Download 5.3 Double Angle and Half Angle Formulas

Section 5.3 Notes Page 1
5.3 Double Angle and Half Angle Formulas
If we have either a double angle 2 or a half angle

2
then these have special formulas:
sin(2 )  2 sin  cos 
cos(2 )  cos 2   sin 2 
cos(2 )  2 cos 2   1
cos(2 )  1  2 sin 2 
tan(2 ) 
sin

cos
tan
2

2

2
There are three formulas for cos(2 )
2 tan 
1  tan 2 

1  cos 
2
You will choose plus or minus depending on what quadrant

1  cos 
2
You will choose plus or minus depending on what quadrant

1  cos 
1  cos 
You will choose plus or minus depending on what quadrant
There are better formulas for tan
tan

2

sin 
,
1  cos 
tan

2


2

2

2

2
is.
is.
is.
that don’t involve a plus or minus:
1  cos 
sin 
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
Section 5.3 Notes Page 2
Now we can get our first 5 trigonometric functions by reading off our triangle:
csc  
sin   0.6
tan  
0.6
 0.75
 0.8
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 ) 
For the half angle formulas I need to determine which quadrant

2
is in. To do this let’s first start with our
given statement 90     180  . If I divide everything by two we get: 45  
in the first quadrant, so sine, cosine, and tangent of
2
2
 90  . This tells us that

2
is
should all be positive.

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)
1.8

sin 

 .9  0.95 .
2
2
2
Now I will find sin






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: sin
by using its formula: cos

sin 
. 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)
Finally I will find tan


. I have three formulas to choose. I will choose: tan


Section 5.3 Notes Page 3
EXAMPLE: Compute sin  , tan  , csc  , sec  , cot  , sin(2 ) , cos(2 ) , tan(2 ) , sin
if you are given cot  

2
, cos

2
, and tan

2
1
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   
1
5

5
5
sec    5
tan   2
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  5  25 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

2
3
5
5
  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(2)
4
 .
2
3
1  (2)
2 tan 
. We already know tan   2 , so we will
1  tan 2 
For the half angle formulas I need to determine which quadrant

2
Section 5.3 Notes Page 4
is in. To do this let’s first start with our
given statement 180     270  . If I divide everything by two we get: 90  
is in the second quadrant, so sine of
2
2
 135  . This tells us that
should be positive, and the cosine and tangent of

2

2
should be negative.

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
Finally I will find tan

. I have three formulas to choose. I will choose: tan


sin 
. We already know
1  cos 
2
2
cos   0.8 , and sin   0.6 so we will substitute in these decimals:
2 5
2 5



5   2 5  5   2 5  5  5   10 5  10   10 5  10   5  1 .
5

tan 
5 5 5
25  5
20
2
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 
. It
We can write this as sin 
2
2
 2 
is positive since 22.5 is in the first quadrant. You get: sin
45 
sin 45 
For tan(22.5  ) we will use: tan


2
1  cos 45 

2

1
2
2
2 
2 2
2

2
2 2

4
2 2
.
2
2
2
2 2 2 2 2
2




 2 1.
2
2 2 2 2 2 2 2
1
2
2
2
2
Section 5.3 Notes Page 5
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 3 2 .
EXAMPLE: Establish the identity:
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
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