Demonstração - Soma de arcos
Enviado: Ter 05 Fev, 2019 15:23
Seja um complexo [tex3]\rho_1 \cdot cis\theta=e^{i \theta}[/tex3]
e [tex3]\rho_2 \cdot cis\alpha=e^{i \alpha}[/tex3]
[tex3]e^{i \theta}\cdot e^{i\alpha}=e^{i(\theta+\alpha)}=\cis(\theta +\alpha)[/tex3]
[tex3]\cis(\theta+\alpha)=\cis \theta\cdot\cis \alpha[/tex3]
[tex3](\cos\theta+ i\sen \theta)(\cos\alpha+i\sen\alpha)=cis(\theta+\alpha)[/tex3]
[tex3]cos\theta \cos \alpha+isen\alpha \cos \theta+i\sen\theta \cos \alpha-\sen \theta\sen \alpha[/tex3]
[tex3](\cos \theta\cos\alpha-\sen\theta\sen \alpha)+i(\sen \alpha \cos \theta + \sen \theta\cos \alpha)=cis(\theta +\alpha)[/tex3]
Logo,
[tex3]\cos(\theta+\alpha)=\cos \theta\cos\alpha-\sen\theta\sen \alpha[/tex3]
[tex3]\sen(\theta+\alpha)=\sen \alpha \cos \theta + \sen \theta\cos \alpha[/tex3]
[tex3]e^{i \theta}\cdot e^{i\alpha}=e^{i(\theta+\alpha)}=\cis(\theta +\alpha)[/tex3]
[tex3]\cis(\theta+\alpha)=\cis \theta\cdot\cis \alpha[/tex3]
[tex3](\cos\theta+ i\sen \theta)(\cos\alpha+i\sen\alpha)=cis(\theta+\alpha)[/tex3]
[tex3]cos\theta \cos \alpha+isen\alpha \cos \theta+i\sen\theta \cos \alpha-\sen \theta\sen \alpha[/tex3]
[tex3](\cos \theta\cos\alpha-\sen\theta\sen \alpha)+i(\sen \alpha \cos \theta + \sen \theta\cos \alpha)=cis(\theta +\alpha)[/tex3]
Logo,
[tex3]\cos(\theta+\alpha)=\cos \theta\cos\alpha-\sen\theta\sen \alpha[/tex3]
[tex3]\sen(\theta+\alpha)=\sen \alpha \cos \theta + \sen \theta\cos \alpha[/tex3]