Is there a series of $e^x$ that only contains $\sin(x)$ in the form of $e^x=\sum\limits_{n=0}^{\infty}c_n\cdot \sin(x)^n$? Hy i'm trying to find a series of the following kind:
$e^x=\sum\limits_{n=0}^{\infty}c_n\cdot \sin(x)^n \ \ \ \ \ \ \ \forall \  x\in \left(a,b \right)$
or maybe
$e^x=\sum\limits_{n=0}^{\infty} \sin(c_n\cdot x)^n \ \ \ \ \ \ \ \forall \  x\in \left(a,b \right)$
or any other variation of the series, so that $e^x$ is only expressed in a series of $\sin(x)$ and it's powers.
(I am aware of the fact that $\sin(x)$ is periodic, so the series can only converge on an intervall of maybe $\left(\frac{-\pi}{2},\frac{\pi}{2} \right)$)
I have two questions:

*

*Do such a series exist?

*How is this subject of study called?

I know about the Fourier Series:
$e^x=\frac{e^\pi-e^{-\pi}}{\pi}+\sum\limits_{n=1}^{\infty}a_n\cdot \sin(n\cdot x)+\sum\limits_{n=1}^{\infty}b_n\cdot \cos(n\cdot x) \ \ \ \ \ \ \forall \ x\in (-\pi,\pi)$
and the Taylor Series:
$e^x=\sum\limits_{n=0}^{\infty}\frac{x^n}{n!} \ \ \ \ \ \ \ \ \forall \ x\in\mathbb{R}$
but both of them are not what i'm searching for.
I'm mainly interested if there exists any work on this subject an how it is called, so that i can read into it.
Just with a similar method to the Taylorseries and with matching up of the coefficients with the derivatives of $e^x$, i was able to produce:
$e^x\approx 1+\sin(x)+\frac{1}{2}\sin(x^2)+\frac{1}{3}\sin(x^3)+\frac{1}{4!}\sin(x^4)+\frac{61}{5!}\sin(x^5)+\cdots$
which looks like:
Function plot
which looks pretty cool in my opinion :)
Any help would be appreciated
 A: You can do it on $[0,a]$ for some $a>0$. The "trick" is to extend the function to $[-a,a]$ as an odd function, i.e. for $x<0$, define $f$ as $f(x)=-\exp(-x)$. Then it will be an odd function on $[-a,a]$, so it will only have $\sin$ terms in the Fourier series. This is called Fourier Sine Series.
A: Let $f(u)=e^{\sin^{-1} u}$. Then $f(u)$ has a Taylor expansion $f(u)=\sum_{n=0}^\infty c_n u^n$ at $u=0$. We therefore have
\begin{align*}
e^x&=f(\sin x)\\
&=\sum_{n=0}^\infty c_n \sin^n x
\end{align*}
which is in your first form.
Using WolframAlpha to do the busy-work for us, we can see that
$$
e^x = 1 + \sin x + \frac{1}{2}\sin^2 x + \frac{1}{3}\sin^3 x + \frac{5}{24}\sin^4 x + \frac{1}{5}\sin^5 x + \dots
$$
in a neighborhood of $x=0$ where $\sin^{-1} \sin x = x$ (i.e., for $x \in \left(-\frac{\pi}{2},\frac{\pi}{2}\right)$).
Here's a graph demonstrating this approximation.
A: The answer is yes (given that you're happy for convergence to be on some subinterval of $[0, \pi]$) by appealing to the Stone-Weierstrass theorem. This doesn't give you a constructive way to find the $c_n$ though.
