Problem Statement
Prove that:
$\frac{\sin 5x - 2\sin 3x + \sin x}{\cos 5x - \cos x} = \tan x$
Verified Solution & Marking Scheme
Group Terms in the Numerator
$\text{Numerator} = (\sin 5x + \sin x) - 2\sin 3x$
Applying the C-D identity $\sin C + \sin D = 2\sin\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)$:
$\sin 5x + \sin x = 2 \sin\left(\frac{5x+x}{2}\right) \cos\left(\frac{5x-x}{2}\right) = 2 \sin 3x \cos 2x$
$\text{Numerator} = 2 \sin 3x \cos 2x - 2 \sin 3x = 2 \sin 3x (\cos 2x - 1)$
Apply C-D Formula to the Denominator
$\text{Denominator} = \cos 5x - \cos x$
Using $\cos C - \cos D = -2\sin\left(\frac{C+D}{2}\right)\sin\left(\frac{C-D}{2}\right)$:
$\cos 5x - \cos x = -2 \sin 3x \sin 2x$
Divide Numerator by Denominator and Cancel
$\frac{\text{Numerator}}{\text{Denominator}} = \frac{2 \sin 3x (\cos 2x - 1)}{-2 \sin 3x \sin 2x} = \frac{\cos 2x - 1}{-\sin 2x} = \frac{1 - \cos 2x}{\sin 2x}$
Using half-angle identities $1 - \cos 2x = 2\sin^2 x$ and $\sin 2x = 2\sin x \cos x$:
$= \frac{2\sin^2 x}{2\sin x \cos x} = \frac{\sin x}{\cos x} = \tan x = \text{RHS}$.