Transcript 5.5 Double-Angle and Half-Angle Formulas

Section 5.5

a)
In the previous sections, we used:
The Fundamental Identities
a) Sin²x + Cos²x = 1
b)
Sum & Difference Formulas
a) Cos (u – v) = Cos u Cos v + Sin u Sin v
Now we will use double angle and half angle
formulas

Double-angle formulas are the formulas used
most often:
Sin 2u  2 Sin u Cos u
2Tan u
Tan 2u 
2
1 - Tan u
Cos 2u  Cos u - Sin u
2
 2Cos u - 1
2
2
 1 - 2Sin u
2

Use the following triangle to find the
following:
Sin 2θ
29
θ
5
2
Cos 2θ
Tan 2θ

Use the following triangle to find the
following:
Sin 2θ = 2Sin θ Cos θ
29
θ
5
2
 2  5 


2 
 29   29 
20

29
Cos 2θ= 2Cos² θ - 1
2
29
θ
5
2
 5 
2
 -1
 29 
 25 
 2   1
 29 
21
 50 
   1 
29
 29 
29
θ
5
2
2Tan 
Tan 2θ 
2
1 - Tan 
2
2 
4
5
5



2
1- 4
2
25
1  
5
4
20
5


21
21
25

Use the following triangle to find the
following:
17
θ
4
1
17
Csc 2θ 
8
17
Sec 2θ 
15
15
Cot 2θ 
8

General guidelines to follow when the
double-angle formulas to solve equations:
1)
Apply the appropriate double-angle formula
2)
Look to factor
3)
Solve the equation using the different
strategies involved in solving equations

Solve the following equation in the interval
[0, 2π)
Sin 2x – Cos x = 0
1. Apply the double-angle formula
2 Sin x Cos x – Cos x = 0
2. Look to factor
Cos x (2 Sin x – 1) = 0
Cos x (2 Sin x – 1) = 0
3. Solve the equation
Cos x = 0
 3
x  ,
2 2
2 Sin x - 1= 0
Sin x = ½
 5
x ,
6 6

Solve the following equation in the interval
[0, 2π)
2 Cos x + Sin 2x = 0
2 Cos x + 2 Sin x Cos x = 0
2 Cos x (1+ Sin x) = 0
2 Cos x = 0
1 + Sin x = 0
2 Cos x = 0
Cos x = 0
 3
x  ,
2 2
1 + Sin x = 0
Sin x = -1
3
x 
2

Solve the following equations for x in the
interval [0, 2π)
a)
Sin 2x Sin x = Cos x
b)
Cos 2x + Sin x = 0
 3  5 7 11
, ,
,
,
x  ,
2 2 6 6 6
6
 7 11
,
x  ,
2 6
6
Sin 2x Sin x = Cos x
2 Sin x Cos x Sin x = Cos x
2 Sin²x Cos x – Cos x = 0
Cos x (2 Sin²x – 1) = 0
Cos x = 0
2 Sin²x – 1 = 0
 3
Sin²x = ½
,
x  ,
Sin
x
=
±
½
2 2
 5 7 11
,
,
x = ,
6 6 6
6
Cos 2x + Sin x = 0
1 – 2Sin² x + Sin x = 0
2Sin² x - Sin x - 1= 0
(2 Sin x + 1) (Sin x – 1) = 0
2 Sin x + 1 = 0
Sin x – 1 = 0
Sin x = ½
Sin x = 1
7 11
,
x=
6
6

x  ,
2
Section 5.5

Evaluating Functions Involving Double Angles
Use the given information to find the following:

12
Sin x 
13
Sin 2x
2
Cos 2x
 x 
Tan 2x

12
Sin x 
13
2
 x 
Sin 2x = 2Sin x Cos x
2
 12  5 
   
 13  13 
120
169
12
13
-5
x

12
Sin x 
13
2
 x 
Cos 2x =2Cos² x - 1

 5
2  
 13 
2
12
13
-5
x
1
50
 25 
- 119
 2
1 
 1 
169
169
 169 
Tan 2x
2Tan x

2
1 - Tan x
12
13
-5
x
 12 
2 
24
24
-5

5
5


2 
- 119
1 - 144
 12 
25
25
1-  
120
 -5

119

Evaluating Functions Involving Double Angles
Use the given information to find the following:
8
Cos x 
17
Sin 2x
3
 x  2
2
Cos 2x
Tan 2x
3
 x  2
2
8
Cos x 
17
Sin 2x = 2Sin x Cos x

 - 15 
2  17 


240
289
 8
 
 17 
x
8
17
-15
3
 x  2
2
8
Cos x 
15
x
8
17
-15
Cos 2x =2Cos² x - 1
2
8
 
 17 
2
1
128
 64 
- 161
 2
1 
 1 
289
289
 289
Tan 2x
2Tan x

2
1 - Tan x
x
8
17
 - 15 
2

- 30
- 30
8 

8
8


2 
- 161
 - 15  1 - 225
64
64
1- 

240
 8 

161
-15

The next (and final) set of formulas we
have are called half-angle formulas.
u
1
Cos
u
Sin  
2
2
1 Cos u
u
Cos  
2
2
1 - Cos u
u
Tan 
2
Sin u
Sin u

1  Cos u
The sign of Sin and Cos depend on what
quadrant u/2 is in

Use the following triangle to find the six trig
functions of θ/2

7
2
Sin 
2
10
25
θ

7 2
Cos 
2
10

1
Tan 
2
7
7
25
24

1 24
2

1 Cos u
Sin  
2
2
θ
25 
1
1
1
1
25 


50 5 2
2
50
2

10
7
25
24

1 24
2

1 Cos u
Cos  
2
2
θ
25 
49
7
7
49
25 


50 5 2
2
50
7 2

10
25
7
24

1 24
7
25
25

1 - Cos u
Tan 
2
Sin u
θ
1
25

7
25
1

7
Find the exact value of the Cos 165º.
165º is half of what angle?
330
Cos 165º = Cos
2
330
1
Cos
330
Cos
 


2
2
1 3
2
2
2 3
2 3  2 3


2

4
2
2
Find the exact value of the Sin 105º.
105º is half of what angle?
210
Sin 105º = Sin
2
210
Sin

2

1 Cos 210

2
2 3
2
2
1  3
2 3

4
2
2
2 3

2
Find the exact value of the Tan 15º.
15º is half of what angle?
30
Tan 15º = Tan
2
1 Cos 30
30
Tan


2
Sin 30
1- 3
1
2
2
 2 3
2 3
2

1
2
Section 5.5
12

Given Sin x 
and
 x   , Find :
13
2
x 3 13
a) Sin

2
13
x
2 13
b) Cos 
2
13
3
x

c) T an
2
2
12
13
x
-5
x
1 Cos u
a) Sin

2
2
12
13
x
1 - 5
13

2
18
13

2
1 5
13

2
-5
3 13
18
9  3



13
13
26
13
x
1 Cos u
b) Cos  
a
2
2
13
12
x
1 - 5
13

2

8
26

8
13
2
-5
2
2 13
4



13
13
13
x
c) T an
2
1 Cos x

Sin x
1 5
1- - 5
13
13


12
12
13
13
18
18
3
13



12
12
2
13
12
13
x
-5
3

Given Tan x 
and 0  x  , Find :
4
2
x
a) Csc
 10
2
x
10
b) Sec 
2
3
x
3
c) Cot
2
5
3
x
4
5
x
1 Cos u
a) Sin

2
2

1 4
2
5

1
5
2
3
x
4
1

10
1

 10
10
5
x
1 Cos u
b) Cos  
a
2
2

1 4
2
5
9
x
4
5
2
9

10
3

10
10

3

3
1 Cos x

Sin x
x
c) T an
2

1- 4
3
5
5
5
1
 5
3
5
3
x
4
1

3
3

1)
2)
Solving Equations using the half-angle
formulas:
Apply the appropriate formula
Use the various methods we have learned to
solve equations
1)
2)
3)
4)
Factor
Combine Like Terms
Isolate the Trig Function
Solve the Equation for an Angle(s)

Solve the following equation for x in the
interval [0, 2π)
2
 1 - Cos x 
x
  Cos x
2 Sin    Cos x  2 

2
2
 


2
 1 - Cos x 
2
  Cos x  1 - Cos x  Cos x
2


5


1
1  2Cos x Cos x 
x ,
2
3 3

Solve the following equation for x in the
interval [0, 2π)
x
2 - Sin x  2Cos  
2
2
2
 1  Cos x 

 2 - Sin x  2

2


2
2
 1  Cos x 
2 - Sin x  2

2


2
 1  Cos x 
2 - Sin x  2

2


2
2 - Sin x  1  Cos x
2
 2 - (1- Cos2 x)  1  Cos x
 2 - 1  Cos x  1  Cos x  Cos x - Cos x  0
2
 Cos x (Cos x - 1)  0
2
Cos x  0 Cos x  1
 3
,
x ,
2 2
0

Solve the following equation for x in the
interval [0, 2π)
x
Sin    Cos x - 1  0
2
1  Cos x

 Cos x - 1  0
2
2
 1  Cos x 
2

 ( 1 - Cos x)
2


2
 1  Cos x 
2

 ( 1 - Cos x)
2


1  Cos x
2
 1 - 2Cos x  Cos x
2
2
1 - Cos x  2 - 4Cos x  2Cos x
2Cos x - 3Cos x  1  0
2
(2Cos x - 1) (Cos x - 1)  0
(2Cos x - 1) (Cos x - 1)  0
2Cos x - 1  0
Cos x  1
2
 5
x ,
3 3
Cos x - 1  0
Cos x  1
x 0
Because we squared both sides, check
your answers!