How to prove the following formulas
$$ \sum_{n= 0}^{\infty} \frac{\cos(nx)}{n!} = e^{\cos(x)} \cos(\sin x) \\ \sum_{n= 0}^{\infty} \frac{\sin(nx)}{n!} = e^{\cos(x)} \sin(\sin x) $$
without using complex numbers ?
These summations can be done with complex numbers, by substituting $e^{inx}$ to $\cos(nx)$ and $\sin(nx)$, and then using the taylor expansion of $e^x$. I am aware that it is possible to do the same thing with matrices, by using the matrix to complex number analogy.
On
Define
$$g(x)=\sum_{n= 0}^{\infty} \frac{\cos(nx)}{n!}$$
$$f(x)=\sum_{n= 0}^{\infty} \frac{\sin(nx)}{n!}$$
We can differentiate $g(x)$ and $f(x)$ term by term (it is a seperate exercise to show why this is allowed) to get
$$g'(x)=\sum_{n= 1}^{\infty} \left(-\frac{n \sin (n x)}{n!}\right)=-\sum_{n=0}^\infty\frac{\sin((n+1)x)}{n!}$$
$$=-\sum_{n=0}^\infty\frac{\sin(nx)\cos(x)+\cos(nx)\sin(x)}{n!}=-\cos(x)f(x)-\sin(x)g(x)$$
By the same logic
$$f'(x)=\sum_{n= 1}^{\infty} \left(\frac{n \cos(n x)}{n!}\right)=\sum_{n=0}^\infty\frac{\cos((n+1)x)}{n!}$$
$$=\sum_{n=0}^\infty\frac{\cos(nx)\cos(x)-\sin(nx)\sin(x)}{n!}=\cos(x)g(x)-\sin(x)f(x)$$
Also, we have initial conditions $g(0)=e$ and $f(0)=0$. There are two unique functions that satisfy these ODEs and initial conditions. As we just showed, these are $g(x)$ and $f(x)$. However, we also know that
$$G(x)=e^{\cos (x)} \cos (\sin (x))$$
$$F(x)=e^{\cos (x)}\sin (\sin (x)) $$
Also satisfy these initial conditions and ODEs. We conclude that $G(x)=g(x)$ and $F(x)=f(x)$.
For the second identity, you can use the Chebyshev polynomial of the second kind $U_n(x)$ since $$\sin(nx)=\sin xU_{n-1}(\cos x).$$ Hence, $$\sum_{n=0}^\infty\frac{\sin(n x)}{n!}=\sin x\sum_{n=0}^\infty\frac{ U_{n-1}(\cos x)}{n!}$$ But a generating function for $U$ is $$\sum_{n=0}^\infty\frac{U_{n-1}(x)}{n!}t^n=\frac{e^{t x} \sin \left(t \sqrt{1-x^2}\right)}{\sqrt{1-x^2}}.$$ So substituting $\cos x$ for $x$ gives $$\frac{1}{\sin x}\sum_{n=0}^\infty\frac{\sin(nx)}{n!}t^n=\frac{e^{t\cos x} \sin \left(t \sin x\right)}{\sin x}.$$ Multiplying by $\sin x$ and letting $t=1$ gives $$\sum_{n=0}^\infty\frac{\sin (nx)}{n!}=e^{\cos x}\sin(\sin x).\tag{1}$$ We can ignore the $n=1$ term since $\sin(0)/0!=0$. Now differentiating with respect to $x$ gives $$\sum_{n=1}^\infty\frac{\cos(nx)}{(n-1)!}=e^{\cos x} \cos x \cos (\sin x)-\sin x \sin (\sin x) e^{\cos x}.$$ The left hand side can be rewritten $$\sum_{n=0}^\infty\frac{\cos(nx+x)}{n!}=\sum_{n=0}^\infty\frac{\cos (x) \cos (n x)-\sin (x) \sin (n x)}{n!},$$ so $$\cos x\sum_{n=0}^\infty\frac{\cos(nx)}{n!}=e^{\cos x} \cos x \cos (\sin x)-\sin x \sin (\sin x) e^{\cos x}+\sin x\sum_{n=0}^\infty\frac{\sin(nx)}{n!}.$$ Using (1) and expanding gives $$\cos x\sum_{n=0}^\infty\frac{\cos(nx)}{n!}=\cos x e^{\cos x}\cos(\sin x).$$ Dividing by $\cos x$ then gives the first identity, $$\sum_{n=0}^\infty\frac{\cos(nx)}{n!}=e^{\cos x}\cos(\sin x).$$
I tried another way, using the Bell numbers, but it involved Euler's identity right at the end. Worth including anyway as it's not the usual way to prove the result. Here goes:
Consider, $$\sum_{k=0}^\infty\frac{x^k}{k!}\sum_{n=0}^\infty\frac{n^k}{n!}.$$ Using the Bell numbers (see A099977 and this Wikipedia page; notation not to be confused with the Bernoulli number notation) then by Dobinski's formula we have, $$e\sum_{k=0}^\infty\frac{x^k}{k!}B_k.$$ Using the generating function for the Bell numbers, $$e\sum_{k=0}^\infty\frac{B_k}{k!}x^n=e\cdot e^{e^x-1}=e^{e^x}.$$ Now let $x=ix$ to obtain $$\sum_{k=0}^\infty\frac{(ix)^k}{k!}\sum_{n=0}^\infty\frac{n^k}{n!}=e^{e^{ix}}=e^{\cos x+i\sin x}=e^{\cos x}(\cos(\sin x)+i\sin(x))\\=e^{\cos x}\cos(\sin x)+ie^{\cos x}\sin(\sin x).$$ The left hand side is equivalent to $$\sum_{n=0}^\infty\frac{1}{n!}\sum_{k=0}^\infty\frac{(-1)^k (xn)^{2k}}{(2k)!}+i\sum_{n=0}^\infty\frac{1}{n!}\sum_{k=0}^\infty\frac{(-1)^k (xn)^{2k+1}}{(2k+1)!}.$$ But the inner sums are the Taylor series for $\cos(nx)$ and $\sin(nx)$, so equating real and imaginary parts gives the result.