Transcript Double Angle Formulas

Double Angle Formulas
By S.V. Cunningham
Three Rivers Community College
?
2
sin( 90) sin( 2  45) 2 sin( 45) 2 
2
?
2
sin( 90) 2 
2
?
1 2
1 2
sin( 2 A)  2 sin A
sin( 2 A)
 sin( A  A)
 sin A  cos A  cos Asin A
 sin Acos A sin Acos A
 2 sin Acos A
sin( 2 A) 2 sin Acos A
Let sinA=1/5 with A in QI.
Find sin(2A).
sin( 2 A)  2 sin Acos A
2
1

2
2
cos
A


1

 
sin A  cos A  1
 5
2
2
1
cos A  1  sin A
cos A  1 
25
2
cos A   1  sin A
25 1
cos A 

25 25
Let sinA=1/5 with A in QI.
Find sin(2A).
sin( 2 A)  2 sin Acos A
2
1

cos
A


1

 
24
5


cos A 
25
1
cos
A

1

24 2 6
25

cos A 
5
5
25 1
cos A 

25 25
Let sinA=1/5 with A in QI.
Find sin(2A).
1 2 6
sin( 2 A)  2 sin Acos A  2  
5 5
4 6
sin( 2 A)
25
24 2 6

cos A 
5
5
Find sin(90°) using a double
angle formula.
sin( 90)
 sin( 2  45)
 2  sin( 45)  cos( 45)
2
2
2 2 ( 2)
 1
 2 

2
2
2 2
Simplify 2sin30°cos30°.
2 sin 30 cos 30
 sin( 2  30)
 sin( 60)
3

2
   

Simplify sin  cos .
8
8










sin  cos 
8 8
1
   

   2 sin  cos  )
2
8 8
 1 2
1
 1


  sin 2     sin   
2
4 2 2
 8 2
2

4
Find cos(2A).
cos( 2 A)
 cos( A  A)
 cos A  cos A sin A  sin A
2
 cos A 
2
sin A
cos( 2 A)
2
2
 cos A  sin A
Simplify
2
2
cos 15°–sin 15°.
2
2
cos 15  sin 15
 cos( 2  15)
 cos( 30)
3

2
Derive an alternative
form of the identity
2
2
cos(2A)=cos A–sin A.
2
2
cos( 2 A)  cos A  sin A
2
2
 (1  sin A)  sin A
2
 1 2 sin A
2
cos( 2 A)  2 sin A
1
Derive an alternative
form of the identity
2
2
cos(2A)=cos A–sin A.
2
2
cos( 2 A)  cos A  sin A
2
2
 cos A  (1  cos A)
2
2
 cos A  1  cos A
2
 2 cos A  1
2
cos( 2 A) 2 cos A 1
The identity with its alternative
forms.
2
2
cos( 2 A)  cos A  sin A
2
cos(2 A)  1  2 sin A
2
cos(2 A)  2 cos A  1
Suppose that cos x =1/ 10 with
xQ IV, find sin(2x) and cos(2x).
sin( 2 x ) 2 sin x cos x
2
2
cos( 2 x ) cos x  sin x
2
cos( 2 x )  1  2 sin x
2
2
1


cos( 2 x )  2 cos x  1 2 
 1
 10 
1
4
1
 2 1  1  
5
5
10
Suppose that cos x =1/ 10 with
xQ IV, find sin(2x) and cos(2x).
sin( 2 x ) 2 sin x cos x
1 

sin x   1  

2
2
sin x  cos x  1
 10 
1
2
2
sin x  1  cos x sin x   1 
10
2
sin x   1  cos x
9  3
sin x  
10
10
2
Suppose that cos x =1/ 10 with
xQ IV, find sin(2x) and cos(2x).
sin( 2 x ) 2 sin x cos x
6
6
3 1


 2
10
10 10 ( 10 )2
3

5
4
3
9
cos( 2 x )  

sin x  
5
10
10
Suppose that cos x =1/ 10 with
xQ IV, find sin(2x) and cos(2x).
2
2
sin ( 2 x )  cos ( 2 x )  1
2
2
sin ( 2 x )  cos ( 2 x )
2
2
3
sin( 2 x )  
5
4
cos( 2 x )  
5
3  4

     
 5  5
9 16 25



25 25 25
1
Simplify
2
2cos 105°–1.
2
2 cos 105  1
 cos( 2  105)
 cos( 210)
  cos 30
3

2
Prove that
(cos x – sin x)(cos x + sin x)=cos(2x)
proof
(cos x  sin x )(cos x  sin x )
2
2
 cos x  sin x
 cos( 2 x )
1  cos 2
2
Prove that sin  
.
proof
2
1  cos( 2 )
2
2 sin 

2 2
2
2
1  (cos   sin  )
2


sin

2
2
2
1  cos   sin 

2
2
2
sin   sin 

2
3
Prove that cos( 3 )  4 cos   3 cos .
proof
cos( 3 )
 cos( 2   )
 cos 2  cos  sin 2  sin
2
 ( 2 cos   1) cos ( 2 sin cos ) sin
2
3
 2 cos   cos  2 sin  cos
2
3
 2 cos   cos  2(1  cos  )cos
2
3
 2 cos   cos  2cos (1  cos  )
3
Prove that cos( 3 )  4 cos   3 cos .
proof
2
3
 2 cos   cos  2cos (1  cos  )
3
3
 2 cos   cos  2 cos  2 cos 
3
 4 cos   3 cos
2
2
sec B csc B
Prove sec(2 B ) 
2
2
proof
csc B  sec B
2
2
sec B csc B
2
2
csc B  sec B
1
1

2
2
cos
B
sin
B

1
1

2
2
sin B cos B
2
2
sec B csc B
Prove sec(2 B ) 
2
2
proof
csc B  sec B
2
2
sec B csc B
2
2
csc B  sec B
1
1

2
2
cos
B
sin
B

1
1

2
2
sin B cos B
2
2
sec B csc B
Prove sec(2 B ) 
2
2
proof
csc B  sec B
2
2
sec B csc B
2
2
csc B  sec B
1
1

2
2
cos
B
sin
B

1
1

2
2
sin B cos B
2
2
sec B csc B
Prove sec(2 B ) 
2
2
proof
csc B  sec B
2
2
sec B csc B
2
2
csc B  sec B
1
1

2
2
cos
B
sin
B

1
1

2
2
sin B cos B
2
2
sec B csc B
Prove sec(2 B ) 
2
2
proof
csc B  sec B
2
2
sec B csc B
2
2
csc B  sec B
1
1

2
2
cos
B
sin
B

1
1

2
2
sin B cos B
2
2
sec B csc B
Prove sec(2 B ) 
2
2
proof
csc B  sec B
2
2
sec B csc B
2
2
csc B  sec B
1
1

2
2
cos
B
sin
B

1
1

2
2
sin B cos B
2
2
sec B csc B
Prove sec(2 B ) 
2
2
proof
csc B  sec B
2
2
sec B csc B
2
2
csc B  sec B
1
1

2
2
cos
B
sin
B

1
1

2
2
sin B cos B
2
2
sec B csc B
Prove sec(2 B ) 
2
2
proof
csc B  sec B
1
1

2
2
cos
B
sin
B

1
1

2
2
sin B cos B
1
2
2
2
2
sin
B
cos
B

2
2
cos B  sin B
sin B cos B
2
2
sec B csc B
Prove sec(2 B ) 
2
2
proof
csc B  sec B
1
1

2
2
cos
B
sin
B

1
1

2
2
sin B cos B
1
2
2
2
2
sin
B
cos
B

2
2
cos B  sin B
sin B cos B
2
2
sec B csc B
Prove sec(2 B ) 
2
2
proof
csc B  sec B
1
1

2
2
cos
B
sin
B

1
1

2
2
sin B cos B
1
2
2
2
2
sin
B
cos
B

2
2
cos B  sin B
sin B cos B
2
2
sec B csc B
Prove sec(2 B ) 
2
2
proof
csc B  sec B
1
1

2
2
cos
B
sin
B

1
1

2
2
sin B cos B
1
2
2
2
2
sin
B
cos
B

2
2
cos B  sin B
sin B cos B
2
2
sec B csc B
Prove sec(2 B ) 
2
2
proof
csc B  sec B
1
2
2
2
2
sin
B
cos
B

2
2
cos B  sin B

sin B cos B
1
2
2
cos B  sin B
2
2
sec B csc B
Prove sec(2 B ) 
2
2
proof
csc B  sec B
1

2
2
cos B  sin B
1

cos( 2 B )
 sec2 B 
Derive a double angle formula
for the tangent function.
tan( 2 A)
 tan( A  A)
tan A  tan A

1  tan A tan A
2 tan A

2
1  tan A
Derive a double angle formula
for the tangent function.
tan( 2 A)
 tan( A  A)
tan A  tan A

1  tan A tan A
2 tan A

2
1  tan A
Derive a double angle formula
for the tangent function.
tan( 2 A)
 tan( A  A)
tan A  tan A

1  tan A tan A
2 tan A

2
1  tan A
Derive a double angle formula
for the tangent function.
tan( 2 A)
 tan( A  A)
tan A  tan A

1  tan A tan A
2 tan A

2
1  tan A
Derive a double angle formula
for the tangent function.
tan( 2 A)
 tan( A  A)
tan A  tan A

1  tan A tan A
2 tan A

2
1  tan A
Derive a double angle formula
for the tangent function.
tan( 2 A)
 tan( A  A)
tan A  tan A

1  tan A tan A
2 tan A

2
1  tan A
Derive a double angle formula
for the tangent function.
tan( 2 A)
 tan( A  A)
tan A  tan A

1  tan A tan A
2 tan A

2
1  tan A
Derive a double angle formula
for the tangent function.
tan( 2 A)
 tan( A  A)
tan A  tan A

1  tan A tan A
2 tan A

2
1  tan A
Derive a double angle formula
for the tangent function.
tan( 2 A)
 tan( A  A)
tan A  tan A

1  tan A tan A
2 tan A

2
1  tan A
Derive a double angle formula
for the tangent function.
tan( 2 A)
 tan( A  A)
tan A  tan A

1  tan A tan A
2 tan A

2
1  tan A
Derive a double angle formula
for the tangent function.
tan( 2 A)
 tan( A  A)
tan A  tan A

1  tan A tan A
2 tan A

2
1  tan A
Derive a double angle formula
for the tangent function.
tan( 2 A)
 tan( A  A)
tan A  tan A

1  tan A tan A
2 tan A

2
1  tan A
Derive a double angle formula
for the tangent function.
tan( 2 A)
 tan( A  A)
tan A  tan A

1  tan A tan A
2 tan A

2
1  tan A
Derive a double angle formula
for the tangent function.
tan( 2 A)
 tan( A  A)
tan A  tan A

1  tan A tan A
2 tan A
2 tan A
tan(
2
A
)


2
2
1  tan A
1  tan A
Given cos  =1/10 and
x  QIV, find tan(2).
2 tan
tan( 2 ) 
2
1  tan 
1
x
y



cos 

10   1
10 r
y   10  1
 3
y


9
2
2
2
x  y r

3
y
2
2
2 tan  
 3
y r x
1
x
2
2
y r x
2
2
Given cos  =1/10 and
x  QIV, find tan(2).
6
2  ( 3)
2 tan

tan( 2 ) 

2 1 9
2
1  tan  1  ( 3)
2
2
1
x


y


10

1

cos 
y   10  1
 3
y


9
2
2
2
x  y r

3
y
2
2
2 tan  
 3
y r x
1
x
2
2
y r x
10
r
Given cos  =1/10 and
x  QIV, find tan(2).
6
2  ( 3)
2 tan

tan( 2 ) 

2 1 9
2
1  tan  1  ( 3)
y    10   1
y   10  1
 3
y


9
2
2
2
x  y r

3
y
2
2
2 tan  
 3
y r x
1
x
2
2
y r x
6 3


8 4
2
2
3 

tan  
8 

Simplify
2 3 
1  tan   
8 
3 
3 


tan  
2 tan  
8   1
 8   1  tan 2  3  


3  2

2 3  2
8
2


1  tan   
1  tan   
8 
8 
1
 3   1  tan(135)  1  (  tan 45
)) )
  tan  
2
2
4  2
1
1
 ( 1)  
2
2
Given csc t = 7 with t in QII,
find sin(2t) and cos(2t).
2
1
1
1




sin t
cos
t


1



7
csc t
 7
sin( 2t ) 2 sin t cos t
1
cos t   1 
7
2
2
sin t  cos t  1
6
6
2
2
cos t  1  sin t cos t    
7
7
2
cos t   1  sin t
Given csc t = 7 with t in QII,
find sin(2t) and cos(2t).
2
1
1
1




sin t
cos
t


1



7
csc t
 7
sin( 2t ) 2 sin t cos t
1   6   2 6
 2  

2
7  7   7 6
6
cos t    
2 6

7
7
7
Given csc t = 7 with t in QII,
find sin(2t) and cos(2t).
1
1

sin t 
7
csc t
2
2
cos( 2t ) cos t  sin t
2 6 1
2
  6   1   

   7
7
7
7


6
6


cos t    
5
7
7

7