Download 7.6 Double Angle and Half Angle Formulas

Section 7.6 Notes Page 1
7.6 Double Angle and Half Angle Formulas
If we have either a double angle 2 or a half angle

then these have special formulas:
2
Double Angle Formulas
sin(2 )  2 sin  cos 
cos(2 )  cos 2   sin 2 
cos(2 )  2 cos 2   1
cos(2 )  1  2 sin 2 
2 tan 
tan(2 ) 
1  tan 2 
There are three formulas for cos(2 )
The above double angle formulas can be manipulated to derive power reducing formulas. These formulas will
be useful primarily in Calculus 2:
Power Reducing Formulas
sin 2  
1  cos2 
2
cos 2  
1  cos2 
2
tan 2  
1  cos2 
1  cos2 
Besides double angle formulas there are also half angle formulas:
Half Angle Formulas
sin
cos
tan

2

2

2

1  cos 
2
You will choose plus or minus depending on what quadrant

is.
2

1  cos 
2
You will choose plus or minus depending on what quadrant

is.
2

1  cos 
1  cos 
You will choose plus or minus depending on what quadrant

is.
2
There are better formulas for tan
tan

2

sin 
,
1  cos 
tan

2


2
that don’t involve a plus or minus:
1  cos 
sin 
Section 7.6 Notes Page 2
EXAMPLE: Compute sin  , tan  , csc , sec , cot  , sin(2 ) , cos(2 ) , tan(2 ) , sin

2
, cos

2
, and tan

2
if you are given cos   0.8 and 90    180 . Round decimals to two decimal places.


We need to draw a triangle for this one. We are given that the triangle should be drawn in the second quadrant.
 0.8
We can rewrite our problem as cos  
. We know the adjacent side is –0.8. The hypotenuse is 1. Once
1
we label our triangle we find the third side by using the Pythagorean theorem.
1
0.6

-0.8
Now we can get our first 5 trigonometric functions by reading off our triangle:
sin   0.6
tan  
0.6
 0.75
 0.8
csc 
1
 1.67
0.6
cot  
sec 
1
 1.25
 0.8
 0.8
 1.33
0.6
Next I will find sin(2 ) by using its formula: sin(2 )  2 sin  cos  . We already know sine and cosine, so we
will substitute in those decimals: sin(2 )  2(0.6)(0.8)  0.96 .
Next I will find cos(2 ) by using its formula: cos(2 )  2 cos 2   1 . Notice I had a choice of three formulas
to use. Any of them would give you the correct answer. We already know cos   0.8 , so we will substitute
in this decimal: cos(2 )  2(0.8) 2  1  0.28 .
2 tan 
. We already know tan  0.75 , so we
1  tan 2 
2(0.75)
 1.5
will substitute in this decimal: tan(2 ) 

 3.43 .
2
0.4375
1  (0.75)
Next I will find tan(2 ) by using its formula: tan(2 ) 

is in. To do this let’s first start with our
2


given statement 90    180 . If I divide everything by two we get: 45   90  . This tells us that
is
2
2

in the first quadrant, so sine, cosine, and tangent of
should all be positive.
2
For the half angle formulas I need to determine which quadrant
Section 7.6 Notes Page 3

1  cos 


Now I will find sin by using its formula: sin 
. We chose a positive because
is in the
2
2
2
2
second quadrant. We already know cos   0.8 , so we will substitute in this decimal:

1  (0.8)
1.8
sin 

 .9  0.95 .
2
2
2
1  cos 

. We chose a positive because
is in the
2
2
2
2
second quadrant. We already know cos   0.8 , so we will substitute in this decimal:

1  (0.8)
0.2
cos 

 .1  0.32 .
2
2
2
Now I will find cos

by using its formula: cos




sin 
Finally I will find tan . I have three formulas to choose. I will choose: tan 
. We already know
2
2 1  cos 

0.6
cos   0.8 , and sin   0.6 so we will substitute in these decimals: tan 
 3.
2 1  (0.8)
EXAMPLE: Compute sin  , tan  , csc , sec , cot  , sin(2 ) , cos(2 ) , tan(2 ) , sin

2
, cos

2
, and tan

2
1
if you are given cot   and 180    270 .
2
We need to draw a triangle first. We are given that the triangle should be drawn in the third quadrant. We
know the adjacent side is 1 and the opposite side is 2. However because we are in the third quadrant we need to
make the 1 and 2 negative. Once we label our triangle we find the third side by using the Pythagorean theorem.
This will give us 5 .
-1

-2
5
Now we can get our first 5 trigonometric functions by reading off our triangle:
sin   
2
csc  
5
2
5

2 5
5
cos   
tan  2
1
5

5
5
sec   5
Section 7.6 Notes Page 4
Next I will find sin(2 ) by using its formula: sin(2 )  2 sin  cos  . We already know sine and cosine, so we
 2 5 
5  20 4
 

will substitute in those fractions: sin(2 )  2 
 5   25  5 .
5



Next I will find cos(2 ) by using its formula: cos(2 )  2 cos 2   1 . Notice I had a choice of three formulas
to use. Any of them would give you the correct answer. We already know cos   
5
, so we will substitute
5
2

5
5
2
3
  1 . This simplifies to: 2   1   1   .
in this fraction: cos(2 )  2 

5
5
 25 
 5 
Next I will find tan(2 ) by using its formula: tan(2 ) 
substitute in this number: tan(2 ) 
2 tan 
. We already know tan  2 , so we will
1  tan 2 
2(2)
4
 .
2
3
1  (2)

is in. To do this let’s first start with our
2


given statement 180    270 . If I divide everything by two we get: 90    135 . This tells us that
2
2


is in the second quadrant, so sine of should be positive, and the cosine and tangent of
should be negative.
2
2
For the half angle formulas I need to determine which quadrant
1  cos 

. We chose a positive because
is in the
2
2
2
2
second quadrant. We already know cos   0.8 , so we will substitute in this decimal:
Now I will find sin

by using its formula: sin

5

1   

5


 
sin 
2
2


5 5
5 5
5

.
2
10
1  cos 

. We chose a negative because
is in the
2
2
2
2
second quadrant. We already know cos   0.8 , so we will substitute in this decimal:
Now I will find cos

by using its formula: cos


5
5 5

1   

5

5 5

 
5
cos  

.
2
2
2
10

Section 7.6 Notes Page 5


sin 
Finally I will find tan . I have three formulas to choose. I will choose: tan 
. We already know
2
2 1  cos 
cos   0.8 , and sin   0.6 so we will substitute in these decimals:
2 5
2 5



5
5   2 5  5   2 5  5  5   10 5  10   10 5  10   5  1 .
tan 

2
5 5 5
25  5
20
2

5 5 5 5
5  5 5

1   

5
 5 
EXAMPLE: Compute sin(22.5 ) and tan(22.5 ) using a half-angle formula.
 45  

1  cos 45
 . Then we know that  is 45 degrees, so now we use: sin 
We can write this as sin 
. It
2
2
2


is positive since 22.5 is in the first quadrant. You get: sin

2

1
2
2
2 
2 2
2

2
2 2

4
2 2
.
2
2
45
sin 45
2
2 2 2 2 2
2
For tan(22.5 ) we will use: tan






 2 1.

2
2
1  cos 45
2 2 2 2 2 2 2
1
2
2


2
2
EXAMPLE: Rewrite 1  2 sin 2  12 using a double angle formula. Then find its exact value.
This one is written in the form 1 2 sin 2  , which is a form of cos 2 . In this problem, we will let    12 .
   

Therefore we will substitute this into the formula cos 2 : cos 2  . Simplifying gives us cos . Looking
6
  12 
at our table we see this has an exact value of
EXAMPLE: Establish the identity:
3 2.
cot   tan 
 cos 2
cot   tan 
First we want to change these into sines and cosines.
cos  sin 

sin  cos   cos 2
cos  sin 

sin  cos 
Now get common denominators on the top and bottom.
 cos   cos  sin   sin  





 cos   sin  cos   sin    cos 2
 cos   cos  sin   sin  





 cos   sin  cos   sin  
Multiply and write over a single denominator
Section 7.6 Notes Page 6
cos 2   sin 2 
cos  sin   cos 2
cos 2   sin 2 
cos  sin 
We will get rid of the double fractions and use cos 2   sin 2   1
cos 2   sin 2   cos 2
The left side is the identity for cos 2 .
cos 2  cos 2
EXAMPLE: 4 sin  cos  1  2 sin 2    sin 4
First we will use the identity 1  2 sin 2   cos 2 . Now our problem becomes:
4 sin  cos  cos 2  sin 4
We know that sin 2  2 sin  cos  . We want to rewrite our problem as the following:
2  2 sin  cos   cos 2  sin 4
Now we can use the identity sin 2  2 sin  cos  .
2 sin 2 cos 2  sin 4
We know that 2 sin 2 cos 2 is the same as sin2  2   sin 4 .
sin 4  sin 4
EXAMPLE: Use power-reducing formulas to rewrite cos 4  as an equivalent expression that does not contain
powers of trigonometric functions greater than 1.
First we can rewrite the original problem as: cos 2   cos 2  . This will allow us to use the power reducing
1  cos2  1  cos2 
formula for cosine. We will use the same formula twice:
. Now multiply across the

2
2
1  2 cos2   cos 2 (2 )
1 1
1
top and bottom to get
. We can split up the fraction:
 cos(2 )  cos 2 (2 ) .
4
4 2
4
1  cos4 
Now we will use another power reducing formula: cos 2 (2 ) 
. When you substitute this in you
2
1 1
1 1  cos4 
1 1
1  cos4 
 cos(2 )  
will get
. Multiply to get  cos(2 ) 
. Now split up the last
4 2
4
2
4 2
8
1 1
1 cos4 
3 1
1
fraction:
. Add the fractions to get the final answer:  cos(2 )  cos(4 )
 cos(2 )  
4 2
8
8
8 2
8