ITA 2023 — 2ª Fase — Questão 05 — Trigonometria
Sejam:
\[
A = \cos(\alpha) + \cos(\beta), \quad
B = sen(\alpha) – sen(\beta),
\]
com \(\alpha, \beta \in \mathbb{R}\).
Calcule \(sen(\alpha – \beta)\) em função de \(A\) e \(B\), sabendo que \(A\) e \(B\) não são ambos nulos.
👀 Solução passo a passo
1) Aplicando identidades de soma para produto:
\[ A = \cos\alpha + \cos\beta = 2\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right), \] \[ B = sen\alpha – sen\beta = 2\cos\left(\frac{\alpha+\beta}{2}\right)sen\left(\frac{\alpha-\beta}{2}\right). \]2) Razão \( \frac{B}{A} \):
\[ \frac{B}{A} = \frac{2\cos\left(\frac{\alpha+\beta}{2}\right)sen\left(\frac{\alpha-\beta}{2}\right)} {2\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)} = tg\left(\frac{\alpha-\beta}{2}\right). \]3) Expressando \(sen(\alpha – \beta)\):
Pela fórmula do seno do ângulo duplo: \[ sen(\alpha – \beta) = 2sen\left(\frac{\alpha-\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right). \] Substituindo: \[ sen(\alpha – \beta) = 2tg\left(\frac{\alpha-\beta}{2}\right)\cos^2\left(\frac{\alpha-\beta}{2}\right). \] Como: \[ \cos^2\theta = \frac{1}{1+tg^2\theta}, \] temos: \[ sen(\alpha – \beta) = \frac{2tg\left(\frac{\alpha-\beta}{2}\right)} {1+tg^2\left(\frac{\alpha-\beta}{2}\right)}. \]4) Substituindo \(tg\left(\frac{\alpha-\beta}{2}\right) = \frac{B}{A}\):
\[ sen(\alpha – \beta) = \frac{ \frac{2B}{A} }{ 1 + \frac{B^2}{A^2} } = \frac{ \frac{2B}{A} }{ \frac{A^2 + B^2}{A^2} } = \frac{ 2AB }{A^2 + B^2}. \]
\[ A = \cos\alpha + \cos\beta = 2\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right), \] \[ B = sen\alpha – sen\beta = 2\cos\left(\frac{\alpha+\beta}{2}\right)sen\left(\frac{\alpha-\beta}{2}\right). \]2) Razão \( \frac{B}{A} \):
\[ \frac{B}{A} = \frac{2\cos\left(\frac{\alpha+\beta}{2}\right)sen\left(\frac{\alpha-\beta}{2}\right)} {2\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)} = tg\left(\frac{\alpha-\beta}{2}\right). \]3) Expressando \(sen(\alpha – \beta)\):
Pela fórmula do seno do ângulo duplo: \[ sen(\alpha – \beta) = 2sen\left(\frac{\alpha-\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right). \] Substituindo: \[ sen(\alpha – \beta) = 2tg\left(\frac{\alpha-\beta}{2}\right)\cos^2\left(\frac{\alpha-\beta}{2}\right). \] Como: \[ \cos^2\theta = \frac{1}{1+tg^2\theta}, \] temos: \[ sen(\alpha – \beta) = \frac{2tg\left(\frac{\alpha-\beta}{2}\right)} {1+tg^2\left(\frac{\alpha-\beta}{2}\right)}. \]4) Substituindo \(tg\left(\frac{\alpha-\beta}{2}\right) = \frac{B}{A}\):
\[ sen(\alpha – \beta) = \frac{ \frac{2B}{A} }{ 1 + \frac{B^2}{A^2} } = \frac{ \frac{2B}{A} }{ \frac{A^2 + B^2}{A^2} } = \frac{ 2AB }{A^2 + B^2}. \]
\(\displaystyle sen(\alpha – \beta) = \frac{2AB}{A^2 + B^2}\)
