$A\sin(wt)$ cancels out on both sides, giving us $$B\sin(wt+s)=B\sin(wt)\cos(s)+B\cos(wt)\sin(s)$$
We then divide both sides by $B$,
$$\sin(wt+s)=\sin(wt)\cos(s)+\cos(wt)\sin(s)$$
Which is the trigonometric identity: $$\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)$$
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Apply the rule: $\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)$
$A\sin(wt)$ cancels out on both sides, giving us $$B\sin(wt+s)=B\sin(wt)\cos(s)+B\cos(wt)\sin(s)$$ We then divide both sides by $B$, $$\sin(wt+s)=\sin(wt)\cos(s)+\cos(wt)\sin(s)$$
Which is the trigonometric identity: $$\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)$$