Question about "Approaching Zero and Limits" in the Intuitive Proof of the Derivative of Sine I'm a high school student hoping to self-study some introductory calculus over the summer. While studying, I came across this intuitive proof of the derivative of the sine function, using trig and the unit circle...

  



As in the picture, as dθ approaches zero, angles A and B will approach 90° -- allowing triangle ABC to be "approaching" similar to triangle BDE, but this would mean you would never get the exact angles for triangle ABC; thus, never the exact ratio of sine and cosine in order to complete the proof. 

Is this small (even negligible) inaccuracy inherent to calculus, or is there a flaw in my understanding?

P.S. Please forgive me if this is a stupid/far-too-basic question.
The inaccuracy is of lower order than the main quantities of interest, and becomes fully accurate in the limit. The key to this question is understanding precisely what the limit means, and for this I direct you towards the rigorous meaning of a limit. Note that an initial course in calculus will include many such "hand-wavy" arguments that seem to be slightly inaccurate, and it is typically in a later course (often called "analysis" or "real analysis") following calculus where this rigorous definition is presented and results in calculus are put on a more solid footing.

For a more concrete instance of this, consider an expression like $x+x^2$ in the limit as $x$ tends to $0$. Now of course, $x+x^2$ does not equal $x$, but now if you imagine substituting a super tiny value for $x$, you will see that $x$ becomes a great approximation for $x(1+x)=x+x^2$. Note that this is stronger than saying that $x$ and $x+x^2$ have the same limit as $x$ goes to $0$, since $x^2$ also has this property - yet $x^2$ is a terrible approximation for $x+x^2$ in the limit when $x$ goes to $0$, since $x^2$ is much, much, much smaller than $x$. (I know this example is somewhat contrived, but it actually includes the main ideas in a particularly simple setting.)
If you look at the computer generated image below your drawing, the angle $\phi$ is equal to angle $\theta$. Which means that $\Delta\sin(\theta) = h \cos(\theta)$. Now if you let $\Delta\theta$ approach 0,  $\Delta\theta$ approaches $h$. So when you divide through, you get: 
$$\Delta\sin(\theta)/\Delta\theta = \cos(\theta)$$
Hope this helps!