Complex numbers bijection when solving a series When looking for a closed form of the series:
$$S=\sum_{n=0}^\infty \frac{\sin{(n)}}{n!}$$
Using complex numbers the solution is very elegant and easy:
$$S=\Im\left(\sum_{n=0}^\infty \frac{e^{in}}{n!}\right)=\Im{\left(\Large{e^{e^i}}\right)}=e^{\cos{1}}\sin(\sin{1})$$
So I wondered how is that complex numbers simplify so much such a problem, so I started trying alternative methods to find the bijection between complex numbers and some operation or transformation. First thing I thought about was that if we call the derivative operator $D$, we have that:
$$D^2\cos{t}=-\cos{t} $$
$$D^2\sin{t}=-\sin{t} $$
Therefore in this context we have that $D^2=-I$, so I started looking for differential equations. I called the functions:
$$y(t)=\sum_{n=0}^\infty \frac{sin{(tn)}}{n!}$$
$$x(t)=\sum_{n=0}^\infty \frac{cos{(tn)}}{n!}$$
Now the problem is to solve the following system with initial conditions $x(0)=e \text{, } \space y(0)=0$ and evaluate $y(t)$ in $t=1$:
$$y'=x\cos{t}-y\sin{t}$$
$$x'=-y\cos{t}-x\sin{t}$$
Using the subs $x=R(t)\cos(\theta (t))$ and $y=R(t)\sin(\theta (t))$ I get the answer:
$$y(t)=e^{\cos{t}}\sin(\sin{t})$$
$$x(t)=e^{\cos{t}}\cos(\sin{t})$$
So my question is regarding another possible bijection that I am thinking of, is the one that pops out of the modulus and argument. So one could find the functions by showing that the following expressions are true:
$$x^2(t)+y^2(t)=e^{2\cos{t}}$$
$$\frac{y(t)}{x(t)}=\tan{(\sin{t})}$$
I tried but failed, and with complex numbers this is so easy to show...so the question is how can we show these expressions without using complex numbers. Any discussion on the topic or alternative methods will be useful and appreciated.
 A: Radial part:
Here is an alternative approach for proving $x^2(t)+y^2(t)=e^{2\cos(t)}$ but it uses some (real) Fourier analysis and Bessel functions; don't know if you're ok with that.
Just by squaring the series, we see that $x^2(t) = \sum_{m,n\geq 0}\frac{\cos(mt)\cos(nt)}{m!n!}$ and $y^2(t) = \sum_{m,n\geq 0}\frac{\sin(mt)\sin(nt)}{m!n!}$ which together with $\cos(mt-nt)=\cos(mt)\cos(nt)+\sin(mt)\sin(nt)$ gives
$$
x^2(t)+y^2(t) = \sum_{m,n\geq 0}\frac{\cos(mt-nt)}{m!n!}=\sum_{k=0}^{\infty}\cos(kt)\sum_{\substack{m,n\geq0\\|m-n|=k}}\frac{1}{m!n!} = \\
=\sum_{n=0}^\infty\frac{1}{(n!)^2}+\sum_{k=1}^{\infty}\cos(kt)2\sum_{n=0}\frac{1}{n!(n+k)!}.$$
But as $t\mapsto e^{2\cos(t)}$ is a $2\pi$-periodic, even function we have that 
$$e^{2\cos(t)}=\frac{a_0}{2}+\sum_{k=1}^{\infty}a_k\cos(kt)
$$
where
$$a_k=\frac{2}{\pi}\int_0^\pi e^{2\cos(t)}\cos(kt)dt
$$
are the cosine Fourier-coefficients. It is therefore enough to prove that
$$
\frac{1}{\pi}\int_0^\pi e^{2\cos(t)}\cos(kt)dt=\sum_{n=0}\frac{1}{n!(n+k)!}
$$
for all $k\geq0$. But as you can see here, we have
$$
\sum_{n=0}\frac{1}{n!(n+k)!}=I_k(2)=\frac{1}{\pi}\int_0^\pi e^{2\cos(t)}\cos(kt)dt
$$
where $I_k$ is a modified Bessel function of the first kind, so we're done.
Argument part: We want to prove that $\frac{y(t)}{x(t)}=\tan(\sin(t))$, or equivalently that $y(t)\cos(\sin(t))=x(t)\sin(\sin(t))$. By drawing some insipiration of the above argument, we calculate the Fourier-series of $\cos(\sin(t))$ and $\sin(\sin(t))$. This can be done by using the general identity
$$2I_k(z)=\frac{1}{\pi}\int_{-\pi}^{\pi}e^{z\cos(\theta)}\cos(k\theta)d\theta\\
=\frac{1}{\pi}\int_{-\pi}^{\pi}e^{z\sin(\theta)}\cos(k\pi/2-k\theta)d\theta
$$
which holds for all $z\in\mathbb{C}$, as well as $J_k(z)=i^kI_k(-iz)$, where $J_k$ is the (unmodified) Bessel function of the first kind. By calculating $2J_k(1)$ using these formulas we get after some computation
$$2J_k(1) = \begin{cases}
\frac{1}{\pi}\int_{-\pi}^{\pi}\cos(\sin(\theta))\cos(k\theta)d\theta && \text{ for } k \text{ even,}\\
\frac{1}{\pi}\int_{-\pi}^{\pi}\sin(\sin(\theta))\sin(k\theta)d\theta && \text{ for } k \text{ odd.}
\end{cases}
$$
Furthermore, we have that for $k$ even, $\theta\mapsto\cos(\sin(\theta))\cos(k\theta)$ is even around $0$ and odd around $\frac{\pi}{2}$, which gives
$$\int_{-\pi}^{\pi}\cos(\sin(\theta))\cos(k\theta)d\theta = 2\int_{0}^{\pi}\cos(\sin(\theta))\cos(k\theta)d\theta = 0,
$$
and similarly we have that $\int_{-\pi}^{\pi}\sin(\sin(\theta))\sin(k\theta)d\theta=0$ for $k$ odd. In conclusion, we just obtained
$$\frac{1}{\pi}\int_{-\pi}^{\pi}\cos(\sin(\theta))\cos(k\theta)d\theta=\begin{cases}
2J_k(1) && \text{ for } k \text{ even,}\\
0 && \text{ for } k \text{ odd,}
\end{cases}
$$
as well as 
$$\frac{1}{\pi}\int_{-\pi}^{\pi}\sin(\sin(\theta))\sin(k\theta)d\theta=\begin{cases}
0 && \text{ for } k \text{ even,}\\
2J_k(1) && \text{ for } k \text{ odd,}
\end{cases}
$$
i.e. we calculated the Fourier coefficients of $\cos(\sin(t))$ and $\sin(\sin(t))$. We thus have
$$
\cos(\sin(t)) = J_0(1) + \sum_{\substack{k\geq 1 \\ k\text{ even}}}2J_k(1)\cos(kt)
$$
and
$$
\sin(\sin(t)) = \sum_{\substack{k\geq 1 \\ k\text{ odd}}}2J_k(1)\sin(kt).
$$
We know come back to the equation we seek to prove, namely $y(t)\cos(\sin(t))\overset{?}{=}x(t)\sin(\sin(t))$, and replace each term by its series representation:
$$\sum_{m\geq 0}\frac{\sin(mt)}{m!}\left(J_0(1) + \sum_{\substack{k\geq 1 \\ k\text{ even}}}2J_k(1)\cos(kt)\right)\overset{?}{=}\sum_{m\geq 0}\frac{\cos(mt)}{m!}\sum_{\substack{k\geq 1 \\ k\text{ odd}}}2J_k(1)\sin(kt)\iff\\
\sum_{m\geq 0}\frac{\sin(mt)}{m!}J_0(1)+\sum_{\substack{m\geq0,k\geq 1 \\ k\text{ even}}}\frac{\sin(mt)}{m!}2J_k(1)\cos(kt)\overset{?}{=}\sum_{\substack{m\geq0,k\geq 1 \\ k\text{ odd}}}\frac{\cos(mt)}{m!}2J_k(1)\sin(kt)
$$
We now try to write the sums on both sides in a similar fashion. By defining
$$
c_{m,k}=\begin{cases}
2\sin(mt)\cos(kt) && \text{ if }k\text{ even,}\\
-2\cos(mt)\sin(kt) && \text{ if }k\text{ odd}
\end{cases}
$$
we have by using $2\sin(x)\cos(y)=\sin(x-y)+\sin(x+y)$ that $c_{m,k}=\sin(mt-kt)+(-1)^k\sin(mt+kt)$ and thus
$$\sum_{m\geq 0}\frac{\sin(mt)}{m!}J_0(1)+\sum_{\substack{m\geq0,k\geq 1 \\ k\text{ even}}}\frac{\sin(mt)}{m!}2J_k(1)\cos(kt)\overset{?}{=}\sum_{\substack{m\geq0,k\geq 1 \\ k\text{ odd}}}\frac{\cos(mt)}{m!}2J_k(1)\sin(kt)\iff\\
\sum_{m\geq 0}\frac{\sin(mt)}{m!}J_0(1)+\sum_{\substack{m\geq0,k\geq 1 \\ k\text{ even}}}\frac{c_{m,k}}{m!}J_k(1)\overset{?}{=}-\sum_{\substack{m\geq0,k\geq 1 \\ k\text{ odd}}}\frac{c_{m,k}}{m!}J_k(1)\iff\\
\sum_{m\geq 0}\frac{\sin(mt)}{m!}J_0(1)+\sum_{m\geq0,k\geq 1}\frac{c_{m,k}}{m!}J_k(1)\overset{?}{=}0\iff\\
\sum_{m\geq 0}\frac{\sin(mt)}{m!}J_0(1)+\sum_{m\geq0,k\geq 1}\frac{\sin(mt-kt)+(-1)^k\sin(mt+kt)}{m!}J_k(1)\overset{?}{=}0.
$$
We now examine for a given integer $n\geq1$ the coefficient of $\sin(nt)$ on the LS. By carefully going through each term of both sums, we find that this coefficient is equal to
$$\frac{J_0(1)}{n!}+\sum_{\substack{m\geq0,k\geq 1 \\ m=k+n}}\frac{J_k(1)}{m!}-\sum_{\substack{m\geq0,k\geq 1 \\ m+n=k}}\frac{J_k(1)}{m!}+\sum_{\substack{m\geq0,k\geq 1 \\ m+k=n}}\frac{(-1)^kJ_k(1)}{m!}=\\
\frac{J_0(1)}{n!}+\sum_{m\geq n+1}\frac{J_{m-n}(1)}{m!}-\sum_{m\geq0}\frac{J_{m+n}(1)}{m!}+\sum_{n>m\geq0}\frac{(-1)^{n-m}J_{n-m}(1)}{m!}=\\
\sum_{m\geq n}\frac{J_{m-n}(1)}{m!}-\sum_{m\geq0}\frac{J_{m+n}(1)}{m!}+\sum_{n>m\geq0}\frac{J_{m-n}(1)}{m!}=\\
\sum_{m\geq 0}\frac{J_{m-n}(1)}{m!}-\sum_{m\geq0}\frac{J_{m+n}(1)}{m!}
$$
where we used $(-1)^{-\nu}J_{-\nu}(z)=J_{\nu}(z)$. But now, as you can see here, we have
$$
\sum_{m\geq 0}\frac{J_{m-n}(1)}{m!}=I_{-n}(1),\quad \sum_{m\geq0}\frac{J_{m+n}(1)}{m!}=I_n(1)
$$
and as $I_{-n}(1)=I_n(1)$ we finally arrive at
$$
\sum_{m\geq 0}\frac{\sin(mt)}{m!}J_0(1)+\sum_{m\geq0,k\geq 1}\frac{\sin(mt-kt)+(-1)^k\sin(mt+kt)}{m!}J_k(1)\overset{?}{=}0\iff\\
\sum_{n\geq1}\sin(nt)(I_{-n}(1)-I_n(1))=0
$$
which now is obvious.
I guess our conclusion can reasonably be that it isn't worth avoiding the direct use of complex numbers for the problem at hand, as things get pretty messy ^^
