Sine of Sine inversed

Let

$x = \arcsin{\frac{\sqrt{63}}{8}} = \arccos{\frac18}$

Using half-angle formula twice:

$\sin{\frac{x}{4}} = \sqrt{\frac{1-\cos{(x/2)}}{2}} = \sqrt{\frac{1-\sqrt{\frac{1+\cos{(x)}}{2}} }{2}}$

From the first equation, we have $\cos{x} = 1/8$

So:

$\sin{\left (\frac14 \arcsin{\frac{\sqrt{63}}{8}} \right )} = \sqrt{\frac{1-\sqrt{\frac{9/8}{2}} }{2}} = \frac1{2 \sqrt{2}} = \frac1{\sqrt{8}}$

Hence, $a=1$ and $b=8$ . $a+b = 9$ .

Let $\theta = \arcsin \frac {\sqrt {63}}8$ , $\implies \sin \left(\frac 14 \arcsin \frac {\sqrt {63}}8 \right) = \sin \frac \theta 4$ and $\implies \sin \theta = \frac {\sqrt {63}}8$ and $\cos \theta =$ $\sqrt{1-\sin^2 \theta}$ $= \sqrt{1-\frac {63}{64}}$ $= \frac 18$ .

$\begin{aligned} 2\cos^2 \frac \theta 2 - 1 & = \cos \theta = \frac 18 \\ 2\cos^2 \frac \theta 2 & = \frac 98 \\ \implies \cos \frac \theta 2 & = \frac 34 \end{aligned}$

$\begin{aligned} 2\cos^2 \frac \theta 4 - 1 & = \cos \frac \theta 4 = \frac 34 \\ 2\cos^2 \frac \theta 4 & = \frac 74 \\ \implies \cos \frac \theta 4 & = \sqrt{\frac 78} \end{aligned}$

$\begin{aligned} \implies \sin \frac \theta 4 & = \sin \left(\frac 14 \arcsin \frac {\sqrt {63}}8 \right) \\ & = \sqrt{1-cos^2 \frac \theta 4} = \sqrt{1-\frac 78} = \frac 1{\sqrt 8} \end{aligned}$

$\implies a+b = 1+8 = \boxed{9}$