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

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

$$\:\:{y}\:=\:\mathrm{sin}\:\mathrm{8}{x}\:\mathrm{cos}\:\mathrm{4}{x}\: \\ $$$$\:\:{y}^{\left({n}\right)} \:=? \\ $$

Commented by blackmamba last updated on 22/Nov/21

y= sin 8x cos 4x ⇒y = (1/2)(sin 12x+sin 4x) y ′=(1/2)(12cos 12x+4cos 4x) y′= 6 sin (12x+(π/2))+2sin (4x+(π/2)) y′′= 72 cos (12x+(π/2))+8cos (4x+(π/2)) y′′=72 sin (12x+((2π)/2))+8sin (4x+((2π)/2)) y′′′=(1/2).12^3 sin (12x+((3π)/2))+(1/2).4^3 sin (4x+((3π)/2)) ⋮ y^((n)) =(1/2).12^n sin (12x+((nπ)/2))+(1/2).4^n sin (4x+((nπ)/2))

$$\:\:\:{y}=\:\mathrm{sin}\:\mathrm{8}{x}\:\mathrm{cos}\:\mathrm{4}{x}\:\Rightarrow{y}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{sin}\:\mathrm{12}{x}+\mathrm{sin}\:\mathrm{4}{x}\right) \\ $$$$\:\:\:{y}\:'=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{12cos}\:\mathrm{12}{x}+\mathrm{4cos}\:\mathrm{4}{x}\right) \\ $$$$\:\:{y}'=\:\mathrm{6}\:\mathrm{sin}\:\left(\mathrm{12}{x}+\frac{\pi}{\mathrm{2}}\right)+\mathrm{2sin}\:\left(\mathrm{4}{x}+\frac{\pi}{\mathrm{2}}\right) \\ $$$$\:\:{y}''=\:\mathrm{72}\:\mathrm{cos}\:\left(\mathrm{12}{x}+\frac{\pi}{\mathrm{2}}\right)+\mathrm{8cos}\:\left(\mathrm{4}{x}+\frac{\pi}{\mathrm{2}}\right) \\ $$$$\:\:{y}''=\mathrm{72}\:\mathrm{sin}\:\left(\mathrm{12}{x}+\frac{\mathrm{2}\pi}{\mathrm{2}}\right)+\mathrm{8sin}\:\left(\mathrm{4}{x}+\frac{\mathrm{2}\pi}{\mathrm{2}}\right) \\ $$$$\:\:{y}'''=\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{12}^{\mathrm{3}} \:\mathrm{sin}\:\left(\mathrm{12}{x}+\frac{\mathrm{3}\pi}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{4}^{\mathrm{3}} \:\mathrm{sin}\:\left(\mathrm{4}{x}+\frac{\mathrm{3}\pi}{\mathrm{2}}\right) \\ $$$$\:\:\vdots \\ $$$$\:\:\:{y}^{\left({n}\right)} =\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{12}^{{n}} \:\mathrm{sin}\:\left(\mathrm{12}{x}+\frac{{n}\pi}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{4}^{{n}} \:\mathrm{sin}\:\left(\mathrm{4}{x}+\frac{{n}\pi}{\mathrm{2}}\right)\: \\ $$

Commented by tounghoungko last updated on 22/Nov/21

⇒2sin A cos B=sin (A+B)+sin (A−B) ⇒sin 8x cos 4x = (1/2){sin 12x+sin 4x}

$$\Rightarrow\mathrm{2sin}\:{A}\:\mathrm{cos}\:{B}=\mathrm{sin}\:\left({A}+{B}\right)+\mathrm{sin}\:\left({A}−{B}\right) \\ $$$$\Rightarrow\mathrm{sin}\:\mathrm{8}{x}\:\mathrm{cos}\:\mathrm{4}{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{sin}\:\mathrm{12}{x}+\mathrm{sin}\:\mathrm{4}{x}\right\}\: \\ $$

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