I'm trying to prove from the following implicit equation:
$$ \frac{\sin (p)-\sin (a)}{p-a}=\cos (p)\land -\frac{\pi }{2}\leq a<\pi \land \pi <p\leq \frac{3 \pi }{2} $$ that $\cos(p)$ is strictly decreasing on $a$.
The obvious first step was to try implicitly differentiating it with respect to $p$, to try and prove $p$ was decreasing on that interval (derivative always $ < 0 $) and therefore $\cos(p)$ would also be strictly decreasing (works since obviously $\cos(p)$ on $\pi <p\leq \frac{3 \pi }{2}$ is increasing), from which I recovered:
$$ \frac{(a-p) \cos (a)-\sin (a)+\sin (p)}{\left((a-p)^2-1\right) \sin (p)+(p-a) \cos (p)+\sin (a)}\leq 0 $$
then
$$ (a-p) \cos (a)-\sin (a)+\sin (p)\leq 0 $$
which is in a very similar form to the first equation:
$$ \cos (a)\leq \frac{\sin (a)-\sin (p)}{a-p} $$
This is where I'm completely stuck, is there some easy way to prove the final relation, or a better way to solve this problem all-together? Thanks!
Thanks to Dr. Sonnhard Graubner for the answer! You can rewrite the final equation in terms of the right greater than or equal to -1 (as that is the minimal value of cos), and the rest is just finding max/min values of the expressions on either side of the inequality to prove it.