General formula for $\sin\left(k\arcsin (x)\right)$ I'm wondering if there's a simple way to rewrite this in terms of $k$ and $x$, especially as a polynomial. It seems to me to crop up every so often, especially for $k=2$, when I integrate with trig substitution. But $k=2$ is not so bad, because I can use the double angle formula; it's the prospect of higher values of $k$ that motivates this question. 
I think the law of sines may help? Or maybe even De Moivre's theorem, to find the length of the hypotenuse as the length of the angle changes, if we think of the right triangle drawn from $\arcsin(x)$ with side1 = $x$, hypotenuse = $1$, and side2 = $\sqrt{1 - x^2}$ as a complex number, though I'm not sure how that might work. 

I would try to write $A[k](x)=\sin (k.\arcsin(x) )$ and $B[k](x)=\cos (k.\arcsin (x) )$ 
and then $A[k+1](x) = \sin ( \arcsin(x) + k.\arcsin (x) ) = x B[k](x) + \sqrt{1-x^2} A[k](x)$
You could then either use the same recurrence form for $B[k+1](x)$ and have a double recurrence relation, or use $B[k](x)=\sqrt{1-A[k](x)^2}$ and have a simple one
