સાબિત કરો કે: $\sin 3x + \sin 2x - \sin x = 4 \sin x \cos \frac{x}{2} \cos \frac{3x}{2}$

$L.H.S. = \sin 3x + \sin 2x - \sin x$
$= \sin 3x + (\sin 2x - \sin x)$
$= \sin 3x + 2 \cos \left( \frac{2x + x}{2} \right) \sin \left( \frac{2x - x}{2} \right) \quad [\text{સૂત્ર } \sin A - \sin B = 2 \cos \frac{A+B}{2} \sin \frac{A-B}{2} \text{ નો ઉપયોગ કરતા}]$
$= \sin 3x + 2 \cos \frac{3x}{2} \sin \frac{x}{2}$
$= 2 \sin \frac{3x}{2} \cos \frac{3x}{2} + 2 \cos \frac{3x}{2} \sin \frac{x}{2} \quad [\text{સૂત્ર } \sin 2A = 2 \sin A \cos A \text{ નો ઉપયોગ કરતા}]$
$= 2 \cos \frac{3x}{2} \left( \sin \frac{3x}{2} + \sin \frac{x}{2} \right)$
$= 2 \cos \frac{3x}{2} \left[ 2 \sin \left( \frac{\frac{3x}{2} + \frac{x}{2}}{2} \right) \cos \left( \frac{\frac{3x}{2} - \frac{x}{2}}{2} \right) \right] \quad [\text{સૂત્ર } \sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2} \text{ નો ઉપયોગ કરતા}]$
$= 2 \cos \frac{3x}{2} \cdot 2 \sin x \cos \frac{x}{2}$
$= 4 \sin x \cos \frac{x}{2} \cos \frac{3x}{2} = R.H.S.$