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Without using a calculator, calculate the value of each of the following:

(1) $\cos 320^{\circ} \cdot \cos 20^{\circ}-\sin 140^{\circ} \cdot \sin 200^{\circ}$

(2) $\cos 10^{\circ} \cdot \sin 160^{\circ}-\sin 10^{\circ} \cdot \sin 110^{\circ}$
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(b) (1)
\begin{aligned} & \cos 320^{\circ} \cdot \cos 20^{\circ}+\sin 140^{\circ} \cdot \sin 200^{\circ} \\ & =\cos \left(360^{\circ}-40^{\circ}\right) \cdot \cos 20^{\circ}+\sin \left(180^{\circ}-40^{\circ}\right) \cdot \sin \left(180^{\circ}+20^{\circ}\right) \\ & =\cos 40^{\circ} \cdot \cos 20^{\circ}+\sin 40^{\circ} \cdot\left(-\sin 20^{\circ}\right) \quad \text { (Reduce the angles to acute angles and } \\ & =\cos 40^{\circ} \cdot \cos 20^{\circ}-\sin 40^{\circ} \cdot \sin 20^{\circ} \quad \text { then apply the appropriate compound } \\ & =\cos \left(40^{\circ}+20^{\circ}\right) \\ & \text { angle formula). } \\ & =\cos 60^{\circ}=\frac{1}{2} \\ & \end{aligned}
(2)
\begin{aligned} & \cos 10^{\circ} \cdot \sin 160^{\circ}-\sin 10^{\circ} \cdot \sin 110^{\circ} \\ & =\cos 10^{\circ} \cdot \sin \left(180^{\circ}-20^{\circ}\right)-\sin 10^{\circ} \cdot \sin \left(180^{\circ}-70^{\circ}\right) \\ & =\cos 10^{\circ} \cdot \sin 20^{\circ}-\sin 10^{\circ} \cdot \sin 70^{\circ} \\ & =\cos 10^{\circ} \cdot \sin 20^{\circ}-\sin 10^{\circ} \cdot \sin \left(90^{\circ}-20^{\circ}\right) \\ & =\cos 10^{\circ} \cdot \sin 20^{\circ}-\sin 10^{\circ} \cdot \cos 20^{\circ} \\ & =\sin 20^{\circ} \cdot \cos 10^{\circ}-\cos 20^{\circ} \cdot \sin 10^{\circ} \\ & =\sin \left(20^{\circ}-10^{\circ}\right) \quad(\sin \mathrm{A} \cdot \cos \mathrm{B}-\cos \mathrm{A} \cdot \sin \mathrm{B}=\sin (\mathrm{A}-\mathrm{B})) \\ & =\sin 10^{\circ} \end{aligned}
by Diamond (71,587 points)

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