46. We rewrite Eq. 39-9 as
h
mλ
−
h
mλ
cos φ =
v
1
− (v/c)
2
cos θ ,
and Eq. 39-10 as
h
mλ
sin φ =
v
1
− (v/c)
2
sin θ .
We square both equations and add up the two sides:
h
m
2
1
λ
−
1
λ
cos φ
2
+
1
λ
sin φ
2
=
v
2
1
− (v/c)
2
,
where we use sin
2
θ + cos
2
θ = 1 to eliminate θ. Now the right-hand side can be written as
v
2
1
− (v/c)
2
=
−c
2
1
−
1
1
− (v/c)
2
,
so
1
1
− (v/c)
2
=
h
mc
2
1
λ
−
1
λ
cos φ
2
+
1
λ
sin φ
2
+ 1 .
Now we rewrite Eq. 39-8 as
h
mc
1
λ
−
1
λ
+ 1 =
1
1
− (v/c)
2
.
If we square this, then it can be directly compared with the previous equation we obtained for [1
−
(v/c)
2
]
−1
. This yields
h
mc
1
λ
−
1
λ
+ 1
2
=
h
mc
2
1
λ
−
1
λ
cos φ
2
+
1
λ
sin φ
2
+ 1 .
We have so far eliminated θ and v. Working out the squares on both sides and noting that sin
2
φ+cos
2
φ =
1, we get
λ
− λ = ∆λ =
h
mc
(1
− cos φ) .