Well, I'll take a crack at it, although I think chris identified the nub of the problem in the first comment. We need to tighten up the language from the comments. First, write
$$F(\theta)=F(\theta,k)=\sin \theta \int_{-L}^{+L}h(z)\,e^{-ikz\cos \theta} \,dz=\sin \theta \int_{-\infty}^{+\infty}\tilde{h}(z)\,e^{-ikz\cos \theta} \,dz$$
where
$$\tilde{h}(z)=\cases{h(z) & $-L\le z \le L$ \cr 0 & otherwise \cr}$$
Technically, then, substituting $w = k\cos\theta$, $F(\theta,k)\,/\sin \theta$ is given by the inverse Fourier transform of $\tilde{h}$,
$${F(\theta,\,w\,/\cos\theta) \over \sin\theta} = \int_{-\infty}^{+\infty}\tilde{h}(z)\,e^{-iwz} \,dz$$
The OP indicates that $k$ is the wave number, which suggests a discrete spectrum. But given that the length $2L$ is arbitrary and not a multiple of the period of the exponential factor, I think we have to fall back on the continuous integral transform. They're related anyway, so it's should be no great concern.
Consequently we have
$$\tilde{h}(z) = {1 \over 2\pi} \int_{-\infty}^{+\infty}{F(\theta,\,w\,/\cos\theta)\over\sin\theta}\,e^{iwz} \,dz$$
or, in Mathematica,
FourierTransform[1/Sqrt[2 Pi] F[t, w/Cos[t]] / Sin[t], w, z]
where the factor 1/Sqrt[2 Pi] is needed to balance the default coefficients of the transforms in Mathematica.
Example: Take h[z_] = z, L = 3.
F[t_, k_] = Sin[t] Integrate[z Exp[-I k z Cos[t]], {z, -3, 3}]
(* (2 I (3 k Cos[3 k Cos[t]] - Sec[t] Sin[3 k Cos[t]]) Tan[t]) / k^2 *)
Recover h:
Assuming[0 < t < Pi && z ∈ Reals,
FourierTransform[1/Sqrt[2 Pi] F[t, w/Cos[t]]/Sin[t], w, z] //
Simplify // PiecewiseExpand
]
Of course, we get the midpoints at the discontinuities, which agrees with the theory of Fourier integrals.
Of course, one can do the integral with NIntegrate (or Integrate) when appropriate.
