Let $\displaystyle \rho := \frac{\cos(\pi x)}{1 + \sin(\pi x)}$ and $\displaystyle \theta := -\frac{\pi}{2}$. How can I see, that
I mean, obviously $\displaystyle \cos(\theta - \varphi) = -\sin(\varphi)$, but I somehow mess up with the simplification.
Define $\rho=\frac{a}{1+b}\ $ i.e. $a=\cos (\pi x)$ and $b=\sin(\pi x)$. Then your LHS equals
\begin{align*} \frac{(b+1)^2-a^2}{(b+1)^2-2a(1+b)\cos(\pi/2+\phi)+a^2} \end{align*}
Since $a^2+b^2=1$ and $\cos(\pi/2+\phi)=-\sin(\phi)$, we have
\begin{align*} \frac{b^2+2b+1-a^2}{2 + 2b + 2a(1+b)\sin(\phi)} &= \frac{2b^2+2b}{2 + 2b + 2a(1+b)\sin(\phi)}\\ &= \frac{b^2+b}{1 + b + a(1+b)\sin(\phi)}\\ &= \frac{b}{1 + a\sin(\phi)}\\ \end{align*}