danjr5 Escreveu:Calcule \(\mathcal{L} = \left \{ \sin t \cdot \mu \left ( t - \frac{\pi}{2} \right ) \right \}\).
Tendo em vista que \(\sin \ t = \sin (t - 2\pi)\), temos:
\(\\ \sin \ t = \sin (t - 2\pi) \\\\\\ \sin \ t = \sin \left ( t - \frac{\pi}{2} - \frac{3\pi}{2} \right ) \\\\\\ \sin \ t = \sin \left [\left ( t - \frac{\pi}{2} \right ) - \frac{3\pi}{2} \right ] \\\\\\ \sin \ t = \sin \left ( t - \frac{\pi}{2} \right ) \cdot \cos \frac{3\pi}{2} - \sin \frac{3\pi}{2} \cdot \cos \left ( t - \frac{\pi}{2} \right ) \\\\\\ \sin \ t = \cos \left ( t - \frac{\pi}{2} \right )\)
Daí,
\(\\ \mathcal{L} \left \{ \sin \ t \cdot \mu \left ( t - \frac{\pi}{2} \right ) \right \} = \mathcal{L} \left \{ \cos \left ( t - \frac{\pi}{2} \right ) \cdot \mu \left ( t - \frac{\pi}{2} \right ) \right \} \\\\\\ \mathcal{L} \left \{ \sin \ t \cdot \mu \left ( t - \frac{\pi}{2} \right ) \right \} = e^{- \frac{\pi}{2} s} \cdot \mathcal{L} \left \{ \cos \ t \right \} \\\\\\ \fbox{\mathcal{L} \left \{ \sin \ t \cdot \mu \left ( t - \frac{\pi}{2} \right ) \right \} = e^{- \frac{\pi}{2} s} \cdot \frac{s}{s^2 + 1}}\)
Não consigo perceber nada de errado em minha resolução, mas de acordo com o gabarito do livro a resposta final é: \(\mathcal{L} \left \{ \sin \ t \cdot \mu \left ( t - \frac{\pi}{2} \right ) \right \} = e^{- s\pi} \cdot \frac{s}{s^2 + 1}\).
Qualquer ajuda será bem recebida!
Desde já agradeço.