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. 
 A: Funny coincidence: just saw this
formula of Euler's
on page 240 of the current
(August 2014)
Fibonacci Quarterly:
$\sin(ma)
=\sum_{n=0}^{\infty}\dfrac{m}{(2n+1)!}(-1)^n\sin^{2n+1}(a) \prod_{k=1}^{n}(m^2-(2k-1)^2) 
$.
Putting
$a = \sin^{-1}(x)$,
so $x = \sin(a)$,
this becomes
$\sin(m\sin^{-1}x)
=\sum_{n=0}^{\infty}\dfrac{m}{(2n+1)!}(-1)^nx^{2n+1} \prod_{k=1}^{n}(m^2-(2k-1)^2) 
$.
Note that there are
only a finite number of terms
if $m$ is an odd integer.
A quite reasonable proof
of Euler's formula
is given in the paper.
(added later)
Here is the proof:
Let 
$f(a) = \sin(m \sin^{-1}(\sin(a)))$
or
$f(x) = \sin(m \sin^{-1}(\sin(x)))$
.
Differentiating w.r.t. $x$,
we get
$f'(x)
=m(1-x^2)^{-1/2} \cos(m \sin^{-1}(x))
$
and
$f''(x)
=m x (1-x^2)^{-3/2}\cos(m\sin^{-1}(x))-m^2(1-x^2)^{-1}\sin(m\sin^)-1)(x))
$
$
=x(1-x^2)^{-1}f'(x) - m^2(1-x^2)^{-1}f(x)
$.
In other
words,
$f''(x)-P(x)f'(x)+Q(x)f(x) = 0$,
where $P(x) = x/(1-x^2)$
and
$Q(x) = m^2/(1-x^2)$.
Since the denominators of
$P$ and $Q$
are non-zero at $0$,
these are analytic there.
Thus,
$x=0$ 
is an ordinary point of the
differential equation
$$(1-x^2)f''(x) - xf'(x) + m^2 f(x) = 0
$$
so there is a
power series solution
of the form
$f(x)
=\sum_{n=0}^{\infty} u_n x^n
$.
Substituting this,
we get
$$u_{n+2}
=-\dfrac{m^2-n^2}{(n+1)(n+2)}u_n
, n \ge 0
.
$$
Since
$u_0
=f(0)
=\sin(m  \sin^{-1}(0))
= 0
$,
$u_{2n} = 0$
for all $n$.
Since
$u_1
=f'(0)
=m(1-0^2)^{-1/2}
= m
$,
$u_{2n+1} 
$
is given by the formula
for all $n$.
A: We have:
$$\arcsin(x)=\frac{\pi}{2}-\arccos x$$
and:
$$ \sin(m\arccos(x))=\sqrt{1-x^2} U_{m-1}(x), \qquad \cos(m\arccos x)=T_m(x), $$
where $T_m$ and $U_{m-1}$ are Chebyshev polynomials of the first and second kind, respectively.
So we simply have:

$$\sin(m\arcsin x)=\left\{\begin{array}{rcl}T_m(x)&\text{ if }&m\equiv 1\pmod{4},\\ \sqrt{1-x^2} U_{m-1}(x)&\text{ if }&m\equiv 2\pmod{4},\\-T_m(x)&\text{ if }&m\equiv 3\pmod{4},\\ -\sqrt{1-x^2} U_{m-1}(x)&\text{ if }&m\equiv 0\pmod{4}.  \end{array}\right.$$

A: 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
