What is the value of sin(630 + A) + cosA?
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Reduce \(630^\circ = 360^\circ + 270^\circ\), so \(\sin(630^\circ+A)=\sin(270^\circ+A)\).
Using \(\sin(270^\circ+A)=\sin 270^\circ\cos A+\cos 270^\circ\sin A = -\cos A\).
Therefore \(\sin(630^\circ+A)+\cos A = -\cos A+\cos A = \mathbf{0}\).
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