How can I solve a difference of inverse sines? I'm trying to solve the equation
$$\arcsin(\lambda_1 / d) - \arcsin(\lambda_2 / d) = \pi / 6$$
for $d$, where $\lambda_1$ and $\lambda_2$ are known.
I don't really know where to start here.
I don't know any identities for inverse trigonometric functions akin to the ones like $\sin(u + v) = \sin u \cos v + \cos u \sin v$.
I see via Wikipedia that these have complex logarithmic forms, but I don't see how that might help (and I don't know any complex analysis).
Mathematica and WolframAlpha each give me a closed-form expression, but refuse to show how:

What's the best/cleanest way to go about solving this?
 A: Note that if $\sin x=y$, then since
$$\sin^2 x+\cos^2 x=1$$
we have that $\cos x=\pm \sqrt{1-y^2}$. Use this together with the sum formula.
A: Matt Samuel's method did end up working, but it got pretty ugly in the end.
Instead, I found a way to do it by letting $\theta_1 = \arcsin(\lambda_1 / d)$ and $\theta_2 = \arcsin(\lambda_2 / d)$, converting this into a $3 \times 3$ system:
$$
\begin{align}\theta_1 - \theta_2 &= \phi \tag{1} \\
d \sin \theta_1 &= \lambda_1 \tag{2} \\
d \sin \theta_2 &= \lambda_2, \tag{3}
\end{align}
$$
where $\phi = \pi / 6$ for brevity.
Then, rearranging $(1)$, we have $\theta_1 = \theta_2 + \phi$, so, substituting into $(2)$:
$$\begin{align*}
d \sin (\phi + \theta_2) &= \lambda_1 \\
d \sin \phi \cos \theta_2 + d \cos \phi \sin \theta_2 &= \lambda_1 \\
d \sin \phi \cos \theta_2 + \lambda_2 \cos \phi &= \lambda_1 \\
d \sin \phi \sqrt{1 - \sin^2 \theta_2} &= \lambda_1 - \lambda_2 \cos \phi \\
d \sin \phi \sqrt{1 - \left( \frac{\lambda_2}{d} \right)^2} &= \lambda_1 - \lambda_2 \cos \phi \\
d \sin \phi \sqrt{\frac{d^2 - \lambda_2^2}{d^2}} &= \lambda_1 - \lambda_2 \cos \phi \\
\sin \phi \sqrt{d^2 - \lambda_2^2} &= \lambda_1 - \lambda_2 \cos \phi \\
d^2 &= \lambda_2^2 + \frac{1}{\sin^2 \phi} (\lambda_1 - \lambda_2 \cos \phi)^2  \\
d &= \sqrt{\lambda_2^2 + \frac{1}{\sin^2 \phi} (\lambda_1 - \lambda_2 \cos \phi)^2}. 
\end{align*}$$
This yields the same answer, but only one square root (instead of the difference of two), no squares of trinomials, and no quartic equations!
(Another nice feature is that the original physics formulation was in terms of $(1)$, $(2)$, and $(3)$, so this attacks the problem closer to its heart.)
