Inverse sine simplifying problem

Question

It's a simple but not popular problem and there isn't a lot to say here, the problem is:

simplify the following function:

\begin{align} \arcsin(b\sin(x)) \end{align}

And for example, if b=2 what is the resulting simplified function?

Answer 1

You don't just simplify $\sin^{-1}(b \sin(x))$, unless $|b|<1 $... and even if you only consider $-1<b<1$, I doubt you'll get anything pretty...

Answer 2

Use power series to get the result $$ \arcsin(b\sin(x)) = (b)\frac{x^1}{1!} + (b^3-b)\frac{x^3}{3!} + (9b^5-10b^3+b)\frac{x^5}{5!} +\cdots. \tag{1} $$ This series converges for certain values of $\,b\,$ and $\,x.\,$ Except for $\,b=\pm 1\,$ the series is very unlikely to be able to be simplified.

Note that the (unsigned) polynomial coefficients in equation $(1)$ are given by OEIS sequence A008596.

Note the superficially similar result $$ \sin(b\arcsin(x)) = (b)\frac{x^1}{1!} + (-b^3+b)\frac{x^3}{3!} + (b^5-10b^3+9b)\frac{x^5}{5!} +\cdots. \tag{2} $$ Here, if $\,b\,$ is an integer, then this does simplify. For example, $$\sin(2\arcsin(x)) = 2x\sqrt{1-x^2},\; \sin(3\arcsin(x)) = 3x-4x^3, \dots. \tag{3}$$ This case is closely related to the Chebyshev polynomials of the second kind.