Ensino Médio ⇒ Trigonometria: Arco Metade Tópico resolvido

[kildo]

Se [tex3]\text{tg} \frac{\pi}{4} = 1,[/tex3] determine o valor de [tex3]\text{tg} \frac{\pi}{8}.[/tex3]

[bigjohn]

• [tex3]\text{tg}\frac{x}{2} =\sqrt{\frac{1-\cos x}{1+\cos x}}[/tex3]
  
  [tex3]\text{tg}\frac{\pi}{8} =\sqrt{\frac{1-\cos \frac{\pi}{4}}{1+\cos \frac{\pi}{4}}}=\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{1+\frac{\sqrt{2}}{2}}}=\sqrt{\frac{2-\sqrt{2}}{2+\sqrt{2}}\cdot\frac{2-\sqrt{2}}{2-\sqrt{2}} }=\sqrt{\frac{(2-\sqrt{2})^2}{2}}=\frac{2-\sqrt{2}}{\sqrt{2}}=\sqrt{2}-1[/tex3]