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प्रश्न
Prove that \[\sin 3x + \sin 2x - \sin x = 4 \sin x \cos\frac{x}{2} \cos\frac{3x}{2}\]
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उत्तर
\[LHS = \sin3x + \sin2x - \text{ sin } x\]
\[ = \sin3x + 2\sin\left( \frac{2x - x}{2} \right)\cos\left( \frac{2x + x}{2} \right) \left[ \because \text{ sin } A - \text{ sin } B = 2\sin\left( \frac{A - B}{2} \right)\cos\left( \frac{A + B}{2} \right) \right]\]
\[ = \sin3x + 2\sin\left( \frac{x}{2} \right)\cos\left( \frac{3x}{2} \right)\]
\[ = 2\sin\left( \frac{3x}{2} \right)\cos\left( \frac{3x}{2} \right) + 2\sin\left( \frac{3x}{2} \right)\cos\frac{x}{2} \left[ \because \sin2A = 2\text{ sin } A\text{ cos } A \right]\]
\[ = 2\cos\left( \frac{3x}{2} \right)\left[ \sin\left( \frac{3x}{2} \right)\cos\left( \frac{x}{2} \right) \right]\]
\[ = 2\cos\left( \frac{3x}{2} \right)\left[ 2\sin\left( \frac{\frac{3x}{2} + \frac{x}{2}}{2} \right)\cos\left( \frac{\frac{3x}{2} - \frac{x}{2}}{2} \right) \right] \left[ \because \text{ sin } A + \text{ sin } B = 2\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right) \right]\]
\[ = 2\cos\left( \frac{3x}{2} \right)\left[ 2\sin\left( x \right)\cos\left( \frac{x}{2} \right) \right]\]
\[ = 4\sin\left( x \right)\cos\left( \frac{x}{2} \right)\cos\left( \frac{3x}{2} \right) = RHS\]
\[\text{ Hence proved } . \]
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