# 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!

• You have in the second term $\cos\alpha\sin 90$, not $\cos\alpha\sin(-90)$ – Bernard Sep 6 '18 at 19:32
• So do I take the absolute value 90 instead of -90 simply because angles are never negative? – Pregunto Sep 6 '18 at 19:36
• When matching "$\alpha-\beta$" with "$\alpha-90^\circ$", we have $\beta = 90^\circ$. The value of $\beta$ is whatever's being subtracted from $\alpha$; that's simply $90^\circ$. On the other hand, if you had wanted to use the formula $\sin(\alpha+\beta)$, then you'd want to match "$\alpha+\beta$" with $\alpha-90^\circ$. Here, $\beta$ is something added to $\alpha$, so we reinterpret $\alpha-90^\circ$ as $\alpha+(^{-}90^\circ)$, so that $\beta = -90^\circ$. – Blue Sep 6 '18 at 19:41
• @Pregunto: note at all: if α – β = α – 90, then β = 90, that's all. – Bernard Sep 6 '18 at 19:49
• @Pregunto Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details HERE – user Oct 23 '18 at 21:13

$$\sin(\alpha - 90) = \sin\alpha\cos 90 - \sin 90\cos\alpha$$ $$= \sin \alpha \cdot 0 - 1\cdot\cos \alpha = -\cos \alpha.$$
• In the formula, the minus is not attached to the $90$. See that $\beta$ travels by itself, leaving the minus behind. – B. Goddard Sep 6 '18 at 19:38
$$\sin(\alpha-90^\circ)=-\sin(90°-\alpha)=-\cos \alpha$$