Proving $-\cos\alpha = \sin(\alpha-90^\circ)$ from $\sin (\alpha-\beta) = \sin\alpha\cdot\cos\beta-\cos\alpha\cdot\sin\beta$

Question

$$-\cos\alpha = \sin(\alpha-90^\circ)$$ but
$$\sin (\alpha-\beta) = \sin\alpha\cdot\cos\beta-\cos\alpha\cdot\sin\beta$$

How does this work out?

$$\begin{align} \sin\alpha\cdot\cos\beta &=\phantom{-}0\\ \cos\alpha\cdot\sin(-90^\circ)&=-1 \end{align}$$ so I get $$\sin(\alpha-90^\circ) = \cos\alpha$$ not $-\cos\alpha$

The only way to make this work is by taking the absolute value of $\beta$, but how is that logical if we are dealing with $-90^\circ$?

Thanks!

Answer 1

Write

$$\sin(\alpha - 90) = \sin\alpha\cos 90 - \sin 90\cos\alpha $$ $$= \sin \alpha \cdot 0 - 1\cdot\cos \alpha = -\cos \alpha.$$

Answer 2

As an alternative, recall that by the basic definitions and reflection properties

$$\sin(\alpha-90^\circ)=-\sin(90°-\alpha)=-\cos \alpha$$