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Prove that: ( sin 3 x + sin x ) sin x + ( cos 3 x − cos x ) cos x = 0

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Question

Prove that: \[\left( \sin 3x + \sin x \right) \sin x + \left( \cos 3x - \cos x \right) \cos x = 0\]

Numerical
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Solution

\[LHS = \left( \sin3x + \text{ sin } x \right) \text{ sin } x + \left( \cos3x - \text{ cos } x \right)\text{ cos } x\]

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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Values of Trigonometric Functions at Multiples and Submultiples of an Angle
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Chapter 9: Values of Trigonometric function at multiples and submultiples of an angle - Exercise 9.1 [Page 28]

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R.D. Sharma Mathematics [English] Class 11
Chapter 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.1 | Q 15 | Page 28

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