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

Question

(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.

Answer 1

(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 \(\cos \left(60^{\circ}+\theta\right)-\cos \left(60^{\circ}-\theta\right)=-\sqrt{3} \sin \theta\)
\[
\begin{aligned}
& \therefore \theta=45^{\circ} \\
& \therefore \cos \left(60^{\circ}+45^{\circ}\right)-\cos \left(60^{\circ}-45^{\circ}\right) \\
& =-\sqrt{3} \sin 45^{\circ}=-\sqrt{3}\left(\frac{\sqrt{2}}{2}\right)=\frac{-\sqrt{6}}{2}
\end{aligned}
\]