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

Step 1: Rewrite the product of sines as a sum of cosines.
Use the identity \( \sin A\sin B = \dfrac{1}{2}[\cos(A-B) - \cos(A+B)] \).
With \( A=2t \) and \( B=4t \): \( \sin 2t\sin 4t = \dfrac{1}{2}[\cos(-2t) - \cos(6t)] = \dfrac{1}{2}[\cos 2t - \cos 6t] \), since cosine is an even function.

Step 2: Take the Laplace transform term by term.
Recall the standard result \( L\{\cos(at)\} = \dfrac{s}{s^2+a^2} \).
So \( L\{f(t)\} = \dfrac{1}{2}\left[\dfrac{s}{s^2+4} - \dfrac{s}{s^2+36}\right] \).

Step 3: Combine the two fractions.
\( \dfrac{s}{s^2+4} - \dfrac{s}{s^2+36} = \dfrac{s(s^2+36) - s(s^2+4)}{(s^2+4)(s^2+36)} = \dfrac{32s}{(s^2+4)(s^2+36)} \).

Step 4: Multiply by the leading half.
\( L\{f(t)\} = \dfrac{1}{2} \times \dfrac{32s}{(s^2+4)(s^2+36)} = \dfrac{16s}{(s^2+4)(s^2+36)} \).

Final Answer:
This matches the transform listed as option A. \[ \boxed{L\{f(t)\} = \dfrac{16s}{(s^2+4)(s^2+36)}} \]