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A 2 sin i
(1
sin α
A 1 cos 2 α + A 2 cos 2 i sin 2 α
tan B =
(16.38)
E 2 )
versus
tan α = 1
E 2 co s 2 i
1
cos( L
E 2 )sin i tan B +cos i sin( L
1
Ω ) [(1
Ω )] .
(16.39)
E 2
End of Corollary.
Proof.
x 3 = z =0
x 1 = x 1 sin Ω + x 2 cos Ω sin i
x 1 cos Ω − x 2 sin Ω cos i = A 1 cos α sin Ω + A 2 sin α cos Ω cos i
tan L = x 2
A 1 cos α cos Ω − A 2 sin α sin Ω cos i
( A 1 cos α cos Ω
A 2 sin α sin Ω cos i )sin L
=( A 1 cos α sin Ω + A 2 sin α cos Ω cos i )cos L
(16.40)
A 1 cos α (cos Ω sin L
sin Ω cos L )= A 2 sin α cos i (cos Ω cos L +sin Ω sin L )
A 1 cos α sin( L
Ω )= A 2 sin α cos i cos( L
Ω )
A 1 cos i tan α = 1
Ω )= A 2
tan( L
E 2 cos i tan α.
End of Proof.
Proof.
x 3 = z =0
1
x 3
( x 1 ) 2 +( x 2 ) 2 =
tan B =
(16.41)
1
E 2
1
E 2 [ x 2 sin i ][( x 1 cos Ω
x 2 sin Ω cos i ) 2
=
1
+( x 1 sin Ω + x 2 cos Ω cos i ) 2 ] 1 / 2 ,
( x 1 ) 2 +( x 2 ) 2 cos i = A 1 cos 2 α + A 2 sin 2 α cos i,
x 2 sin i = A 2 E sin α sin i
 
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