Let us assume that $x$ is an angle that lies in the second quadrant i.e. $\dfrac{\pi}{2} < x < \pi$.
We have to prove that $\dfrac{d(\sin x)}{dx} = \cos x$. I will use the unit circle to prove this. The method will be like the one used by Grant Sanderson of 3Blue1Brown in this video, which is a part of his Essence of Calculus series.
The angles are measured in radians. In the diagram below, I have marked the angle $x$ and $dx$ on the unit circle. The angle $dx$ approaches $0$, so it is very very small but for the sake of clarity, I have made it considerably large.
Now, since $dx$ is actually really small, we can approximate arc $AB$ as a straight line approximately perpendicular to $OA$. We are measuring the angles in radians and we have a unit circle, so its radius is $1 \text{ units}$. Hence, the length of arc (now line segment) $AB$ is $\dfrac{\theta}{r}$, where $\theta$ is $\angle AOB$ i.e. $dx$ and $r = 1 \text{ units}$. So, $AB = \dfrac{dx}{1} = dx$.
Now, $d(\sin x) = \sin(x+dx)-\sin x$ which is the change in the ordinate of $A$ and $B$.
Now, $AP = d(\sin x)$ and $AB = dx$. Also, $\triangle APB \sim \triangle AOQ$. So, $\angle BAP = \angle OAQ = \pi - x$. $\cos(\angle BAP) = \dfrac{AP}{AB} = \dfrac{d(\sin x)}{dx}$. And $\cos(\angle BAP) = \cos (\pi - x) = -\cos x$
So, $\dfrac{d(\sin x)}{dx} = -\cos x$ which is not the case at all, since $\dfrac{d(\sin x)}{dx}$ will be negative as $\sin(x+dx) < \sin x$ but the sign of $-\cos x$ will be positive as $\cos x < 0$.
So, what mistake did I make here?
According to me, the mistake was in assuming that $AP = d(\sin x)$. I think that $AP$ should be $|d(\sin x)|$. And as $d(\sin x) < 0 \implies |d(\sin x)| = -d(\sin x)$. This fixes everything but I still want to verify if this indeed is the cause of the error.
Thanks!
PS : Let me know if I should justify why $\triangle APB \sim \triangle OQA$ to make the question clearer.
PPS : It is necessary to prove that differentiating $\sin x$ with respect to $x$ gives $\cos x$ for all 4 quadrants when proving using the unit circle, right?
Note that $$d(\sin x)= \sin(x+dx)-\sin x \lt 0$$as $\sin x$ is decreasing in $\left(\frac{\pi}{2},\pi\right)$. You’re absolutely right in spotting your mistake. When you claim $$AP=d(\sin x)$$ , you’re assigning a negative value to a length, and so you do need an absolute sign there. $$AP=|d(\sin x)| =-d(\sin x)$$