Ensino Médio ⇒ Arcos trigonométricos Tópico resolvido

[ElAxo]

Seja um arco do quarto quadrante tal que tg(x)=-5/12.

Qual o valor de sen(2x)?


a.
1

b.
119/169

c.
119/13

d.
-120/169

e.
2/169

Resposta

---

D

[NathanMoreira]

[tex3]\tg x=-\frac{5}{12}[/tex3]
[tex3]\frac{\sen x}{\cos x}=-\frac{5}{12}[/tex3]
[tex3]\sen x=-\frac{5.\cos x}{12}[/tex3]

Relação Fundamental da Trigonometria:
[tex3]\sen^2 x+\cos^2x=1[/tex3]
[tex3]\left(-\frac{5.\cos x}{12}\right)^2+\cos^2x=1[/tex3]
[tex3]\frac{25.\cos^2x}{144}+\cos^2x=1[/tex3]
[tex3]\frac{169.\cos^2x}{144}=1[/tex3]
[tex3]\cos^2x=\frac{144}{169}[/tex3]
[tex3]\cos x=\frac{12}{13}[/tex3]

[tex3]\sen x=-\frac{5.\cos x}{12}[/tex3]
[tex3]\sen x=-\frac{5.\left(\frac{12}{13}\right)}{12}[/tex3]
[tex3]\sen x=-\frac{5}{13}[/tex3]

[tex3]\sen(2x)=2.\sen x.\cos x[/tex3]
[tex3]\sen(2x)=2.\left(-\frac{5}{13}\right).\left(\frac{12}{13}\right)[/tex3]
[tex3]{\color{red}\boxed{\sen(2x)=-\frac{120}{169}}}[/tex3]

[petrasMOD]

ElAxo,

Outra resolução:

[tex3]\mathsf{\frac{senx}{cosx}=-\frac{5}{12}\\
\frac{senx \cdot2cosx}{cosx.2cosx}=-\frac{5}{12}\rightarrow \frac{sen2x}{2cos^2x}=-\frac{5}{12}\\
mas~\boxed{2cos^2x = cos2x + 1}\rightarrow \frac{sen2x}{cos2x+1}=-\frac{5}{12}\\
mas~\boxed{cos2x = \frac{1-tg^2x}{1+tg^2x}}=\frac{1-\frac{25}{144}}{1+\frac{25}{144}}=\frac{119}{169}\\
\therefore\frac{sen2x}{\frac{119}{169}+1}=-\frac{5}{12}\rightarrow \frac{sen2x}{\frac{288}{169}}=-\frac{5}{12}\\
sen2x = -\frac{5}{\cancel{12}}\cdot\frac{\cancel{288}_{24}}{169} \therefore\boxed{\mathsf{\color{red}sen2x = -\frac{120}{169}}}




}[/tex3]