Question

Let \( f(t) \) be a function of \( t \) defined for all positive values of \( t \). The Laplace transform of \( f(t) \), denoted by \( L\{f(t)\} = \int_0^{\infty} e^{-st}f(t)\,dt \), exists for a parameter \( s \) that may be real or complex, provided the integral exists. The Laplace transform of \( f(t) = \sin 2t\sin 4t \) is

• A. \( \dfrac{16s}{(s^2+4)(s^2+36)} \)
• B. \( \dfrac{32s}{(s^2+4)(s^2+36)} \)
• C. \( \dfrac{16}{(s^2+4)(s^2+36)} \)
• D. \( \dfrac{32}{(s^2+4)(s^2+36)} \)

Hint:

Turn the product sin 2t sin 4t into a sum of cosines first, then use the standard cosine Laplace transform.

Answer 1

The Correct Option is A

A second route writes both sine terms using complex exponentials and multiplies them out directly, without reaching for the product to sum identity.

$\sin 2t = \dfrac{e^{i2t}-e^{-i2t}}{2i}$ and $\sin 4t = \dfrac{e^{i4t}-e^{-i4t}}{2i}$. Multiplying: $\sin 2t\sin 4t = \dfrac{-1}{4}\left[e^{i6t} - e^{-i2t} - e^{i2t} + e^{-i6t}\right] = \dfrac{-1}{4}\left[2\cos 6t - 2\cos 2t\right] = \dfrac{1}{2}(\cos 2t - \cos 6t)$, the same combination as before.

Using $L\{\cos at\} = \dfrac{s}{s^2+a^2}$ gives $L\{f(t)\} = \dfrac{1}{2}\left(\dfrac{s}{s^2+4} - \dfrac{s}{s^2+36}\right) = \dfrac{16s}{(s^2+4)(s^2+36)}$.

1. $\dfrac{16s}{(s^2+4)(s^2+36)}$: matches our result, this is correct.
2. $\dfrac{32s}{(s^2+4)(s^2+36)}$: double our value, comes from forgetting the leading one half factor.
3. $\dfrac{16}{(s^2+4)(s^2+36)}$: missing the $s$ in the numerator, cannot be right since cosine transforms carry an $s$ on top.
4. $\dfrac{32}{(s^2+4)(s^2+36)}$: has both mistakes at once, wrong scale and missing $s$.

The exponential expansion confirms option A as the transform of $\sin 2t\sin 4t$.