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ITA 2024 — 1ª Fase — Questão 39 — Trigonometria
Determine o valor de
\[
\cos\!\Big(2\,\operatorname{arctg}\!\big(\tfrac{4}{3}\big)\Big)
\;+\;
sen\!\Big(2\,\operatorname{arctg}\!\big(\tfrac{4}{3}\big)\Big).
\]
a) \(\dfrac{17}{25}\) b) \(\dfrac{4}{5}\) c) \(\dfrac{24}{25}\) d) \(\dfrac{28}{25}\) e) \(\dfrac{31}{25}\)
a) \(\dfrac{17}{25}\) b) \(\dfrac{4}{5}\) c) \(\dfrac{24}{25}\) d) \(\dfrac{28}{25}\) e) \(\dfrac{31}{25}\)
👀 Solução Passo a Passo
Seja \(\alpha=\arctan\!\big(\tfrac{4}{3}\big)\). Então \(\tan\alpha=\tfrac{4}{3}\).
Num triângulo retângulo com catetos \(4\) e \(3\), temos a hipotenusa \(5\). Logo,
\[
sen\alpha=\frac{4}{5},\qquad \cos\alpha=\frac{3}{5}.
\]
Usando as fórmulas do ângulo duplo:
\[
sen(2\alpha)=2sen\alpha\cos\alpha
=2\cdot\frac{4}{5}\cdot\frac{3}{5}
=\frac{24}{25},
\]
\[
\cos(2\alpha)=\cos^2\alpha-sen^2\alpha
=\left(\frac{3}{5}\right)^2-\left(\frac{4}{5}\right)^2
=\frac{9}{25}-\frac{16}{25}
=-\frac{7}{25}.
\]
Portanto,
\[
\cos(2\alpha)+sen(2\alpha)
=-\frac{7}{25}+\frac{24}{25}
=\boxed{\frac{17}{25}}.
\]
Resposta: a) \(\dfrac{17}{25}\)
