Discussion
Let \(f(t)\) be a function of \(t\) defined for all positive values of \(t\). The Laplace transform of \(f(t)\) denoted by \(L\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t)\, dt\), provided that the integral exists where \(s\) is a parameter which may be a real or complex number.
The Laplace transform of \(f(t) = \sin 2t \sin 4t\) is
Answer: A
The transform is \(\frac{16s}{(s^{2}+4)(s^{2}+36)}\).
Use \(\sin A\sin B = \frac{1}{2}[\cos(A - B) - \cos(A + B)]\): \(\sin 2t\sin 4t = \frac{1}{2}(\cos 2t - \cos 6t)\). With \(L\{\cos at\} = s/(s^{2} + a^{2})\):
\[ \begin{aligned} F(s) &= \frac{1}{2}\left(\frac{s}{s^{2}+4} - \frac{s}{s^{2}+36}\right) \\ &= \frac{1}{2}\cdot\frac{32s}{(s^{2}+4)(s^{2}+36)} = \frac{16s}{(s^{2}+4)(s^{2}+36)} \end{aligned} \]
\(32s/\ldots\) forgets the factor \(\frac{1}{2}\). The options without \(s\) in the numerator would be transforms of sine terms; a product of two sines reduces to cosines, which bring the factor \(s\).
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