Trigonometric ratios of angles greater than 90

Question

So, essentially, according to the right triangle definition of trigonometry, trigonometric ratios of acute angles are the ratios of the necessary sides. Then, historically speaking, we derived the sum and difference angle formulae, i.e.,

                            Sin(A±B) = SinACosB ± SinBCosA

Then, perhaps while playing with the equation, we obtain Sinɸ such that ɸ > 90°.

Now, as in the case of acute angles, the ratio of the particular sides seemed to be a constant for a given angle. How do we know that the value of Sin of obtuse angles is a constant, i.e., how did we realise that it is fixed and unchanging like that of acute angles?

Please do not give a unit circle explanation as I know that in it, the trig operators for all angles are expressed as ratios. Is there any proof of my question in the right triangle method?

Answer 1

Consider the obtuse $\triangle$ABC where the line segment $\overline{AB}$ is horizontal and the $\angle$ BAC is obtuse.

Extend the line segment $\overline{AB}$ to the left to point $D$ so that $\angle$ BDC is $90^\circ$.

By convention, $\sin(\angle\text{BAC}) = \sin(\angle\text{DAC})$ and $\cos(\angle\text{BAC}) = - \cos(\angle\text{DAC})$.

Therefore, since $\angle\text{DAC}$ is acute, the previously assumed result that $\sin(\angle\text{DAC})$ is constant kicks in.