The core problem was simple:
Determine the interior angles of triangle ABC given side $a = 12.34cm$ and hypotenuse $c = 35.32cm$.
Simple. Clearly, the solution is:
$\sin A = \frac{12.34cm}{35.32cm}; A \approx 20.45º$
$\cos B = \frac{12.34cm}{35.32cm}; B \approx 69.55º$
Now, his argument was that if we round $A$ and $B$ to the same number of significant figures as our input (4 significant figures), in some cases, $A + B \neq 90.00º$ due to "rounding error". As a result, you should calculate $A$, then calculate $B = 90º - A$ "to ensure $A + B = 90º$".
He claims he's graded assignments in which this happened, but I'm skeptical. What would be the point of sig figs if they only work part of the time? I'm trying to find a way to formally disprove his statement. I understand the intuition behind how significant figures would prevent this from happening.
I have a feeling that this would generalize to a proof that for any $a$, $b$, and $z$ such that
$z = a + b$
the following statement (with errors $\epsilon$ and $\omega$ determined by the number of significant figures) will hold:
$z + (\epsilon + \omega) = (x + \epsilon) + (y + \omega)$
Edit to clarify:
Clearly $\arcsin x + \arccos x = 1$ exactly.
The professor is saying that, given inputs with, say, 4 significant figures, calculating $\arcsin x$ would give you, say, $59.34º$. Then, calculating $\arccos x$ could give you something like $30.67º$ (obviously this doesn't work, but he claims some such $x$ exists). Thus, he recommended that we instead just calculate $\arcsin x$ and then subtract it from 90º to calculate $\arccos x$.