Download Using Sum/Difference and Double/Half Angle Identities

Using Sum/Difference and Double/Half Angle Identities
Given that sin  5 and cos  13 and that  and  are in Quadrant I,
3
5
find the exact value for each of the following. **NO DECIMALS**
1.
Problem
sin   
a
f
2.
sin   
a
f
3.
cos   
a
f
4.
cos   
a
f
5.
tan   
a
f
6.
tan   
a
f
7.
sin2
8.
sin2
9.
cos2
Apply Identity

Substitute Values

Simplify
10. cos2
11. tan2
12. tan2
13. tan 2 
1
14. tan 2 
1
a
15. tan 2  3
f
Using Sum/Difference and Double/Half Angle Identities KEY
Given that sin  5 and cos  13 and that  and  are in Quadrant I,
3
5
5

find the exact value for each of the following. **NO DECIMALS**
3
4
13
12

5
Problem
1. sin   
a
a
2. sin   
a
f
a
a
5. tan   
a
6. tan   
Substitute Values
Simplify
sin 𝜃 cos 𝛾 + cos 𝜃 sin 𝛾
3 5
4 12
( )( ) + ( )( )
5 13
5 13
63
65
sin 𝛾 cos 𝜃 − cos 𝛾 sin 𝜃
(
12 4
5 3
)( ) − ( )( )
13 5
13 5
33
65
cos 𝛾 cos 𝜃 − sin 𝛾 sin 𝜃
(
5 4
12 3
)( ) − ( )( )
13 5
13 5
cos 𝜃 cos 𝛾 + sin 𝜃 sin 𝛾
4 5
3 12
( )( ) + ( )( )
5 13
5 13
f
3. cos   
4. cos   
Apply Identity
f
f
f
f
tan 𝜃 + tan 𝛾
1 − tan 𝜃 tan 𝛾
tan 𝜃 − tan 𝛾
1 + tan 𝜃 tan 𝛾
7. sin2
3 12
+
4 5 =
3 12
1− ( )
4 5
3 12
−
4 5 =
3 12
1+ ( )
4 5
63
20
16
−
20
33
−
20
56
20
−
16
65
56
65
−
63
16
−
33
56
2 sin 𝜃 cos 𝜃
2 3 4
( )( )
1 5 5
24
25
2 sin 𝛾 cos 𝛾
2 12 5
( )( )
1 13 13
120
169
4 2
3 2
( ) −( )
5
5
7
25
8. sin2
9. cos2
2
2
cos 𝜃 − sin 𝜃
10. cos2
2
5 2
12 2
( ) −( )
13
13
2
cos 𝛾 − sin 𝛾
11. tan2
2 3
6
(
)
1 4 = 4
7
3 2
1 − (4)
16
2 12
24
(
)
1 5
5
2 =
119
12
−
1−( )
25
5
2 tan 𝜃
1 − tan2 𝜃
12. tan2
2 tan 𝛾
1 − tan2 𝛾
13. tan 2 
1
±√
1 − cos 𝜃
1 + cos 𝜃
1
1 − cos 𝛾
±√
1 + cos 𝛾
8
8
4
13
√
=√
=√ =√
18
5
18
9
1 + 13
13
a
f
Note: 𝛾 is in Quadrant I therefore 2 is
in Quadrant I. Choose the positive.
tan 2𝜃 + tan 3𝛾
1 − tan 2𝜃 tan 3𝛾
24 828
+
7 2035
24 828
1− (
7 2035)
Note: Now you have to substitute for
tan 2𝜃 and split 3𝛾 into 𝛾 + 2𝛾. Then
finally substitute for 2𝛾.
From #11 tan 2𝜃 =
Therefore tan 3𝛾 =
24
7
and from #12 tan 2𝛾 = −
tan 𝛾+tan 2𝛾
1−tan 𝛾 tan 2𝛾
=
12 −120
+
5
119
12 −120
1− 5 ( 119 )
=
120
119
828
595
407
119
.
=
828
2035
120
119
1
3
5
1 − 13
𝛾
15. tan 2  3
−
4
1
5 = √ 5 = √1
√
4
9
9
1+
5
5
𝜃
119
169
24
7
1−
Note: 𝜃 is in Quadrant I therefore 2 is
in Quadrant I. Choose the positive.
14. tan 2 
−
2
3
−
54636
5627