Simplifying $\sin(a\arcsin(x))$

Question

I am trying to simplify: $$f_a(x)=\sin(a\arcsin(x))$$ We can write: $$f_a(x)=\frac{1}{2} i \left(\sqrt{1-x^2}+i x\right)^{-a}-\frac{1}{2} i \left(\sqrt{1-x^2}+i x\right)^a$$ I tried using the binomial expansion: $$f_a(x)=\frac{i}{2} (i x)^{-a} \sum _{n=0}^{\infty } \left(\frac{\sqrt{1-x^2}}{i x}\right)^n \binom{-a}{n}-\frac{i}{2} (i x)^a \sum _{n=0}^{\infty } \left(\frac{\sqrt{1-x^2}}{i x}\right)^n \binom{a}{n}$$ but it does not seem to lead anywhere… Does this expression relate somehow to some special functions?

EDIT:

Supported by Mathematica I found the following pattern: $$f_a(x)\stackrel{?}{=}\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!}a\prod_{j=0}^{n-1} (a^2-(2j+1)^2) $$

Answer 1

Theorem. For $\alpha\in\mathbb{C}$ and $|x|<1$, we have \begin{equation}\label{sin-arcsin-ser}\tag{T} \sin(\alpha\arcsin x)=\alpha\sum_{k=0}^{\infty} \Biggl(\prod_{\ell=1}^{k}\bigl[(2\ell-1)^2-\alpha^2\bigr]\Biggr) \frac{x^{2k+1}}{(2k+1)!}. \end{equation}

Proof. Let $f_\alpha(x)=\sin(\alpha\arcsin x)$. Then \begin{gather*} f_\alpha'(x)=\frac{\alpha}{\sqrt{1-x^2}\,}\cos(\alpha\arcsin x)=\frac{\alpha}{\sqrt{1-x^2}\,}\sqrt{1-f_\alpha^2(x)}\,\\ \bigl(1-x^2\bigr)\bigl[f_\alpha'(x)\bigr]^2-\alpha^2\bigl[1-f_\alpha^2(x)\bigr]=0,\\ \bigl(1-x^2\bigr)f_\alpha''(x)-xf_\alpha'(x)+\alpha^2f_\alpha(x)=0,\\ \bigl(1-x^2\bigr)f_\alpha^{(3)}(x)-3xf_\alpha''(x)+\bigl(\alpha^2-1\bigr)f_\alpha'(x)=0,\\ \bigl(1-x^2\bigr)f_\alpha^{(4)}(x)-5xf_\alpha^{(3)}(x)+\bigl(\alpha^2-4\bigr)f_\alpha''(x)=0, \end{gather*} and, inductively, \begin{equation}\label{n-deriv-gen+alpha-sin}\tag{P} \bigl(1-x^2\bigr)f_\alpha^{(k+2)}(x)-(2k+1)xf_\alpha^{(k+1)}(x)+\bigl(\alpha^2-k^2\bigr)f_\alpha^{(k)}(x)=0, \quad k\ge0. \end{equation} Letting $x\to0$ in \eqref{n-deriv-gen+alpha-sin} results in \begin{equation}\label{n-deriv-x=0+alpha-sin}\tag{Q} f_\alpha^{(k+2)}(0)+\bigl(\alpha^2-k^2\bigr)f_\alpha^{(k)}(0)=0, \quad k\ge0. \end{equation} Recusing the relation \eqref{n-deriv-x=0+alpha-sin} and considering $f_\alpha(0)=0$ and $f_\alpha'(0)=\alpha$ arrive at \begin{equation*} f^{(2k)}(0)=0 \quad\text{and}\quad f^{(2k+1)}(0)=\alpha \prod_{\ell=1}^{k}\bigl[(2\ell-1)^2-\alpha^2\bigr] \end{equation*} for $k\ge0$. Consequently, the series expansion \eqref{sin-arcsin-ser} is valid.

Remark. For details of the above result and its proof, please refer to the equation (3.22) in Lemma 3.3 in the paper [1] below. For more information on related results, please refer to the papers [2,3,4,5] below.

References

1. F. Qi, Taylor's series expansions for real powers of functions containing squares of inverse (hyperbolic) cosine functions, explicit formulas for special partial Bell polynomials, and series representations for powers of circular constant, arXiv (2021), available online at https://arxiv.org/abs/2110.02749v2.
2. B.-N. Guo, D. Lim, and F. Qi, Maclaurin's series expansions for positive integer powers of inverse (hyperbolic) sine and tangent functions, closed-form formula of specific partial Bell polynomials, and series representation of generalized logsine function, Appl. Anal. Discrete Math. 16 (2022), in press; available online at https://doi.org/10.2298/AADM210401017G.
3. B.-N. Guo, D. Lim, and F. Qi, Series expansions of powers of arcsine, closed forms for special values of Bell polynomials, and series representations of generalized logsine functions, AIMS Math. 6 (2021), no. 7, 7494--7517; available online at https://doi.org/10.3934/math.2021438.
4. F. Qi, Explicit formulas for partial Bell polynomials, Maclaurin's series expansions of real powers of inverse (hyperbolic) cosine and sine, and series representations of powers of Pi, Research Square (2021), available online at https://doi.org/10.21203/rs.3.rs-959177/v3.
5. F. Qi and M. D. Ward, Closed-form formulas and properties of coefficients in Maclaurin's series expansion of Wilf's function, arXiv (2021), available online at https://arxiv.org/abs/2110.08576v1.

Answer 2

This is based on my answer here: https://math.stackexchange.com/q/4157803. Let $$ g(x) = e^{ia\arcsin x} = \sum\limits_{n = 0}^\infty {g_n x^n } . $$ Then $g(x)$ satisfies the non-linear ODE $(1 - x^2 )(g'(x))^2 = -a^2g^2 (x)$. Differentiating this equation and dividing through by $2g'(x)$ yields $$ (1 - x^2 )g''(x) - xg'(x) + a^2 g(x) = 0. $$ Substituting the power series into this equation gives $g_0 = 1$, $g_1 = ia$ (you can see from the definition that $g(0) = 1$, $g'(0) = ia$) and $$ g_{n + 2} = \frac{{n^2-a^2}}{{(n + 1)(n + 2)}}g_n $$ for $n\geq 0$. The power series expansion then follows since $$ f_a(x) = \Im g(x). $$ Your function can also be written in terms of the Gauss hypergeometric function: $$ f_a (x) = ax \cdot {}_2F_1\! \left( {1 - a,1+a;\tfrac{3}{2};\tfrac{1}{2}(1 - \sqrt {1 - x^2 } )} \right). $$