Pré-Vestibular ⇒ (UFPB - 1978) Trigonometria: Arco Duplo

[ALDRINMOD]

Da fórmula [tex3]\cos 2a=1-2\text{sen}^2a ,[/tex3] concluímos que [tex3]\text{sen}\frac{\pi}{12}[/tex3] vale:

a) [tex3]\sqrt{2-{\sqrt3}}.[/tex3]
b) [tex3]\frac{1}{2}\sqrt{2+{\sqrt3}}.[/tex3]
c) [tex3]\sqrt{2+{\sqrt3}}.[/tex3]
d) [tex3]\frac{1}{4}\sqrt{2-{\sqrt3}}.[/tex3]
e) [tex3]\frac{1}{2}\sqrt{2-{\sqrt3}}.[/tex3]

[Thadeu]

• [tex3]\cos\,2a=\cos\,2\cdot \left(\frac{\pi}{12}\right)=\cos\,\frac{\pi}{6}=\frac{\sqrt{3}}{2}[/tex3]

Substituindo:

• [tex3]\frac{\sqrt{3}}{2}=1-2\text{sen}\left(\frac{\pi}{12}\right)^2\,\Rightarrow\,2\text{sen}\left(\frac{\pi}{12}\right)^2=1-\frac{\sqrt{3}}{2}\,\Rightarrow\,2\text{sen}\left(\frac{\pi}{12}\right)^2=\frac{2-\sqrt{3}}{2}\\
  \Rightarrow\,\text{sen}\left(\frac{\pi}{12}\right)^2=\frac{2-\sqrt{3}}{4}\,\Rightarrow\,\text{sen}\left(\frac{\pi}{12}\right)=\sqrt{\frac{2-\sqrt{3}}{4}}\,\Rightarrow\,\text{sen}\left(\frac{\pi}{12}\right)=\frac{1}{2}\sqrt{2-\sqrt{3}}[/tex3]

Resposta: (e).