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From Fig. 9.10 we compute the sin and cos of θ and φ in terms of a , b and c , and
then compute their equivalent sin 2
and cos 2
values:
1
cos 2 θ
b 2
b 2
cos θ
=
=
1
sin 2 θ = b 2
sin θ = b
a/ 1
a 2 / 1
b 2
cos 2 φ
b 2
cos φ
=
=
c/ 1
c 2 / 1
b 2 .
sin 2 φ
b 2
sin φ
=
=
To find a 11 :
cos 2 φ cos 2 θ
cos 2 φ sin 2 θ cos α
sin 2 φ cos α
a 11 =
+
+
a 2 b 2
1
c 2
= a 2
+
cos α +
cos α
b 2
b 2
1
c 2
cos α
a 2 b 2
+
a 2
=
+
b 2
1
but
a 2
b 2
c 2
c 2
a 2
b 2
+
+
=
=
1
1
1
cos α
a 2
b 2
a 2 b 2
+
a 2
a 11 =
+
1
b 2
( 1
cos α
a 2 )( 1
b 2 )
a 2
=
+
b 2
1
= a 2 ( 1
cos α) +
cos α.
Let
K
=
1
cos α
then
a 2 K
a 11 =
+
cos α.
To find a 12 :
a 12 =
cos φ cos θ sin θ
cos φ sin θ cos θ cos α
sin φ cos θ sin α
b 2 1
b 2 b 1
b 2 1
a
a
c
=
1
b 2 b
1
b 2 cos α
1
b 2 sin α
= ab ab cos α c sin α
=
ab( 1
cos α)
c sin α
a 12 =
abK
c sin α.
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