LESSON 12 SUM AND DIFFERENCE FORMULAS
The sum and difference formulas for the cosine function:
cos (    )  cos  cos   sin  sin 
cos (    )  cos  cos   sin  sin 
The sum and difference formulas for the sine function:
sin (    )  sin  cos   cos  sin 
sin (    )  sin  cos   cos  sin 
The sum and difference formulas for the tangent function:
tan (    ) 
tan   tan 
1  tan  tan 
tan (    ) 
tan   tan 
1  tan  tan 
Since 75   30   45  , then we can find the exact value of the cosine, sine, and
tangent of 75  using the respective sum formula with 30  and 45  . Since
2
3
5




30  
45


75






and
, then
. Of course, we
6
4
6
4
12
12
12
could have obtained this by converting 75  to units of radians using the conversion
factor
5
.
12

. That is 75 
180 
=
75  

180 
=
75 



= 15 
= 5 
=
180
36
12
The cosine of 75  or
5
:
12
cos 75   cos ( 30   45  )  cos 30  cos 45   sin 30  sin 45  =
 2
3  2 

  1

2  2 
2  2  =
cos
6
4

6 
2
4
=
2
4
…… (a)
5
 





 cos 
   cos cos
 sin sin
=
12
4
6
4
6
4
 6
 2
3  2 

  1

2  2 
2  2  =
The sine of 75  or
6
4

6 
2
4
=
2
4
…… (b)
5
:
12
sin 75   sin ( 30   45  )  sin 30  cos 45   cos 30  sin 45  =
1
2
 2 

 
 2 


sin
1
2
3  2



2  2  =
2
4

6 
6
4
=
2
4
…… (c)
5
 





 sin 
   sin cos
 cos sin
=
12
4
6
4
6
4
 6
 2 

 
 2 


3  2



2  2  =
2
4

6 
6
4
=
4
2
…… (d)
The tangent of 75  or
5
:
12
3
 1
3
tan 30   tan 45 
tan 75   tan ( 30   45 ) 
=
=
3
1  tan 30  tan 45 
1 
(1)
3
3
3
 1
 1
3
3  3
3  3
3
3

3 = 3  3 = 3  3 =
3
3 =
1 
1 
3
3
3 
3
3 
3
6(2 
6

3 
3 
3)
(3  3 ) 2
9  6 3  3
12  6 3
=
=
=
=
9  3
6
6
3
3
= 2 
3 …… (e)
3


 tan
 1
5
 

3
6
4
tan
 tan    
=

 =
12
4
3
 6
1  tan tan
1 
(1)
6
4
3
tan
3
 1
3
3 =  = 2 
1 
3
3 …… (f)
Since 75  is the reference angle for  75  ,  105  ,  255  , and  285  , then we
will be able to find the exact value of the six trigonometric functions for these
angles.
5
7
17 
19 
5
is the reference angle for 
, 
, 
, and 
, then we
12
12
12
12
12
will be able to find the exact value of the six trigonometric functions for these
angles.
Since
Examples Use a reference angle to find the exact value of the six trigonometric
functions of the following angles.
1.
 
7
12
(This
is the 105  angle in units of degrees.)
7
is in the II quadrant. The reference angle of the angle
12
7
5
 
is the angle  ' 
.
12
12
The angle  
Since cosine is negative in the II quadrant and cos
6 
5
=
12
2
by (b)
4
above, then
cos
7
5

cos
=
= 
12
12
sec
7
=
12
4( 2 
2  6
6 
2 
6
=
2 
4( 2 
6)
=
=
4
4
4
4
6)
2 
2
6

above, then
4
2 
6
2 
6
= ( 2 
Since sine is positive in the II quadrant and sin
6
5
12
=
6)
6 
=
4
2
by (d)
sin
7
5
sin
=
=
12
12
csc
7
=
12
4( 6 
6  2
6 
2
4
4
4
6 
2
=
6 
4( 6 
2)
=
2

2)
=
4
6 
2
6 
2
6 
Since tangent is negative in the II quadrant and tan
=
2
5
= 2 
12
3 by (f)
above, then
tan
7
5
=  tan
=  (2 
12
12
cot
2 
1
1
7



=
=
12
2  3
2 
2  3

2.
2 
1
  285 
3
=  (2 
(This
3) =
is the
3)
3
3
= 
2 
3
=
4  3
3  2
19 
angle in units of radians.)
12
The angle   285  is in the IV quadrant. The reference angle of the angle
  285  is the angle  '  75  .
Since cosine is positive in the IV quadrant and cos 75  =
(a) above, then
6 
4
2
by
6 
cos 285  = cos 75  =
4
4
4
sec 285  =
4( 6 
2
6 
6 
4( 6 
2)
6  2
=
2
=
2
2)
=
4

6 
2
6 
2
6 
=
2
Since sine is negative in the IV quadrant and sin 75  =
6 
2
by (c)
4
above, then
sin 285  =  sin 75  = 
csc 285  = 

4( 6 
6  2
2 
4
6 
2)
= 
2
6 
2
4
= 
4( 6 
4
6 
2)
4
2

6 
2
6 
2
= ( 6 
=
2) =
6
Since tangent is negative in the IV quadrant and tan 75  = 2 
above, then
tan 285  =  tan 75  =  ( 2 
3)
3 by (e)
cot 285  = 

3.
2 
3
1
1
2 
= 
3
1
2 
=  (2 
3) =
(This
is the 
   105 

3
2 
3
2 
3
= 
2 
3
=
4  3
3  2
7
angle in units of radians.)
12
The angle    105  is in the III quadrant. The reference angle of the
angle    105  is the angle  '  75  .
6 
Since cosine is negative in the III quadrant and cos 75  =
2
by
4
(a) above, then
cos (  105  ) =  cos 75  = 
sec (  105  ) = 

4( 6 
6  2
2)
4
6 
= 
2
6 
=
4
4
= 
4( 6 
2 
2
6 
2)
4
2
4

= ( 6 
Since sine is negative in the III quadrant and sin 75  =
above, then
sin (  105  ) =  sin 75  = 
6 
4
2
6
6 
2
6 
2
=
2)
6 
4
2
by (c)
csc (  105  ) = 

4( 6 
2)
6  2
2 
4
6 
= 
2
4
= 
4( 6 
6 
2)
2

6 
2
6 
2
= ( 6 
4
2) =
6
Since tangent is positive in the III quadrant and tan 75  = 2 
above, then
tan (  105  ) = tan 75  = 2 
cot (  105  ) =
2 
3
1
=
= 2 
2 
3
1
1
3
=
2 
3 by (e)
3

2 
2 
2 
3
3
=
3
4  3
=
3
Since 15   60   45  , then we can find the cosine, sine, and tangent of 15 
using the respective difference formula with 60  and 45  . Since 60  
4
3








, then 15  
.
4
3
4
12
12
12
obtained this by converting 15  to units of radians



15


15

15

. That is
=
=
= 5 
180 
180 
180
45  

and
3
Of course, we could have
using the conversion factor



1

=
=
.
60
12
12
The cosine of 15  or

:
12
cos 15   cos ( 60   45  )  cos 60  cos 45   sin 60  sin 45  =
1
2
 2 

 
 2 


cos
1
2
3  2


2  2  =
2
4

6 
6
4
=
2
4
…… (g)

 





 cos 
   cos cos
 sin sin
=
12
4
3
4
3
4
 3
 2 

 
 2 


The sine of 15  or
3  2


2  2  =
2
4

6 
6
4
=
2
4
…… (h)

:
12
sin 15   sin ( 60   45  )  sin 60  cos 45   cos 60  sin 45  =
 2
3  2 

  1




2  2 
2  2  =
sin
6
4

6 
2
4
=
2
4
…… (i)

 





 sin 
   sin cos
 cos sin
=
12
4
3
4
3
4
 3
 2
3  2 

  1




2  2 
2  2  =
6
4

6 
2
4
=
4
2
…… (j)
The tangent of 15  or

:
12
tan 15   tan ( 60   45 ) 
3 1
1 
3
3 1
=
3  1
3  2 3 1
3 1
=
3  1
4  2 3
=
2
3 1
tan 60   tan 45 
=
=
1  tan 60  tan 45 
1  3 (1)
2

3 1
3 1
2( 2 
=
2
3)
( 3  1) 2
=
3 1
= 2 
=
3 …… (k)


 tan

 

3 1
3
4
tan
 tan    

 = 1  3 (1) =
12
4
 3
1  tan tan
3
4
tan
3 1
1 
3
3 1
=
3  1
=  = 2 
3 …… (l)
NOTE: We also have that 15   45   30  . So, you can find the exact value of
the cosine, sine, and tangent of 15  using the respective difference formula with
45  and 30  . Of course, you will obtain the same values that were obtained above.
Since 15  is the reference angle for  15  ,  165  ,  195  , and  345  , then we
will be able to find the exact value of the six trigonometric functions for these
angles.
Since
11
13
23 


is the reference angle for  , 
, 
, and 
, then we will
12
12
12
12
12
be able to find the exact value of the six trigonometric functions for these angles.