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If $\sin A = \frac{1}{2}$ and $\tan B = \sqrt{3}$, then verify that $\cos (A + B) = \cos A \cos B - \sin A \sin B$.
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Solution: $\sin A = \frac{1}{2} \Rightarrow A = 30^\circ$, $\tan B = \sqrt{3} \Rightarrow B = 60^\circ$ (1 Mark)
LHS = $\cos (30^\circ + 60^\circ) = \cos 90^\circ = 0$ (1/2 Mark)
RHS = $\cos 30^\circ \cos 60^\circ - \sin 30^\circ \sin 60^\circ$
= $\frac{\sqrt{3}}{2} \times \frac{1}{2} - \frac{1}{2} \times \frac{\sqrt{3}}{2} = 0$ (1/2 Mark)
... LHS = RHS
LHS = $\cos (30^\circ + 60^\circ) = \cos 90^\circ = 0$ (1/2 Mark)
RHS = $\cos 30^\circ \cos 60^\circ - \sin 30^\circ \sin 60^\circ$
= $\frac{\sqrt{3}}{2} \times \frac{1}{2} - \frac{1}{2} \times \frac{\sqrt{3}}{2} = 0$ (1/2 Mark)
... LHS = RHS