[ASK] Limit of Trigonometry Function

[Monox D. I-Fly]

[tex]\lim_{x\to0}\frac{sin2x+sin6x+sin10x-sin18x}{3sinx-sin3x=}[/tex]

A. 0

B. 45

C. 54

D. 192

E. 212


Either substituting or using L'Hopital gives [tex]\frac00[/tex]. Is there any way to simplify it and make the result a real number?

[nathi123]

[tex]A=\lim_{x \to 0}\frac{sin2x+sin6x+sin10x-sin18x}{3sin3x-sin3x} ; sin3x=3sinx-4sin^{3}x\Rightarrow 3sin3x-sin3x=4sin^{3} ; sin2x+sin6x=2sin4xcos2x[/tex]
[tex]sin10x - sin18x=(-2)cos14xsin4x\Rightarrow A = \lim_{x \to 0}\frac{2sin4xcos2x-2sin4xcos14x}{4sin^{3} }=\lim_{x \to 0}\frac{2sin4xsin6xsin8x}{4sin^{3} } because[/tex]
[tex]cos2x-cos14x=(-2)sin8xsin(-6x)=2sin8xsin6x \Rightarrow A=\lim_{x \to 0}\frac{sin4x}{sinx}.\lim_{x \to 0}\frac{sin6x}{sinx}.\lim_{x \to 0}\frac{sin8x}{sinx}[/tex]
[tex]\lim_{x \to 0}\frac{sin6x}{sinx}=\lim_{x \to 0}\frac{6x}{sinx}.\lim_{x \to 0}\frac{sin6x}{6x}=1.6.\lim_{x \to 0}\frac{x}{sinx}=6 ; \lim_{x \to 0}\frac{sinx}{x}=\lim_{x \to 0}\frac{x}{sinx}=1 = \lim_{x \to 0}\frac{sin6x}{6x}[/tex].
Similarly [tex]\lim_{x \to 0}\frac{sin8x}{sinx}=8 ; \lim_{x \to 0}\frac{sin4x}{sinx}=4\Rightarrow A=6.4.8=192[/tex].

[Monox D. I-Fly]

How do you get [tex]sin3x=3sinx-4sin^3x[/tex]?

[nathi123]

[tex]sin3x = sin(2x+x)=sin2xcosx+cos2xsinx=2sinxcosx.cosx+(1-sin^{2}x-sin^{2}x)sinx=sinx(2-2sin^{2}x+1-2sin^{2}x)=sinx(3-4sin^{2}x)[/tex]
[tex]\Rightarrow sin3x=3sinx-4sin^{3}x[/tex].

[Monox D. I-Fly]

Ah, I see. Thank you.