Question

Determine os valores das demais funções trigonométricas de um arco X quando:
a) sen x = –1/2 e π < x < 3π/2
b) cos x = 1/3 e 0 < x < π/2
c) cos sec x = –√2 e π < x < 3π/2
d) tg x = √3 e 0 < x < π/2

Conteúdo: relação fundamental da trigonometria
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Answer 1

$\large\boxed{\begin{array}{l}\tt a)~\sf sen(x)=-\dfrac{1}{2},x\in~\pi<x<\dfrac{3\pi}{2}\\\\\sf sen(x)=-\dfrac{1}{2}\implies sen^2(x)=\dfrac{1}{4}\\\\\sf cos^2(x)=\dfrac{4}{4}-\dfrac{1}{4}=\dfrac{3}{4}\\\\\sf cos(x)=-\sqrt{\dfrac{3}{4}}=-\dfrac{\sqrt{3}}{2}\\\\\sf tg(x)=\dfrac{-\frac{1}{\backslash\!\!\!2}}{-\frac{\sqrt{3}}{\backslash\!\!\!2}}=\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}\\\\\sf sec(x)=-\dfrac{2}{\sqrt{3}}=-\dfrac{2\sqrt{3}}{3}\\\\\sf cossec(x)=-2\\\\\sf cotg(x)=\sqrt{3}\end{array}}$

$\large\boxed{\begin{array}{l}\tt b)~\sf cos(x)=\dfrac{1}{3}~x\in~0<x<\dfrac{\pi}{2}\\\\\sf cos(x)=\dfrac{1}{3}\implies cos^2(x)=\dfrac{1}{9}\\\\\sf sen^2(x)=\dfrac{9}{9}-\dfrac{1}{9}=\dfrac{8}{9}\\\\\sf sen(x)=\sqrt{\dfrac{8}{9}}=\dfrac{2\sqrt{2}}{3}\\\\\sf tg(x)=\dfrac{\frac{2\sqrt{2}}{\backslash\!\!\!3}}{\frac{1}{\backslash\!\!\!3}}=2\sqrt{2}\\\\\sf sec(x)=3\\\\\sf cossec(x)=\dfrac{3}{2\sqrt{2}}=\dfrac{3\sqrt{2}}{4}\\\\\sf cotg(x)=\dfrac{1}{2\sqrt{2}}=\dfrac{\sqrt{2}}{4}\end{array}}$

$\large\boxed{\begin{array}{l}\tt c)~\sf cossec(x)=-\sqrt{2}~x\in~\pi<x<\dfrac{3\pi}{2}\\\\\sf cossec(x)=-\sqrt{2}\implies cossec^2(x)=2\\\sf cotg^2(x)=2-1=1\\\\\sf cotg(x)=\sqrt{1}=1\\\\\sf tg(x)=1\\\sf sen(x)=-\dfrac{1}{\sqrt{2}}=-\dfrac{\sqrt{2}}{2}\\\\\sf cos(x)=\dfrac{-\frac{\sqrt{2}}{2}}{1}=-\dfrac{\sqrt{2}}{2}\\\\\sf sec(x)=-\dfrac{2}{\sqrt{2}}=-\dfrac{\backslash\!\!\!2\sqrt{2}}{\backslash\!\!\!2}=-\sqrt{2}\end{array}}$

$\large\boxed{\begin{array}{l}\tt d)~\sf tg(x)=\sqrt{3}~x\in~0<x<\dfrac{\pi}{2}\\\sf tg(x)=\sqrt{3}\implies tg^2(x)=3\\\sf sec^2(x)=1+3=4\\\sf sec(x)=\sqrt{4}=2\\\sf cos(x)=\dfrac{1}{2}\\\\\sf sen(x)=cos(x)\cdot tg(x)\\\sf sen(x)=\dfrac{1}{2}\cdot\sqrt{3}=\dfrac{\sqrt{3}}{2}\\\\\sf cossec(x)=\dfrac{2}{\sqrt{3}}=\dfrac{2\sqrt{3}}{3}\\\\\sf cotg(x)=\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}\end{array}}$