Proving two formulae involving two trigonometric series of Euler using Euler identities

I have recently read the book "Тригонометрические ряды от Эйлера до Лебега" (Trigonometric series from Euler to Lebesgue) and on page 12, there are 2 intriguing formlae derived by Euler in 1748. The book doesn't specify from which publication these two series come. Further up on that page, there is a mention of the "введения в анализ бесконечных" (Introduction to the analysis of the infinite), which was published also in 1748. Thus I suspect that these two formulae may come from there.

The above two formulae are as followed:

$$\dfrac{1-r\cos(x)}{1-2r\cos(x)+r^2}=1+r\cos(x)+r^2\cos(2x)...=1+\sum_{n=1}^{+\infty}r^{n}\cos\left(nx\right)\tag{1}$$

$$\dfrac{r\sin(x)}{1-2r\cos(x)+r^2}=r\sin(x)+r^2\sin(2x)...=1+\sum_{n=1}^{+\infty}r^{n}\sin\left(nx\right)\tag{2}$$

In the above book, the author states that Euler didn't provide the radius of convergence for these series, he wrote: "никаких указаний на области применимости этих и подобных им разложений у Эйлера нет" (No indications of range of applications are provided by Euler for these and similar expressions) (page 13)

The radius of convergence should be $|r|<1$ 

At $r=1$ and $r=-1$, Euler obtained:

$$-\dfrac{1}{2}=\cos(x)+\cos(2x)+\cos(3x)+...\tag{3}$$

$$\dfrac{1}{2}=\cos(x)-\cos(2x)+\cos(3x)+...\tag{4}$$

Both of these trigonometric series diverge $\forall x \in \mathbb{R}$

I wish to provide a derivation of these two formulae. This is derived from an answer of Student A Level, a member of Mathstackexchange, the site is here

We proceed by working with the RHS of the equation $(1)$ and $(2)$ and try to turn it into the LHS.

We denote the series in $(1)$ as $C$, and the series in $(2)$ as B

$$C=1+r\cos x+r^2\cos2x+r^3\cos3x+\cdots\\$$

$$S=r\sin x+r^2\sin2x+r^3\sin3x+\cdots\\$$

We will use the Euler identity $\color{green}{e^{ix}=\cos(x)+i\sin(x)}$ and $\color{green}{e^{inx}=\cos(nx)+i\sin(nx)}$

We multiply $S$ with imaginary number $i=\sqrt{-1}$, and obtain the new series:

$$iS=ri\sin(x)+r^2i\sin(2x)+r^3i\sin(3x)...$$

Summing these two series, we have

$$C+iS=1+r[\cos(x)+i\sin(x)]+r^2[(\cos(2x)+i\sin(2x)]+r^3[\cos(3x)+i\sin(3x)]...$$

Using Euler identity, we have:

$$C+iS=1+re^{ix}+(re^{ix})^2+(re^{ix})^{3}+(re^{ix})^4+...$$

This is, in fact, a geometric series, and the general formula is:

$$\dfrac{1}{1-kt}=1+kt+(kt)^2+(kt)^3+(kt)^4+...$$

Let $r=k$ and $t=e^{ix}$

We have $$\dfrac{1}{1-re^{ix}}$$

Let $r=k$ and $t=e^{ix}$

We have $$\dfrac{1}{1-re^{ix}}=1+re^{ix}+(re^{ix})^2+(re^{ix})^{3}+(re^{ix})^4+...$$

On the LHS: we mutiply the fraction with $1-re^{-ix}$, so we will have

$$\dfrac{1-re^{-ix}}{(1-re^{ix})(1-re^{-ix})}=\dfrac{1-r(\cos(x)-ir\sin(x))}{1-re^{-ix}-re^{ix}+r^2}$$

This is the tricky part of the proof, we wish to turn

$$1-re^{-ix}-re^{ix}+r^2=1-r(e^{-ix}+e^{ix})+r^2=1-2r\cos(x)+r^2$$

We can work backward and realize that $\cos(x)=\dfrac{e^{ix}+e^{-ix}}{2}$, so we need to multiply $re^{-ix}-re^{ix}$ with $2r\left(\dfrac{e^{-ix}+e^{ix}}{2}\right)$. So we will have:

$$1-re^{-ix}-re^{ix}+r^2=1-2r\left(\dfrac{e^{-ix}+e^{ix}}{2}\right)+r^2=1-2r\cos(x)+r^2$$

Thus we have:

$$\dfrac{1-r(\cos(x)-i\sin(x)}{1-2r\cos(x)+r^2}=\dfrac{1-r\cos(x)+ir\sin(x)}{1-2r\cos(x)+r^2}=\dfrac{1-r\cos(x)}{1-2r\cos(x)+r^2}+\dfrac{ir\sin(x)}{1-2r\cos(x)+r^2}$$

Equating the imaginary and the real part, we thus have:

$$\boxed{\color{green}{A=\dfrac{1-r\cos(x)}{1-2r\cos(x)+r^2}=\sum_{n=1}^{+\infty}r^{n}\cos\left(nx\right)}}$$

$$\boxed{\color{green}{B=\dfrac{r\sin(x)}{1-2r\cos(x)+r^2}=\sum_{n=1}^{+\infty}r^{n}\sin\left(nx\right)}}$$