Question

Prove that : (sin5x + 2sin8x + sin11x)÷(sin8x + 2sin11x + sin14x) = (sin8x÷sin11x)

Answer 1

To Prove:

$\frac{\sin 5x + 2 \sin 8x + \sin 11x}{\sin 8x + 2 \sin 11x + \sin 14x} = \frac{\sin 8x}{\sin 11x}$

We will use the trigonometric identity as below:

$\sin A+ \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$

Consider LHS:

$\frac{\sin 5x + 2 \sin 8x + \sin 11x}{\sin 8x + 2 \sin 11x + \sin 14x}$

= $\frac{2 \sin(\frac{16x}{2}) \cos(\frac{-6x}{2}) + 2 \sin8x}{2 \sin(\frac{22x}{2}) \cos(\frac{-6x}{2}) + 2 \sin11x}$

= $\frac{2 \sin(8x) \cos(-3x) + 2 \sin8x}{2 \sin(11x) \cos(-3x) + 2 \sin11x}$

Using the trigonometric identity $\cos(-x) = \cos x$

= $\frac{2 \sin(8x) \cos(3x) + 2 \sin8x}{2 \sin(11x) \cos(3x) + 2 \sin11x}$

=$\frac{2 \sin(8x)[\cos(3x) + 1]}{2 \sin(11x)[\cos(3x) + 1]}$

= $\frac{\sin(8x)}{\sin(11x)}$

= RHS

Hence, proved.