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Prove That: Sin 5 X = 5 Sin X − 20 Sin 3 X + 16 Sin 5 X

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Question

Prove that:  \[\sin 5x = 5 \sin x - 20 \sin^3 x + 16 \sin^5 x\]

 
Numerical
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Solution

\[LHS = \sin5x\]
\[ = \sin\left( 3x + 2x \right)\]
\[ = \sin3x \times \cos2x + \cos3x \times \sin2x\]
\[ = \left( 3\text{ sin } x - 4 \sin^3 x \right)\left( 1 - 2 \sin^2 x \right) + \left( 4 \cos^3 x - 3 \text{ cos } x \right) \times 2\text{ sin } x \text{ cos } x\]
\[ = 3\text{ sin } x - 6 \sin^3 x - 4 \sin^3 x + 8 \sin^5 x + \left( 8 \cos^4 x - 6 \cos^2 x \right)\text{ sin } x\]
\[ = 3\text{ sin } x - 10 \sin^3 x + 8 \sin^5 x + \left\{ 8\text{ sin } x \left( 1 - \sin^2 x \right)^2 - 6\text{ sin } x\left( 1 - \sin^2 x \right) \right\}\]
\[ = 3\text{ sin } x - 10 \sin^3 x + 8 \sin^5 x + \left\{ 8\text{ sin } x\left( 1 - 2 \sin^2 x + \sin^4 x \right) - 6\text{ sin } x + 6 \sin^3 x \right\}\]
\[ = 3\text{ sin } x - 10 \sin^3 x + 8 \sin^5 x + 8\text{ sin } x - 16 \sin^3 x + 8 \sin^5 x - 6\text{sin } x + 6 \sin^3 x\]
\[ = 5\text{ sin } x - 20 \sin^3 x + 16 \sin^5 x\]
\[ = RHS\]
\[\text{ Hence proved }  .\]

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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 9.2 [Page 36]

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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 9.2 | Q 1 | Page 36

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