratio
first and then the angle. For the secant, cosecant, and cotangent, we want
to use the following facts: sec(θ) =
1
cos(θ)
, csc(θ) =
1
,
sin(θ)
cot(θ) =
1
tan(θ)
Evaluate the following (round to three decimal places):
Ex. 8a
cos(26˚)
Ex. 8b
tan(78.3˚)
Ex. 8c
sin(16 3 ˚)
Ex. 8d
csc(56.11˚)
Ex. 8e
sec(1.2)
Ex. 8f
tan(
2
3π
2
)
Solution:
a)
We put our calculator in degree mode, type 26 and hit the COS
key:
cos(26˚) = 0.8987… ≈ 0.899
b)
We put our calculator in degree mode, type 78.3 and hit the
TAN key: tan(78.3˚) = 4.8289… ≈ 4.829
c)
We put our calculator in degree mode, type 16 3 and hit the
2
SIN key:
d)
2
sin(16 3 ˚) = 0.2868… ≈ 0.287
We put our calculator in degree mode, type 56.11 and hit
SIN key:
sin 56.11˚ = 0.8308…
Now, hit the 1/x key to take the reciprocal:
1
1
csc(56.11˚) =
=
= 1.2046… ≈ 1.205
sin(56.11)
e)
0.8308...
Since there is no degree mark on the angle, then the angle is in
radians. We put our calculator in radian mode, type 1.2 and hit
the COS key:
cos(1.2) = 0.3623…
Now, hit the 1/x key to take the reciprocal:
sec(1.2) =
1
cos(1.2)
=
1
0.3623...
= 2.7597… ≈ 2.760
23
f)
Since there is no degree mark on the angle, then the angle is in
radians. We put our calculator in radian mode, type
the TAN key: tan(
Objective 7:
3π
2
3π
2
and hit
) = Error. This means that it is undefined.
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 (x, y). By the Pythagorean
Theorem, x2 + y2 = r2. Using similar triangles, it can be shown that:
sin(θ) =
csc(θ) =
y uc
1
1
y uc
=
=
y
r
r
y
cos(θ) =
sec(θ) =
xuc
1
1
xuc
=
=
x
r
r
x
tan(θ) =
cot(θ) =
y uc
xuc
xuc
y uc
=
=
y
x
x
y
Theorem
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
Find the six trigonometric values of an angle θ with the following
conditions:
Ex. 9 (– 5, 12) is a point on the terminal side.
Solution:
First, we need to find the radius of the circle:
r2 = x2 + y2
r2 = (– 5)2 + (12)2 = 25 + 144 = 169
r = 169 = 13 (ignore the – answer)
Thus, x = – 5, y = 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