1. Se \operatorname{sen}(x)=\frac{1}{3} e x \in 2^{\circ} quadrante, encontre o valor de \operatorname{sen}\left(\frac{x}{2}\right), \cos \left(\frac{x}{2}\right) e \operatorname{tg}\left(\frac{x}{2}\right)

Passo 1

O enunciado dá \operatorname{sen}(x)=\tfrac{1}{3} e diz que x está no 2 ^\circ quadrante (ou seja, 90^\circ<x<180^\circ). Queremos calcular \operatorname{sen}\bigl(\tfrac{x}{2}\bigr), \cos\bigl(\tfrac{x}{2}\bigr) e \operatorname{tg}\bigl(\tfrac{x}{2}\bigr). Para isso, vamos usar as fórmulas de ângulo metade:
\operatorname{sen}\Bigl(\frac{x}{2}\Bigr) =\pm\sqrt{\frac{1-\cos x}{2}}, \quad \cos\Bigl(\frac{x}{2}\Bigr) =\pm\sqrt{\frac{1+\cos x}{2}}, \quad \operatorname{tg}\Bigl(\frac{x}{2}\Bigr) =\frac{1-\cos x}{\operatorname{sen} x}.

Passo 2

Primeiro, encontramos \cos x. Como \operatorname{sen}^2x+\cos^2x=1 e \operatorname{sen} x=\tfrac{1}{3}, temos
\cos^2x=1-\Bigl(\tfrac{1}{3}\Bigr)^2 =1-\tfrac{1}{9} =\tfrac{8}{9}.
No 2 ^\circ quadrante, o cosseno é negativo, logo
\cos x=-\sqrt{\tfrac{8}{9}} =-\tfrac{2\sqrt{2}}{3}.

Passo 3

Como 90^\circ<x<180^\circ, ao dividirmos por 2 obtemos
45^\circ<\frac{x}{2}<90^\circ,
ou seja, \frac{x}{2} está no 1 ^\circ quadrante. Então \operatorname{sen}(\tfrac{x}{2})>0 e \cos(\tfrac{x}{2})>0, e \operatorname{tg}(\tfrac{x}{2})>0.

Passo 4

Usando as fórmulas de ângulo metade e escolhendo o sinal positivo:
\operatorname{sen}\Bigl(\frac{x}{2}\Bigr) =\sqrt{\frac{1-\cos x}{2}} =\sqrt{\frac{1-(-\tfrac{2\sqrt{2}}{3})}{2}} =\sqrt{\frac{1+\tfrac{2\sqrt{2}}{3}}{2}} =\sqrt{\frac{\tfrac{3+2\sqrt{2}}{3}}{2}} =\sqrt{\frac{3+2\sqrt{2}}{6}},
\cos\Bigl(\frac{x}{2}\Bigr) =\sqrt{\frac{1+\cos x}{2}} =\sqrt{\frac{1+(-\tfrac{2\sqrt{2}}{3})}{2}} =\sqrt{\frac{\tfrac{3-2\sqrt{2}}{3}}{2}} =\sqrt{\frac{3-2\sqrt{2}}{6}}.

Passo 5

Para a tangente metade, podemos usar
\operatorname{tg}\Bigl(\frac{x}{2}\Bigr) =\frac{1-\cos x}{\operatorname{sen} x} =\frac{1-(-\tfrac{2\sqrt{2}}{3})}{\tfrac{1}{3}} =\frac{1+\tfrac{2\sqrt{2}}{3}}{\tfrac{1}{3}} =3\bigl(1+\tfrac{2\sqrt{2}}{3}\bigr) =3+2\sqrt{2}.

Resposta

\operatorname{sen}\bigl(\tfrac{x}{2}\bigr)=\sqrt{\tfrac{3+2\sqrt{2}}{6}} \cos\bigl(\tfrac{x}{2}\bigr)=\sqrt{\tfrac{3-2\sqrt{2}}{6}} \operatorname{tg}\bigl(\tfrac{x}{2}\bigr)=3+2\sqrt{2}

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