Transcript Double-and-Half

Double-Angle and Half-Angle
Identities
Section 5.3
Objectives
• Apply the half-angle and/or double angle
formula to simplify an expression or
evaluate an angle.
• Apply a power reducing formula to
simplify an expression.
Double-Angle Identities
sin(2a )  2 sin(a ) cos( a )
cos( 2a )  cos 2 (a )  sin2 (a )
Half-Angle Identities
a
1  cos a

sin   
2
2
a
1  cos a

cos   
2
2
Power-Reducing Identities
1  cos( 2a )
sin a  
2
2
1  cos( 2a )
cos a  
2
2
Use a half-angle identity to
find the exact
value
 

of sin 
 12 
We will use the half-angle formula for sine
a
1  cos a

sin   
2
2
We need to find out what a is in order to use this formula.
a


2 12

12a  2
a 
2 

12 6
continued on next slide
Use a half-angle identity to
find the exact
value
 

of sin 
 12 
We now replace a with
6
in the formula to get
   
 
1

cos
 
 
6
6
sin     
 2 
2




 3

1  

2
 



sin   
2
 12 
continued on next slide
Use a half-angle identity to
find the exact
value
 

of sin 
 12 
2  3 


 2 




sin    
2
 12 
 
2 3

sin   
4
 12 
 
2 3

sin   
2
 12 
Now all that we have left to do is
determine if the answer should be
positive or negative.
continued on next slide
Use a half-angle identity to
find the exact
value
 

of sin 
 12 
We determine which to use based on what quadrant the original angle is
in. In our case, we need to know what quadrant π/12 is in. This angle
fall in quadrant I. Since the sine values in quadrant I are positive, we
keep the positive answer.
 
2 3

sin  
2
 12 
continued on next slide
If
7
cos(t )  
9
where
3
 t 
2
find the values of the following
trigonometric functions.
cos( 2t )
For this we will need the double angle formula for cosine
cos( 2a )  cos 2 (a )  sin2 (a )
In order to use this formula, we will need the cos(t) and sin(t). We can
use either the Pythagorean identity or right triangles to find sin(t).
continued on next slide
If
7
cos(t )  
9
where
3
 t 
2
find the values of the following
trigonometric functions.
Triangle for angle t
7 2  b 2  92
49  b 2  81
9
b
b 2  32
b   32
length is positive
t
7
b  32
continued on next slide
If
7
cos(t )  
9
where
3
 t 
2
find the values of the following
trigonometric functions.
Triangle for angle t
9
Since angle t is in quadrant III,
the sine value is negative.
32
32
sin(t )  
9
t
7
continued on next slide
If
7
cos(t )  
9
where
3
 t 
2
find the values of the following
trigonometric functions.
cos( 2t )
Now we fill in the values for sine and cosine.
2
32 
 7 

cos( 2t )       
9 
 9 
49 32
cos( 2t ) 

81 81
17
cos( 2t ) 
81
2
continued on next slide
If
7
cos(t )  
9
where
3
 t 
2
find the values of the following
trigonometric functions.
sin(2t )
For this we will need the double angle formula for cosine
sin(2a )  2 sin(a ) cos( a )
We know the values of both sin(t) and cos(t) since we found them for
the first part of the problem.
continued on next slide
If
7
cos(t )  
9
where
3
 t 
2
find the values of the following
trigonometric functions.
sin(2t )
Now we fill in the values for sine and cosine.

32  7 
  
sin(2t )  2 
9  9 

sin(2t ) 
14 32
81
continued on next slide
At this point, we can ask the question “What quadrant is the angle 2t
in?”
This question can be answered by looking at the signs of the sin(2t) and
cos(2t).
14 32
sin(2t ) 
81
is positive
and
17
cos( 2t ) 
81
is positive
The only quadrant where both the sine value and cosine value of
an angle are positive is quadrant I.
If
7
cos(t )  
9
where
3
 t 
2
find the values of the following
trigonometric functions.
t

sin 
2
We will use the half-angle formula for sine
t
1  cost

sin   
2
2
Since we know the value of cos(t), we can just plug that into the formula.
continued on next slide
If
7
cos(t )  
9
where
3
 t 
2
find the values of the following
trigonometric functions. sint2 
t
1  cost

sin   
2
2
 7
1   
t
9


sin   
2
2
 
continued on next slide
If
7
cos(t )  
9
where
3
 t 
2
find the values of the following
trigonometric functions. sint2 
9 7

t

sin    9 9
2
2
 16 
 
t
9


sin   
2
2
 
continued on next slide
If
7
cos(t )  
9
where
3
 t 
2
find the values of the following
trigonometric functions. sint2 
t
16

sin   
18
2
 
Now all that we have left to do is determine if the answer should be
positive or negative. In order to do this, we need to know which quadrant
the angle t/2 falls in.
To do this we will need to use the information we have about the angle t.
continued on next slide
If
7
cos(t )  
9
where
3
 t 
2
find the values of the following
t

trigonometric functions. sin 2 
t
16

sin   
18
2
The information that we have about the angle t is
3
 t 
2
What we need is information about t/2. If we divide each piece of the
inequality, we will get t/2 in the middle of the inequality and bounds for the
angle on the left and right sides.
continued on next slide
If
7
cos(t )  
9
where
3
 t 
2
find the values of the following
t

trigonometric functions. sin 2 
t
16

sin   
18
2
3
2
 3 


2


2
3

4
 t 
t
2 2
 t

2 2


Thus we see that the angle t/2 is in quadrant II.
continued on next slide
If
7
cos(t )  
9
where
3
 t 
2
find the values of the following
t

trigonometric functions. sin 2 
t
16

sin   
18
2
Since the angle t/2 is in quadrant II, the sine value must be positive.
t
16

sin  
18
2
continued on next slide
If
7
cos(t )  
9
where
3
 t 
2
find the values of the following
trigonometric functions.
t

cos  
2
We will use the half-angle formula for sine
t
1  cost

cos   
2
2
Since we know the value of cos(t), we can just plug that into the formula.
continued on next slide
If
7
cos(t )  
9
where
3
 t 
2
find the values of the following
t

trigonometric functions. cos 2 
t
1  cost

cos    
2
2
 7
1   
t
9


cos    
2
2
continued on next slide
If
7
cos(t )  
9
where
3
 t 
2
find the values of the following
trigonometric functions. cost2 
9 7

t

cos     9 9
2
2
2
 
t
9


cos    
2
2
continued on next slide
If
7
cos(t )  
9
where
3
 t 
2
find the values of the following
trigonometric functions. cost2 
t
1
1

cos    

9
3
2
Now all that we have left to do is determine if the answer should be
positive or negative. In order to do this, we need to know which quadrant
the angle t/2 falls in.
To do this we will need to use the information we have about the angle t.
continued on next slide
If
7
cos(t )  
9
where
3
 t 
2
find the values of the following
t

trigonometric functions. cos 2 
t
1

cos    
3
2
The information that we have about the angle t is
3
 t 
2
What we need is information about t/2. If we divide each piece of the
inequality, we will get t/2 in the middle of the inequality and bounds for the
angle on the left and right sides.
continued on next slide
If
7
cos(t )  
9
where
3
 t 
2
find the values of the following
t

trigonometric functions. cos 2 
t
1

cos    
3
2
3
2
 3 


2


2
3

4
 t 
t
2 2
 t

2 2


Thus we see that the angle t/2 is in quadrant II.
continued on next slide
If
7
cos(t )  
9
where
3
 t 
2
find the values of the following
t

trigonometric functions. cos 2 
t
1

cos    
3
2
Since the angle t/2 is in quadrant II, the cosine value must be negative.
t
1

cos    
3
2
Use the power-reducing
formula to simplify the
expression
cos (7x )  sin (7x )
4
4
We need to use the power-reducing identity for the cosine and sine
functions to do this problem.
cos 2 a  
1  cos( 2a )
2
sin2 a  
1  cos( 2a )
2
In our problem the angle a in the formula will be 7x in our problem.
We also need to rewrite our problem.
cos (7x )  sin (7x )
2
2
2
2
continued on next slide
Use the power-reducing
formula to simplify the
expression
cos (7x )  sin (7x )
4
4
Now we just apply the identity to get:
cos
2
(7x )   sin (7x ) 
2
2
2
2
2
 1  cos( 14x )   1  cos( 14x ) 

 

2
2

 

1  2 cos( 14x )  cos 2 (14x ) 1  2 cos( 14x )  cos 2 (14x )

4
4
1  2 cos( 14x )  cos 2 (14x )  1  2 cos( 14x )  cos 2 (14x )
4


continued on next slide
Use the power-reducing
formula to simplify the
expression
cos (7x )  sin (7x )
4
4


1  2 cos( 14x )  cos 2 (14x )  1  2 cos( 14x )  cos 2 (14x )
4
1  2 cos( 14x )  cos 2 (14x )  1  2 cos( 14x )  cos 2 (14x )
4
4 cos( 14x )
4
cos( 14x ) This is much simpler than the original expression and the
power (exponent) is clearly reduced.
continued on next slide
Use the power-reducing
formula to simplify the
expression
cos (7x )  sin (7x )
4
4
Is there another way to simplify this without using a
power-reducing formula?
The answer to this question is yes. The original expression is the
difference of two squares and can be factoring into
cos (7x )  sin (7x )cos (7x )  sin (7x )
2
2
2
2
Now you should notice that the expression in the second set of
square brackets is the Pythagorean identity and thus is equal to 1.
continued on next slide
Use the power-reducing
formula to simplify the
expression
cos 4 (7x )  sin 4 (7x )
Is there another way to simplify this without using a
power-reducing formula?
cos 2 (7x )  sin 2 (7x ) * 1


Now you should notice that what is left is the right side of the
double angle identity for cosine where the angle a is 7x.
cos( 2a )  cos 2 (a )  sin2 (a )
This will allow us to rewrite the expression as
cos( 2(7x ))  cos(14x )