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