f = sin(A sin(B) + C)
(1)
The basic trigonometric identities
sin(x + y) = sin x cos y + cos x sin
2 sin x cos y = sin(x + y) + sin(x − y)
2 cos x sin y = sin(x + y) − sin(x − y)
sin(A sin(B) + C) = sin(A sin(B)) · cos(C) + cos(A sin(B)) · sin(C)
(2)
Taking the first term
sin(A sin(B)) = 2[J1 (A) · sin(B) + J3 (A) · sin(3B) + J5 (A) · sin(5B) + ...] =
∞
X
=2
J2n+1 (A) sin((2n + 1)B)
n=1
where Jn (A) is a Bessel function of first kind, where A is a argument and n is the order.
sin(A sin(B)) cos(C) = 2
∞
X
J2n+1 (A) sin((2n + 1)B) · cos(C) =
n=0
=
∞
X
J2n+1 (A){sin((2n + 1)B + C) + sin((2n + 1)B − C)}
n=0
Simplifing
X
Jk (A) sin(kB + C),
(3)
kodd
where the sum runs over all odd integers.
For the second term
cos(A sin(B)) = J0 (A) + 2[J2 (A) · sin(2B) + J4 (A) · sin(4B) + J6 (A) · sin(6B) + ...] =
∞
X
=2
J2n (A) cos(2nB)
n=0
cos(A sin(B)) sin(C) = 2
∞
X
J2n (A) cos(2nB) · sin(C) =
n=0
∞
X
J2n (A){sin(2nB + C) − sin(2nB − C)}
n=0
X
Jk (A) sin(kB + C),
(4)
keven
where the sum runs over all even integers, positive and negative.
After Combination of the two halves
f=
+∞
X
Jk (A) sin(kB + C)
k=−∞
1
(5)