Show that $C\frac{dC}{dr}\ + S\frac{dS}{dr}\ = (C^2 + S^2)\cos{\theta}$ Given $$C=1+r\cos{\theta}\ +\frac{r^2\cos{2\theta}}{2!}\ + \frac{r^3\cos{3\theta}}{3!}\ + \dotsb$$ and $$S = r\sin{\theta}\ + \frac{r^2\sin{2\theta}}{2!}\ + \frac{r^3\sin{3\theta}}{3!}\ + \dotsb$$
Show the following$$C\frac{dC}{dr}\ + S\frac{dS}{dr}\ = (C^2 + S^2)\cos{\theta}$$
I am currently solving the problems given in Differential Calculus for Beginners by Joseph Edwards. As a beginner I am completely clueless about the approach to the above question. I tried to find the answer keys to this book, but sadly non exist on the internet.
 A: Clearly
$$ C+iS = e^{re^{i\theta}} $$
and hence
$$ C^2+S^2=(C+iS)(C-iS)=e^{re^{i\theta}}e^{re^{-i\theta}}=e^{2r\cos\theta}. $$
So
$$ C\frac{dC}{dr}\ + S\frac{dS}{dr}=\frac12\frac{d}{dr}(C^2+S^2)=\frac12\frac{d}{dr}e^{2r\cos\theta}=e^{2r\cos\theta}\cos\theta=(C^2+S^2)\cos\theta. $$
A: Completing the proof of A-level Student:
$$C\frac{dC}{dr}+S\frac{dS}{dr}=\frac12\frac{d}{dr}(C^2+S^2)=\frac12\frac{d}{dr}(C+iS)(C-iS)\\
=\frac12\frac{d}{dr}(e^{re^{i\theta}}\cdot e^{re^{-i\theta}})=\frac12(e^{i\theta}+e^{-i\theta})(e^{re^{i\theta}}\cdot e^{re^{-i\theta}})=(C^2+S^2)\cos\theta
$$
where I have used the fact that for any power $p$, $\bar{ z^p }=(\bar{z})^p$
A: $$
C\frac{dC}{dr}+S\frac{dS}{dr}=C\sum_{n=1}^\infty\frac{nr^{n-1}\cos n\theta}{n!}+S\sum_{n=1}^\infty\frac{nr^{n-1}\sin n\theta}{n!}\\
=C\sum_{n=1}^\infty\frac{ r^{n-1}\cos n\theta}{(n-1)!}+S\sum_{n=1}^\infty\frac{ r^{n-1}\sin n\theta}{(n-1)!}=C\sum_{n=0}^\infty\frac{ r^{n }\cos (n+1)\theta}{ n !}+S\sum_{n=0}^\infty\frac{ r^{n }\sin (n+1)\theta}{ n !}\\
=C\sum_{n=0}^\infty\frac{ r^ n }{ n !}(\cos n\theta\cos\theta-\sin n\theta\sin\theta )+S\sum_{n=0}^\infty\frac{ r^ n }{ n !}(\sin n\theta\cos\theta+\cos n\theta\sin\theta )\\
= C^2 \cos\theta-C S\sin\theta+SC\sin\theta+S^2 \cos\theta
=(C^2+S^2)\cos\theta
$$
A: We have
$$\begin{align}
C+iS&=1+r(\cos\theta+i\sin\theta)+\frac{r^2}{2!}(\cos2\theta+i\sin2\theta)+\frac{r^3}{3!}(\cos3\theta+i\sin3\theta)+\cdots\\
&=1+re^{i\theta}+\frac{1}{2!}(re^{i\theta})^2+\frac{1}{3!}(re^{i\theta})^3+\cdots\\
&=e^{re^{i\theta}}
\end{align}$$
using De Moivre's Theorem and Euler's relation in the second line and the Maclaurin series expansion for $e^x$ in the final line. Using $C+iS$ is a common approach when finding closed form formulae for trigonometric summations (where $C$ is the series for $\cos$ and $S$ is a similar series for $\sin$), whether they be infinite or finite. I can make the working more complete if you like.

ADDED SECTION
This is also equivalent to
$$\begin{align}
(e^{e^{i\theta}})^r &=(e^{\cos\theta+i\sin\theta})^r\\
&=(e^{\cos\theta}\cdot e^{i\sin\theta})^r\\
&=e^{r\cos\theta}(\cos(\sin\theta)+i\sin(\sin\theta))^r\\
&=e^{r\cos\theta}(\cos(r\sin\theta)+i\sin(r\sin\theta))\\
&=e^{r\cos\theta}\cos(r\sin\theta)+ie^{r\cos\theta}\sin(r\sin\theta)
\end{align}$$
So we have
$$C=e^{r\cos\theta}\cos(r\sin\theta)$$
$$S=e^{r\cos\theta}\sin(r\sin\theta)$$

FINAL SECTION
So, if
$$C=e^{r\cos\theta}\cos(r\sin\theta)$$
$$S=e^{r\cos\theta}\sin(r\sin\theta)$$
then $$\frac{dC}{dr}=e^{r\cos\theta}\cos\theta\cos(r\sin\theta)-e^{r\cos\theta}\sin\theta\sin(r\sin\theta)=e^{r\cos\theta}\cos(\theta+r\sin\theta)$$
$$\frac{dS}{dr}=e^{r\cos\theta}\cos\theta\sin(r\sin\theta)+e^{r\cos\theta}\sin\theta\cos(r\sin\theta)=e^{r\cos\theta}\sin(\theta+r\sin\theta)$$
So
$$\begin{align}
C\frac{dC}{dr}+S\frac{dS}{dr}&=e^{2r\cos\theta}\cos(r\sin\theta)\cos(\theta+r\sin\theta)+e^{2r\cos\theta}\sin(r\sin\theta)\sin(\theta+r\sin\theta)\\
&=e^{2r\cos\theta}\cos\theta.
\end{align}$$
on using the compound angle formulae for Cosine.
And finally,
$$(C^2+S^2)\cos\theta=(e^{2r\cos\theta}\cos^2(r\sin\theta)+e^{2r\cos\theta}\sin^2(r\sin\theta))(\cos\theta)=e^{2r\cos\theta}\cos\theta$$
as required!!
Thanks so much for giving me the opportunity to answer this question, I have hugely enjoyed it!
