y = sin 8x cos 4x y^((n)) =?


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Question Number 159918 by tounghoungko last updated on 22/Nov/21

$y = \sin 8 x \cos 4 x$ $y^{(n)} = ?$
Commented by blackmamba last updated on 22/Nov/21

$y = \sin 8 x \cos 4 x \Rightarrow y = \frac{1}{2} (\sin 12 x + \sin 4 x)$ $y^{'} = \frac{1}{2} (12 \cos 12 x + 4 \cos 4 x)$ $y^{'} = 6 \sin (12 x + \frac{π}{2}) + 2 \sin (4 x + \frac{π}{2})$ $y^{″} = 72 \cos (12 x + \frac{π}{2}) + 8 \cos (4 x + \frac{π}{2})$ $y^{″} = 72 \sin (12 x + \frac{2 π}{2}) + 8 \sin (4 x + \frac{2 π}{2})$ $y^{‴} = \frac{1}{2} . 12^{3} \sin (12 x + \frac{3 π}{2}) + \frac{1}{2} . 4^{3} \sin (4 x + \frac{3 π}{2})$ $⋮$ $y^{(n)} = \frac{1}{2} . 12^{n} \sin (12 x + \frac{n π}{2}) + \frac{1}{2} . 4^{n} \sin (4 x + \frac{n π}{2})$
Commented by tounghoungko last updated on 22/Nov/21

$\Rightarrow 2 \sin A \cos B = \sin (A + B) + \sin (A - B)$ $\Rightarrow \sin 8 x \cos 4 x = \frac{1}{2} {\sin 12 x + \sin 4 x}$
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