Obtaining Fourier series of function without calculating the Fourier coefficients
Yes, this is possible. The way to do is by analogy: With rational functions (that don't have a pole at $x=0$), one may always use a partial fraction's decomposition to rewrite a function in terms as a linear combination of the functions $\frac{1}{1-cx}$ (or powers thereof) and a polynomial. Then we use $\frac{1}{1-x}=1+x+x^2+x^3+\ldots$ (for $|x|<1$) to get a sum. So, for instance, one can go from a rational function back to a series with rational functions like this: $$\frac{1}{x^2+1}=\frac{1/2}{1-ix}+\frac{1/2}{1+ix}=\frac{1}2\sum_{n=0}^{\infty}(ix)^n + (-ix)^n=\sum_{n=0}^{\infty}(-1)^nx^{2n}.$$ This is heavily related to the idea of generating functions - in particular, the coefficient of $x^n$ in a rational function will always obey some homogenous linear recurrence relation after a point.
Now, let's say we've been given the ratio of two trigonometric polynomials - that is a rational function in $e^{ix}$ such as $$f(x)=\frac{10\sin(x)}{101-20\cos(x)}=\frac{5i(e^{-ix}-e^{ix})}{101-10(e^{ix}+e^{-ix})}=\frac{5i(1-e^{2ix})}{101e^{ix}-10(e^{2ix}+1)}.$$ We then continue rearranging into a convenient form, where the denominator has constant term $1$: $$f(x)=\frac{-\frac{1}2i(1-e^{2ix})}{e^{2ix}-\frac{101}{10}e^{ix}+1}$$ Then, to begin partial fraction decomposition, we factor the denominator into terms of the form $(1-ce^{ix})$ which gives $e^{2ix}-\frac{101}{10}e^{ix}+1=(1-10e^{ix})(1-\frac{1}{10}e^{ix})$. Then we want to find the constants so that $$f(x)=c_0+\frac{c_1}{1-\frac{1}{10}e^{ix}}+\frac{c_2}{1-10e^{ix}}$$ which can be done by putting everything over one numerator and equating the numerator of this with that of our previous expression (other techniques for finding the coefficients of a partial fraction expansion work here too, though). One finds that the form of $f(x)$ is $$f(x)=\frac{i}2+\frac{-i/2}{1-\frac{1}{10}e^{ix}}+\frac{-i/2}{1-10e^{ix}}.$$ Now, we proceed with a little care, since naively expanding these terms gives only terms of the form $e^{ix}$ (so we wouldn't recover real valued numbers in the sum) and also wants to use the series $1+10e^{ix}+100e^{2ix}+1000e^{3ix}+\ldots$ which obviously doesn't converge. To get around this, notice that we can actually expand $\frac{1}{1-cx}$ using two identities: $$\frac{1}{1-x}=1+x+x^2+x^3+\ldots$$ $$1- \frac{1}{1-x}=1+\frac{1}x+\frac{1}{x^2}+\frac{1}{x^3}+\ldots$$ where the latter is valid when $|x|>1$ and the former when $|x|<1$. Given that $|e^{ix}|=1$, we expand the term $\frac{-i/2}{1-\frac{1}{10}e^{ix}}$ using the first identity and the term $\frac{i}2-\frac{i/2}{1-10e^{ix}}$ using the latter to get: $$f(x)=\frac{-i}2\left(1+\frac{1}{10}e^{ix}+\frac{1}{100}e^{2ix}+\frac{1}{1000}e^{3ix}+\ldots\right)+\frac{i}2\left(1+\frac{1}{10}e^{-ix}+\frac{1}{100}e^{-2ix}+\frac{1}{1000}e^{-3ix}+\ldots\right)$$ and putting them together into one sum: $$f(x)=\sum_{n=0}^{\infty}\frac{-i}{2}\cdot \frac{e^{nix}}{10^n}+\frac{i}{2}\cdot \frac{e^{-nix}}{10^n}=\sum_{n=0}^{\infty}\frac{\sin(nx)}{10^n}.$$
One might note that repeated factors need special handling since they can result in non-linear denominators in the partial fractions decomposition - but the identity like: $$\frac{1}{(1-x)^n}={n-1 \choose n-1}+{n \choose n-1}x+{n+1 \choose n-1}x^2+{n+2 \choose n-1}x^3+\ldots$$ where, for fixed $n$, the expression ${n+c-1\choose n-1}$ is simply a polynomial in $c$ of degree $n-1$ - so you might get sums like $\sum_{n=0}^{\infty}\frac{n\sin(n)}{10^n}$ at the end.
Another restriction is that $(1-ce^{ix})$ shouldn't be a factor of the denominator for $|c|=1$ since then we can use neither expansion to get a convergent power series. I don't see any easy patch for this, but it holds for any function with no poles.
Since $\sin(0x)=0$, we may start the index at $n=0$. $$\sum_{n=0}^\infty \frac{\sin(nx)}{10^n} = \frac1{2i} \sum_{n=0}^\infty \frac{e^{inx} - e^{-inx}}{10^n} = \frac{1}{2i} \left(\frac{1}{1-\frac{e^{ix}}{10}} - \frac{1}{1-\frac{e^{-ix}}{10}}\right)$$
$$=\frac{1}{2i} \left( \frac{10}{10-e^{ix}} - \frac{10}{10-e^{-ix}}\right)$$
$$=\frac{10}{2i} \left(\frac{10-e^{-ix} - 10 + e^{ix}}{(10-e^{ix})(10-e^{-ix})} \right)$$
$$=\frac{10}{2i} \left(\frac{e^{ix}-e^{-ix}}{100+1-10e^{ix}-10e^{-ix}} \right)$$
$$=\frac{10\sin(x)}{101 - 20 \cos(x)}$$