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प्रश्न
Prove that: \[\left( \sin 3x + \sin x \right) \sin x + \left( \cos 3x - \cos x \right) \cos x = 0\]
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उत्तर
\[LHS = \left( \sin3x + \text{ sin } x \right) \text{ sin } x + \left( \cos3x - \text{ cos } x \right)\text{ cos } x\]
Using the identities
\[\text{ sin } C + \text{ sin } D = 2\sin\frac{C + D}{2}\cos\frac{C - D}{2} \text{ and } \text{ cos } C - \text{ cos } D = - 2\sin\frac{C + D}{2}\sin\frac{C - D}{2}\] , we get
\[LHS = \left( 2\sin\frac{3x + x}{2} \times \cos\frac{3x - x}{2} \times \text{ sin } x \right) + \left( - 2\sin\frac{3x + x}{2} \times \sin\frac{3x - x}{2} \right)\text{ cos } x\]
\[ = \left( 2\sin2x \times \text{ cos } x \times \text{ sin } x \right) - \left( 2\sin2x \times \text{ sin } x \text{ cos } x \right)\]
\[ = 0 = RHS\]
\[\text{ Hence proved } .\]
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