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[
A
1
(1
− E
2
)
2
+
E
2
A
2
sin
2
i
sin
2
α
0
][
A
1
+(
A
2
cos
2
i − A
1
)sin
2
α
0
]sin
α
0
+
+2[
A
1
+(
A
2
cos
2
i
A
1
)sin
2
α
0
]
E
2
A
2
sin
2
i
sin
α
0
cos
2
α
0
+
−
A
1
)sin
α
0
cos
2
α
0
.
+[
A
1
(1
E
2
)
2
+
E
2
A
2
sin
2
i
sin
2
α
0
](
A
2
cos
2
i
−
−
Box 16.5 (Longitude
L
(
α
) as a function of meta-longitude
α
).
Power series expansion
Δl
=
N
r
=1
l
r
Δα
r
up to order
N
=2:
A
1
A
2
cos
i
l
1
:= +
A
1
cos
2
α
0
A
2
cos
2
i
sin
2
α
0
,
(16.58)
2
A
1
A
2
cos
i
(
A
2
cos
2
i
A
1
)sin
α
0
cos
α
0
(
A
1
cos
2
α
0
+
A
2
cos
2
i
sin
2
α
0
)
2
−
2
l
2
:=
−
.
The relations (
16.44
) together with the relations (
16.50
) lead to the relations (
16.51
). Let us prove
the other central relations here.
Proof ((
16.47
):
b
1
).
A
2
sin
α
A
1
+(
A
2
cos
2
i
tan
B
=
1
− E
2
sin
i
A
1
)sin
2
α
,
−
A
1
cos
α
[
A
1
+(
A
1
cos
2
i
A
2
dtan
B
d
α
1
cos
2
B
d
B
d
α
=
=
E
2
sin
i
A
1
)sin
2
α
]
3
/
2
,
1
−
−
d
B
d
α
=cos
2
B
dtan
B
1
1+tan
2
B
dtan
B
d
α
=
=
(16.59)
d
α
=[
A
2
A
1
(1
E
2
)sin
i
cos
α
][
A
1
+(
A
2
cos
2
i
A
1
)sin
2
α
]
−
1
/
2
−
−
×
×
[
A
1
(1
− E
2
)
2
+
E
2
A
2
sin
2
i
sin
2
α
]
−
1
⇒
b
1
:=
d
B
d
α
(
α
0
)
.
End of Proof.
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