Answer: see proof below
Step-by-step explanation:
Use the following Sum to Product Identities:
[tex]\sin x+\sin y=2\sin\bigg(\dfrac{x+y}{2}\bigg)\cos \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\cos x-\cos y=-2\sin\bigg(\dfrac{x+y}{2}\bigg)\sin \bigg(\dfrac{x-y}{2}\bigg)[/tex]
Proof LHS → RHS
[tex]\text{LHS:}\qquad \qquad \qquad \qquad \dfrac{\sin \theta-\sin 3\theta+\sin 5\theta -\sin 7\theta}{\cos \theta -\cos 3\theta -\cos 5\theta +\cos 7\theta}[/tex]
[tex]\text{Regroup:}\qquad \qquad \qquad \dfrac{(\sin 5\theta+\sin \theta)-(\sin 7\theta +\sin 3\theta)}{-(\cos 5\theta -\cos \theta) +(\cos 7\theta -\cos 3\theta)}[/tex]
[tex]\text{Sum to Product:}\quad \dfrac{2\sin\bigg(\dfrac{5\theta +\theta}{2}\bigg)\cos \bigg(\dfrac{5\theta -\theta}{2}\bigg)-2\sin \bigg(\dfrac{7\theta + 3\theta}{2}\bigg)\cos \bigg(\dfrac{7\theta - 3\theta}{2}\bigg)}{2\sin\bigg(\dfrac{5\theta +\theta}{2}\bigg)\sin \bigg(\dfrac{5\theta -\theta}{2}\bigg)-2\sin \bigg(\dfrac{7\theta + 3\theta}{2}\bigg)\sin \bigg(\dfrac{7\theta - 3\theta}{2}\bigg)}[/tex]
[tex]\text{Simplify:}\qquad \qquad \dfrac{2\sin 3\theta \cos 2\theta-2\sin 5\theta \cos 2\theta}{2\sin 3\theta \sin 2\theta-2\sin 5\theta \cos 2\theta}[/tex]
[tex]\text{Factor:}\qquad \qquad \dfrac{\cos 2\theta(2\sin 3\theta -2\sin 5\theta \cos 2\theta)}{\sin 2\theta(2\sin 3\theta -2\sin 5\theta \cos 2\theta)}[/tex]
[tex]\text{Simplify:}\qquad \qquad \dfrac{\cos 2\theta}{\sin 2\theta}\\\\.\qquad \qquad \qquad =\cot 2\theta[/tex]
LHS = RHS: cot 2Ф = cot 2Ф [tex]\checkmark[/tex]
