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(a) Show that $\cos \left(60^{\circ}+\theta\right)-\cos \left(60^{\circ}-\theta\right)=-\sqrt{3} \sin \theta$

(b) Hence evaluate $\cos 105^{\circ}-\cos 15^{\circ}$ without using a calculator.
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(a) $\cos 60^{\circ} \cdot \cos \theta-\sin 60^{\circ} \cdot \sin \theta-\left(\cos 60^{\circ} \cdot \cos \theta+\sin 60^{\circ} \cdot \sin \theta\right)$
\begin{aligned} & =\left(\frac{1}{2}\right) \cdot \cos \theta-\left(\frac{\sqrt{3}}{2}\right) \cdot \sin \theta-\left(\frac{1}{2}\right) \cdot \cos \theta-\left(\frac{\sqrt{3}}{2}\right) \cdot \sin \theta \\ & =\frac{\cos \theta}{2}-\frac{\sqrt{3} \sin \theta}{2}-\frac{\cos \theta}{2}-\frac{\sqrt{3} \sin \theta}{2} \\ & =\frac{-2 \sqrt{3} \sin \theta}{2}=-\sqrt{3} \sin \theta \end{aligned}

(b) The word hence implies that you must use what you have previously shown or proven. In this example, rewrite the angles $105^{\circ}$ and $15^{\circ}$ as $60^{\circ}+\ldots$ and $60^{\circ}-\ldots$ respectively.
\begin{aligned} & \therefore \cos 105^{\circ}-\cos 15^{\circ} \\ & =\cos \left(60^{\circ}+45^{\circ}\right)-\cos \left(60^{\circ}-45^{\circ}\right) \end{aligned}
It has been shown that $\cos \left(60^{\circ}+\theta\right)-\cos \left(60^{\circ}-\theta\right)=-\sqrt{3} \sin \theta$
\begin{aligned} & \therefore \theta=45^{\circ} \\ & \therefore \cos \left(60^{\circ}+45^{\circ}\right)-\cos \left(60^{\circ}-45^{\circ}\right) \\ & =-\sqrt{3} \sin 45^{\circ}=-\sqrt{3}\left(\frac{\sqrt{2}}{2}\right)=\frac{-\sqrt{6}}{2} \end{aligned}
by Diamond (71,587 points)

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