Estude! Com Prof. Caju

Sendo [tex3]\text{sen}\alpha = \frac{9}{10},[/tex3] com [tex3]0 < \alpha < \frac{\pi}{2},[/tex3] tem-se

a) [tex3]\text{sen}\alpha < \text{sen}\frac{\pi}{3} < \text{sen}2\alpha[/tex3]
b) [tex3]\text{sen}\frac{\pi}{3} < \text{sen}\alpha< \text{sen}2\alpha[/tex3]
c) [tex3]\text{sen}\alpha<\text{sen}2\alpha< \text{sen}\frac{\pi}{3}[/tex3]
d) [tex3]\text{sen}2\alpha < \text{sen}\frac{\pi}{3} < \text{sen}\alpha[/tex3]
e) [tex3]\text{sen}2\alpha < \text{sen}\alpha < \text{sen}\frac{\pi}{3}[/tex3]

[tex3]\text{sen} \alpha=\frac{9}{10},[/tex3] [tex3]\text{sen}^2 \alpha+\cos^2\alpha=1[/tex3] e [tex3]0 < \alpha < \frac{\pi}{2}[/tex3] implica em [tex3]\cos \alpha =\frac{\sqrt{19}}{10}.[/tex3]

Como [tex3]\text{sen} 2\alpha=2\cdot \text{sen} \alpha\cdot \cos\alpha,[/tex3] temos [tex3]\text{sen} 2\alpha=\frac{18\sqrt{19}}{100}.[/tex3]

Além disso, [tex3]\text{sen}\frac{\pi}{3}=\frac{\sqrt{3}}{2}.[/tex3]

Sabendo que para quaisquer [tex3]a[/tex3] e [tex3]b[/tex3] reais positivos vale a propriedade

• [tex3]a^2<b^2\Longleftrightarrow a<b,[/tex3]

temos:

• [tex3]\text{sen}^2 \alpha=\frac{81}{100}=\frac{8100}{10000}[/tex3]
  
  [tex3]\text{sen}^2 2\alpha=\frac{324\cdot 19}{10000}=\frac{6156}{10000}[/tex3]
  
  [tex3]\text{sen}^2\frac{\pi}{3}=\frac{3}{4}=\frac{7500}{10000}[/tex3]

Portanto,

• [tex3]\text{sen}^2 2\alpha<\text{sen}^2\frac{\pi}{3}<\text{sen}^2 \alpha\Longrightarrow \text{sen} 2\alpha<\text{sen}\frac{\pi}{3}<\text{sen} \alpha[/tex3]