<s>[INST] Proving $-\cos\alpha = \sin(\alpha-90^\circ)$ from $\sin (\alpha-\beta) = \sin\alpha\cdot\cos\beta-\cos\alpha\cdot\sin\beta$ $$-\cos\alpha = \sin(\alpha-90^\circ)$$ 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!
 [/INST] Write
$$\sin(\alpha - 90) = \sin\alpha\cos 90 - \sin 90\cos\alpha $$ $$= \sin \alpha \cdot 0 - 1\cdot\cos \alpha = -\cos \alpha.$$
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