Sejam \(\alpha,\beta \in \left[\dfrac{\pi}{2},\,\dfrac{3\pi}{2}\right]\) tais que
\[ \sin\alpha – \sin\beta = \frac14 \qquad\text{e}\qquad \sin\alpha – 2\sin\beta + \cos\beta = \frac34. \]Calcule o valor de \(\sin(\alpha+\beta)\).
👀 Solução passo a passo
Do primeiro enunciado: \(\ \sin\alpha = \sin\beta + \dfrac14.\) Subtraindo a 1ª da 2ª equação:
\[ (\sin\alpha – 2\sin\beta + \cos\beta) – (\sin\alpha – \sin\beta) = \frac34 – \frac14 \ \Rightarrow\ \cos\beta – \sin\beta = \frac12. \tag{A} \]De (A), escreva \( \cos\beta = \sin\beta + \dfrac12\). Elevando (A) ao quadrado e usando \(2\sin\beta\cos\beta=\sin 2\beta\):
\[ (\cos\beta – \sin\beta)^2 = \frac14 \ \Rightarrow\ 1 – \sin 2\beta = \frac14 \ \Rightarrow\ \sin 2\beta = \frac34. \]Substituindo \(\cos\beta = \sin\beta + \tfrac12\) em \(2\sin\beta\cos\beta=\tfrac34\):
\[ 2\sin\beta\big(\sin\beta+\tfrac12\big)=\frac34 \ \Rightarrow\ 8\sin^2\beta + 4\sin\beta – 3 = 0. \] \[ \sin\beta = \frac{-1 \pm \sqrt{7}}{4}. \]Como \(\beta\in\left[\dfrac{\pi}{2},\dfrac{3\pi}{2}\right]\) (quadrantes II ou III), \(\cos\beta\le 0\). Para \(\sin\beta=\dfrac{-1+\sqrt{7}}{4}\) teríamos \(\cos\beta=\sin\beta+\tfrac12>0\) (descarta). Logo:
\[ \sin\beta=\frac{-1-\sqrt{7}}{4},\qquad \cos\beta=\sin\beta+\frac12=\frac{1-\sqrt{7}}{4}. \]Da 1ª equação, \(\sin\alpha=\sin\beta+\dfrac14=\dfrac{-\sqrt{7}}{4}\). Como \(\alpha\in\left[\dfrac{\pi}{2},\dfrac{3\pi}{2}\right]\), \(\cos\alpha\le 0\). Usando \(\sin^2\alpha+\cos^2\alpha=1\):
\[ \cos\alpha = -\frac{3}{4}. \]
