Sabe-se que x+y=\frac{\pi}{4} e \operatorname{sen}(x)=\frac{5}{13}, com 0 \leq x<\leq \frac{\pi}{2}.
Encontre \cos (x), \operatorname{tg}(x), \operatorname{sec} (x), \operatorname{cossec}(x) e \cot (x), \operatorname{sen}(2 x), \cos (2 x), \operatorname{tg}(2 x), \operatorname{sen}\left(\frac{x}{2}\right).

Passo 1

O enunciado diz que \sin(x)=\tfrac{5}{13} e que 0\le x\le \tfrac{\pi}{2}. Estamos em quadrante I, portanto todos os valores trigonométricos serão positivos. Vamos usar o teorema de Pitágoras e as fórmulas de ângulo duplo e meio ângulo para encontrar: \cos(x), \tan(x), \operatorname{sec}(x), \csc(x), \cot(x), \sin(2x), \cos(2x), \tan(2x) e \sin\bigl(\tfrac{x}{2}\bigr).

Passo 2

Usamos o teorema de Pitágoras no triângulo retângulo associado:
\cos(x) = \sqrt{1-\sin^2(x)} = \sqrt{1-\bigl(\tfrac{5}{13}\bigr)^2} = \sqrt{1-\tfrac{25}{169}} = \sqrt{\tfrac{144}{169}} = \tfrac{12}{13}.

Passo 3

Agora calculamos as outras razões do mesmo ângulo:
\tan(x) = \frac{\sin(x)}{\cos(x)} = \frac{\tfrac{5}{13}}{\tfrac{12}{13}} = \tfrac{5}{12},
\operatorname{sec}(x) = \frac{1}{\cos(x)} = \frac{1}{\tfrac{12}{13}} = \tfrac{13}{12},
\csc(x) = \frac{1}{\sin(x)} = \frac{1}{\tfrac{5}{13}} = \tfrac{13}{5},
\cot(x) = \frac{\cos(x)}{\sin(x)} = \frac{\tfrac{12}{13}}{\tfrac{5}{13}} = \tfrac{12}{5}.

Passo 4

Usamos as fórmulas de ângulo duplo:
\sin(2x) = 2\sin(x)\cos(x) = 2\cdot\tfrac{5}{13}\cdot\tfrac{12}{13} = \tfrac{120}{169},
\cos(2x) = \cos^2(x)-\sin^2(x) = \bigl(\tfrac{12}{13}\bigr)^2-\bigl(\tfrac{5}{13}\bigr)^2 = \tfrac{144-25}{169} = \tfrac{119}{169},
\tan(2x) = \frac{\sin(2x)}{\cos(2x)} = \frac{\tfrac{120}{169}}{\tfrac{119}{169}} = \tfrac{120}{119}.

Passo 5

Para a meia-ângulo usamos
\sin\bigl(\tfrac{x}{2}\bigr) = \sqrt{\frac{1-\cos(x)}{2}} = \sqrt{\frac{1-\tfrac{12}{13}}{2}} = \sqrt{\frac{\tfrac{1}{13}}{2}} = \sqrt{\tfrac{1}{26}} = \frac{\sqrt{26}}{26}.

Resposta

\cos(x)=\tfrac{12}{13} \tan(x)=\tfrac{5}{12} \operatorname{sec}(x)=\tfrac{13}{12} \csc(x)=\tfrac{13}{5} \cot(x)=\tfrac{12}{5} \sin(2x)=\tfrac{120}{169} \cos(2x)=\tfrac{119}{169} \tan(2x)=\tfrac{120}{119} \sin\bigl(\tfrac{x}{2}\bigr)=\tfrac{\sqrt{26}}{26}

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