Prove that: sin 3x + sin 2x - sin x = 4 sin x | vedclass

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$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}

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$L.H.S. = \sin 3x + \sin 2x - \sin x$$= \sin 3x + (\sin 2x - \sin x)$$= \sin 3x + 2 \cos \left( \frac{2x + x}{2} ight) \sin \left( \frac{2x - x}{2} ight) \quad [	ext{Using } \sin A - \sin B = 2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}]$$= \sin 3x + 2 \cos \frac{3x}{2} \sin \frac{x}{2}$$= 2 \sin \frac{3x}{2} \cos \frac{3x}{2} + 2 \cos \frac{3x}{2} \sin \frac{x}{2} \quad [	ext{Using } \sin 2A = 2 \sin A \cos A]$$= 2 \cos \frac{3x}{2} \left( \sin \frac{3x}{2} + \sin \frac{x}{2} ight)$$= 2 \cos \frac{3x}{2} \left[ 2 \sin \left( \frac{\frac{3x}{2} + \frac{x}{2}}{2} ight) \cos \left( \frac{\frac{3x}{2} - \frac{x}{2}}{2} ight) ight] \quad [	ext{Using } \sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}]$$= 2 \cos \frac{3x}{2} \cdot 2 \sin x \cos \frac{x}{2}$$= 4 \sin x \cos \frac{x}{2} \cos \frac{3x}{2} = R.H.S.$

Prove that: $\sin 3x + \sin 2x - \sin x = 4 \sin x \cos \frac{x}{2} \cos \frac{3x}{2}$

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