Discussion
Laplace transform of function f(t)=sin(wt-θ) is
Answer: D
Expand the argument first:
\[ \sin(\omega t - \theta) = \sin \omega t \cos\theta - \cos \omega t \sin\theta \]
The transform is linear, and \( \cos\theta \) and \( \sin\theta \) are constants, so each term transforms separately:
\[ \mathcal{L}\{\sin(\omega t - \theta)\} = \cos\theta\left(\frac{\omega}{s^{2}+\omega^{2}}\right) - \sin\theta\left(\frac{s}{s^{2}+\omega^{2}}\right) \]
The minus sign carries over from the expansion; a phase of \( +\theta \) would make it a plus.
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