Topic: Engineering Mathematics
Laplace transform of $\sin^{3} 2t$ is
- A. $\frac{24}{(s^{2}+4)(s^{2}+36)}$
- B. $\frac{1}{(s^{2}+4)(s^{2}+64)}$
- C. $\frac{48}{(s^{2}+4)(s^{2}+36)}$
- D. $\frac{64}{(s^{2}+4)(s^{2}+36)}$
Answer: C. $\frac{48}{(s^{2}+4)(s^{2}+36)}$
Explanation: Using the trigonometric identity $\sin^{3}\theta=\frac{1}{4}(3\sin\theta-\sin 3\theta)$, we have $\sin^{3}(2t)=\frac{1}{4}(3\sin 2t-\sin 6t)$. Taking the Laplace transform: $L\{\sin^{3}(2t)\}=\frac{1}{4}\left(\frac{3\times2}{s^{2}+4}-\frac{6}{s^{2}+36}\right)$. Simplifying this gives $\frac{6}{4}\left(\frac{s^{2}+36-(s^{2}+4)}{(s^{2}+4)(s^{2}+36)}\right)=\frac{6}{4}\times\frac{32}{(s^{2}+4)(s^{2}+36)}=\frac{48}{(s^{2}+4)(s^{2}+36)}$.