Evaluate coordinate of turning point : $y={1-\sin x\over 1+\cos x}$ How do I find the coordinate of the turning of $(1)$
$$y={1-\sin x\over 1+\cos x}={u\over v}\tag1$$
Using the Quotient rule:
$${dy\over dx}={vu^{'}-v^{'}u\over v^2}$$
$u=1-\sin x$, $u^{'}=-\cos x$
$v=1+\cos x$, $v^{'}=-\sin x$
$${dy\over dx}={-\cos x(1+\cos x)+(1-\sin x)\sin x\over (1+\cos x)^2}$$
$$={-1-\cos x+\sin x\over (1+\cos x)^2}$$
Turning point happens ${dy\over dx}=0$.
$$1+\cos x=\sin x\tag2$$
How do I solve for $x$?
@King tut: 
$$\sqrt{2}\sin(x-{\pi\over 4})=1$$
$$x={\pi\over 2}$$
$$y={1-\pi\over 1+\pi}$$
Turning point $({\pi\over 2},{1-\pi\over 1+\pi})$

$y={1-\sin x\over 1+\cos x} = \frac{1}{1+\cos{x}}-\frac{\sin{x}}{1+\cos{x}} = \frac{1}{2} \left(1+\tan^2{\frac{x}{2}} \right) - \tan{\frac{x}{2}}$
$y' = \frac{1}{2}\tan{\frac{x}{2}}(1+\tan^2{\frac{x}{2}}) - \frac{1}{2}(1+\tan^2{\frac{x}{2}}) \stackrel{!}{=} 0 \Rightarrow \tan{\frac{x}{2}} = 1 \rightarrow x_t = \frac{\pi}{2}+2k\pi \; k \in \mathbb{Z}$
$y''(x) = \left( \frac{1}{2}(1+\tan^2{\frac{x}{2}})(\tan{\frac{x}{2}}-1) \right)' = \frac{1}{2}(\ldots)(\tan{\frac{x}{2}}-1) + \frac{1}{2}(1+\tan^2{\frac{x}{2}})\frac{1}{2}(1+\tan^2{\frac{x}{2}})$
$\Rightarrow y''(x_t) \gt 0$
