Ensino Médio ⇒ Trigonometria: Funções Trigonométricas Inversas Tópico resolvido

[triplebig]

Calcule:

• [tex3]\text{tg}\,\left( \text{arc sen}\,\left(-\Large\frac{2}{3}\large \right)\,+\,\text{arc sen}\,\Large\frac{1}{4}\large \right)[/tex3]

Resposta:

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[tex3]\Large\frac{\sqrt{5}\,(-6\,+\,\sqrt{3})}{15\,+\,2\sqrt{3}}[/tex3]

[adrianotavares]

Olá, triplebig.

A função [tex3]\text{arcsen}[/tex3] é definida no intervalo [tex3]\left[-\frac{\pi}{2},\frac{\pi}{2}\right].[/tex3]

Façamos

• [tex3]a = \text{arcsen} -\frac{2}{3} \Rightarrow \text{sen}a = -\frac{2}{3}[/tex3] e [tex3]{-}\frac{\pi}{2}<a<\frac{\pi}{2}\Rightarrow {-}\frac{\pi}{2}<a<-\frac{\pi}{4}.[/tex3]

e

• [tex3]b = \text{arcsen} \frac{1}{4} \Rightarrow \text{sen}b = \frac{1}{4}[/tex3] e [tex3]{-}\frac{\pi}{2}<b<\frac{\pi}{2}\Rightarrow 0<b<\frac{\pi}{4}.[/tex3]

• [tex3]\text{tg}\left(\text{arcsen} -\frac{2}{3} + \text{arcsen} \frac{1}{4}\right) = \text{tg}(a+b) = \frac{\text{tg}a + \text{tg}b}{1-\text{tg}a \cdot \text{tg}b}[/tex3]

• [tex3]\cos a = \sqrt{1-\text{sen}^2a} \sqrt{1-\left(-\frac{2}{3}\right)^2}=\frac{\sqrt{5}}{3}[/tex3]
  
  [tex3]\text{tg}a = \frac{\text{sen}a}{\cos a} =- \frac{2\sqrt{5}}{5}[/tex3]
  
  [tex3]\cos b = \sqrt{1-\text{sen}^2b} = \sqrt{1-\left(\frac{1}{4}\right)^2} = \frac{\sqrt{15}}{4}[/tex3]
  
  [tex3]\text{tg}b = \frac{\text{sen}b}{\cos b}= \frac{\sqrt{15}}{15}[/tex3]

• [tex3]\frac{\text{tg}a + \text{tg}b}{1-\text{tg}a \cdot \text{tg}b} = \frac{-\frac{2\sqrt{5}}{5} + \frac{\sqrt{15}}{15}}{1 - \left(-\frac{2 \sqrt{5}}{5}\cdot \frac{\sqrt{15}}{15}\right)}= \frac{\frac{\sqrt{5}(\sqrt{3}-6)}{15}}{1+ \frac{2\sqrt{3}}{15}}=\frac{\sqrt{5}(\sqrt{3}-6)}{15+2\sqrt{3}}.[/tex3]