TITLE: Show that $C\frac{dC}{dr}\ + S\frac{dS}{dr}\ = (C^2 + S^2)\cos{\theta}$
QUESTION [5 upvotes]: 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.

REPLY [6 votes]: 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
$$ C\frac{dC}{dr}\ + S\frac{dS}{dr}=\frac12\frac{d}{dr}(C^2+S^2)=\frac12\frac{d}{dr}e^{2r\cos\theta}=e^{2r\cos\theta}\cos\theta=(C^2+S^2)\cos\theta. $$

REPLY [3 votes]: 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 $p$, $\bar{ z^p }=(\bar{z})^p$