$\cos(x)$ strictly increasing on $(-\pi,0)$ How can one show that $f(x) =\cos(x)$ is strictly increasing on $(-\pi,0)$? That is, how can one show that $\sin(x)&lt;0$ for all $x \in (-\pi,0)$ (as $f'(x)=-\sin(x)$) rigorously? Although I know that the image of $\sin$ on this interval is $(-1,0)$ and this should suffice, I'm not sure how to prove that this is the image.
(continuing from the other answerer's post having gotten to the question of how to show $\sin(x)$ is postive on the interval $(0,\pi)$)

Using the complex analysis definition of $\sin$ and power series expansions of $e^u$,

Let $x\in(0,\pi)$

Then $\sin(x) = \dfrac{e^{ix}-e^{-ix}}{2i} = \sum\limits_{n=0}^\infty\dfrac{(-1)^n}{(2n+1)!}x^{2n+1} = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+c_9(x)~~$ where $c_9(x)\in(-.09,.09)$.  Expand further if you wish to make the error range smaller.  The error range $c_n$ is $O(\frac{1}{n!})$.

For whatever $x$ you pick in the range, you will be able to find an $n$ such that the error range is smaller than the partial sum up to that point, showing that it is in fact positive on the entire interval.
Let $x'= -x$, then since $x \in (-\pi,0)$, $x' \in (0,\pi)$, and $\sin x' \in (0,1)$. Hence:

$\sin x = -\sin x' \in (-1,0)$.