Ensino Médio ⇒ Funções Trigonométricas: Período Tópico resolvido

[triplebig]

Determine o período da função:

• [tex3]\text{sen}^6\,x\,+\,cos^6\,x[/tex3]

Resposta:

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[tex3]\frac{\pi}{2}[/tex3]

[Karl Weierstrass]

• [tex3]\text{sen}^6\,x\,+\,cos^6\,x\,=\,\underbrace{(\text{sen}^2\,x\,+\,\cos^2\,x)}_1\,\cdot\,(\text{sen}^4\,x\,-\,\text{sen}^2\,x\,\cdot\,\cos^2\,x\,+\,\cos^4\,x)[/tex3]

• [tex3]\text{ }=\,(1\,-\,\cos^2\,x)^2\,-\,(1\,-\,\cos^2\,x)\,\cdot\,\cos^2\,x\,+\,\cos^4\,x[/tex3]
  
  [tex3]\text{ }=\,1\,-\,2\,\cdot\,\cos^2\,x\,+\,\cos^4\,x\,-\,\cos^2\,x\,+\,\cos^4\,x\,+\,\cos^4\,x[/tex3]
  
  [tex3]\text{ }=\,1\,-\,3\,\cdot\,\cos^2\,x\,+\,3\,\cdot\,\cos^4\,x\,[/tex3]
  
  [tex3]\text{ }=\,1\,-\,3\,\cdot\,\left(\Large\frac{1\,+\,\cos\,2x}{2}\large\right)\,+\,3\,\cdot\,\left(\Large\frac{1\,+\,\cos\,2x}{2}\large\right)^2[/tex3]
  
  [tex3]\text{ }=\,1\,-\,\Large\frac{3}{2}\large\,-\,\Large\frac{3\,\cdot\,\cos\,2x}{2}\large\,+\,\Large\frac{3}{4}\large\,+\,\Large\frac{3\,\cdot\,\cos\,2x}{2}\large\,+\,\Large\frac{3\,\cdot\,\cos^2\,2x}{4}\large[/tex3]
  
  [tex3]\text{ }=\,\Large\frac{1}{4}\large\,+\,\Large\frac{3}{4}\large\,\cdot\,\Large\frac{1\,+\,\cos\,4x}{2}\large[/tex3]
  
  [tex3]\text{ }=\,\Large\frac{5}{8}\large\,+\,\Large\frac{3}{8}\large\,\cdot\,\cos\,4x[/tex3]

• [tex3]P=\Large\frac{P_0}{|m|}\large \,=\,\Large\frac{2\pi}{4}\large\,=\,\Large\frac{\pi}{2}\large.[/tex3]

[tex3]\text{ }(*)\, \text{sen}^2\,x\,+\,\cos^2\,x\,=\,1[/tex3]

[tex3]\,\,(**)\,\cos\,a\,=\,\sqrt{\Large\frac{1\,+\,\cos\,2a}{2}\large}[/tex3]

[tex3]\,\,(***)\, f(x)\,=\,a\,+\,b\,\cdot\,\cos\,(mx\,\pm\,c)[/tex3]

[tex3]\hspace{70pt}P\,=\,\Large\frac{P_0}{|m|}\large,[/tex3]

onde [tex3]P[/tex3] é o período da função, [tex3]P_0[/tex3] é o período fundamental [tex3](P_0\,=\,2\pi)[/tex3] e [tex3]m[/tex3] é um real não nulo.