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Algumas identidades trigonométricas

Dados [tex]\alpha,\,\beta\in\mathbb{R},[/tex] valem:

• Seno da soma:

     [tex]\mathrm{sen}(\alpha)\cos(\beta)+\cos(\alpha)\,\mathrm{sen}(\beta)=\mathrm{sen}(\alpha+\beta)[/tex]

• Cosseno da soma:

     [tex]\cos(\alpha)\cos(\beta)-\mathrm{sen}(\alpha)\,\mathrm{sen}(\beta)=\cos(\alpha+\beta)[/tex]

Um lema prévio

[Lema] Sejam [tex]k\in\mathbb{N}[/tex] e [tex]x\in\mathbb{R}.[/tex] Então,

     [tex]\mathrm{sen}(kx)\cos((k+1)x)+\mathrm{sen}(x)\cos(2(k+1)x)=\mathrm{sen}((k+1)x)\cos((k+2)x)[/tex]

     [Demonstração]

Mostremos que o lado direito é igual ao lado esquerdo:

     [tex]\mathrm{sen}((k+1)x)\cos((k+2)x)\\\\ =\mathrm{sen}(kx+x)\cos((k+1)x+x)[/tex]

     [tex]=[\mathrm{sen}(kx)\cos(x)+\cos(kx)\,\mathrm{sen}(x)]\cdot [\cos((k+1)x)\cos(x)-\mathrm{sen}((k+1)x)\,\mathrm{sen}(x)][/tex]

Desenvolvendo o produto:

     [tex]\begin{array}{rl} =&\mathrm{sen}(kx)\cos((k+1)x)\cos^2(x)-\mathrm{sen}(x)\cos(x)\,\mathrm{sen}(kx)\,\mathrm{sen}((k+1)x)\\\\ &+\,\mathrm{sen}(x)\cos(x)\cos(kx)\cos((k+1)x)-\cos(kx)\,\mathrm{sen}((k+1)x)\,\mathrm{sen}^2(x)\end{array}[/tex]

Substitua [tex]\cos^2(x)=1-\mathrm{sen}^2(x):[/tex]

     [tex]\begin{array}{rl} =&\mathrm{sen}(kx)\cos((k+1)x)\cdot [1-\mathrm{sen}^2(x)]-\mathrm{sen}(x)\cos(x)\,\mathrm{sen}(kx)\,\mathrm{sen}((k+1)x)\\\\ &+\,\mathrm{sen}(x)\cos(x)\cos(kx)\cos((k+1)x)-\cos(kx)\,\mathrm{sen}((k+1)x)\,\mathrm{sen}^2(x)\\\\\\ =&\mathrm{sen}(kx)\cos((k+1)x)-\mathrm{sen}(kx)\cos((k+1)x)\,\mathrm{sen}^2(x)-\mathrm{sen}(x)\cos(x)\,\mathrm{sen}(kx)\,\mathrm{sen}((k+1)x)\\\\ &+\,\mathrm{sen}(x)\cos(x)\cos(kx)\cos((k+1)x)-\cos(kx)\,\mathrm{sen}((k+1)x)\,\mathrm{sen}^2(x) \end{array}[/tex]

Reagrupando as parcelas:

     [tex]\begin{array}{rl} =&\mathrm{sen}(kx)\cos((k+1)x)-\mathrm{sen}(kx)\cos((k+1)x)\,\mathrm{sen}^2(x)-\cos(kx)\,\mathrm{sen}((k+1)x)\,\mathrm{sen}^2(x)\\\\ &-\,\mathrm{sen}(x)\cos(x)\,\mathrm{sen}(kx)\,\mathrm{sen}((k+1)x)+\mathrm{sen}(x)\cos(x)\,\cos(kx)\cos((k+1)x)\end{array}[/tex]

Coloque os fatores comuns em evidência:

     [tex]\begin{array}{rl} =&\mathrm{sen}(kx)\cos((k+1)x)-\mathrm{sen}^2(x)\cdot [\mathrm{sen}(kx)\cos((k+1)x)+\cos(kx)\,\mathrm{sen}((k+1)x)]\\\\ &+\,\mathrm{sen}(x)\cos(x)\cdot [-\,\mathrm{sen}(kx)\,\mathrm{sen}((k+1)x)+\cos(kx)\cos((k+1)x)]\end{array}[/tex]

     [tex]=\mathrm{sen}(kx)\cos((k+1)x)-\mathrm{sen}^2(x)\,\mathrm{sen}(kx+(k+1)x)+\,\mathrm{sen}(x)\cos(x)\cos(kx+(k+1)x)\\\\ =\mathrm{sen}(kx)\cos((k+1)x)-\mathrm{sen}^2(x)\,\mathrm{sen}((2k+1)x)+\,\mathrm{sen}(x)\cos(x)\cos((2k+1)x)[/tex]

Coloque [tex]\mathrm{sen}(x)[/tex] em evidência:

     [tex]=\mathrm{sen}(kx)\cos((k+1)x)+\mathrm{sen}(x)\cdot [-\,\mathrm{sen}(x)\,\mathrm{sen}((2k+1)x)+\cos(x)\cos((2k+1)x)]\\\\ =\mathrm{sen}(kx)\cos((k+1)x)+\mathrm{sen}(x)\cos(x+(2k+1)x)\\\\ =\mathrm{sen}(kx)\cos((k+1)x)+\mathrm{sen}(x)\cos((2k+2)x)\\\\ =\mathrm{sen}(kx)\cos((k+1)x)+\mathrm{sen}(x)\cos(2(k+1)x)\qquad\qquad\square[/tex]

A soma de cossenos de arcos em P.A.

[Proposição] Seja [tex]\theta\in\mathbb{R}\setminus\{2k\pi:~k\in\mathbb{Z}\}.[/tex] Então,

     [tex]\cos(\theta)+\cos(2\theta)+\cos(3\theta)+\cdots+\cos(n\theta)=\dfrac{\mathrm{sen}(\frac{n\theta}{2})\cos(\frac{(n+1)\theta}{2})}{\mathrm{sen}(\frac{\theta}{2})}[/tex]

para todo [tex]n\in\mathbb{N},[/tex] [tex]n\ge 1.[/tex]

     [Demonstração]

Será feita por indução sobre [tex]n.[/tex]

• O caso base é verificado para [tex]n=1,[/tex] pois

     [tex]\cos(1\theta)=\dfrac{\mathrm{sen}(\frac{1\theta}{2})\cos(\frac{(1+1)\theta}{2})}{\mathrm{sen}(\frac{\theta}{2})}.[/tex]

• Suponha por hipótese de indução que a fórmula para o somatório é valida para algum [tex]k\in\mathbb{N},[/tex] [tex]k\ge 1,[/tex] ou seja,

     [tex]\displaystyle\sum_{j=1}^k \cos(j\theta)=\frac{\mathrm{sen}(\frac{k\theta}{2})\cos(\frac{(k+1)\theta}{2})}{\mathrm{sen}(\frac{\theta}{2})}\qquad\mathsf{(H.I.)}[/tex]

• Passo indutivo:

Segue que

     [tex]\displaystyle\sum_{j=1}^{k+1} \cos(j\theta)\\\\ =\sum_{j=1}^k\cos(j\theta)+\cos((k+1)\theta)\\\\ =\frac{\mathrm{sen}(\frac{k\theta}{2})\cos(\frac{(k+1)\theta}{2})}{\mathrm{sen}(\frac{\theta}{2})}+\cos((k+1)\theta)\qquad(\mathsf{pela~H.I.})[/tex]

Reduza as parcelas ao mesmo denominador comum:

     [tex]=\dfrac{\mathrm{sen}(\frac{k\theta}{2})\cos(\frac{(k+1)\theta}{2})+\mathrm{sen}(\frac{\theta}{2})\cos((k+1)\theta)}{\mathrm{sen}(\frac{\theta}{2})}\\\\ =\dfrac{\mathrm{sen}(\frac{k\theta}{2})\cos(\frac{(k+1)\theta}{2})+\mathrm{sen}(\frac{\theta}{2})\cos(\frac{2(k+1)\theta}{2})}{\mathrm{sen}(\frac{\theta}{2})}[/tex]

Pelo lema anterior, para [tex]x=\dfrac{\theta}{2},[/tex] a expressão acima fica

     [tex]=\dfrac{\mathrm{sen}(\frac{(k+1)\theta}{2})\cos(\frac{(k+2)\theta}{2})}{\mathrm{sen}(\frac{\theta}{2})}[/tex]

Portanto, pelo Princípio da Indução Finita, concluímos que

     [tex]\displaystyle\sum_{j=1}^n \cos(j\theta)=\frac{\mathrm{sen}(\frac{n\theta}{2})\cos(\frac{(n+1)\theta}{2})}{\mathrm{sen}(\frac{\theta}{2})}[/tex]

para todo [tex]n\in\mathbb{N},[/tex] [tex]n\ge 1.[/tex]          ■

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