The Laplace transform of sin^3t is equal to 6/[(s2+1)(s2+9)]. In this article, let us learn to find the Laplace of sin cube t. The Laplace transform formula of sin3t is given below:
L{sin3t} = $\dfrac{6}{(s^2+1)(s^2+9)}$.
Table of Contents
Find the Laplace Transform of sin3t
Answer: The Laplace transform of sin cube t is L{sin3t} = 6/[(s2+1)(s2+9)].
Proof:
From the theory of trigonometry, we know that sin3t=3sint – 4sin3t. Hence, sin3t can be written as follows
sin3t = $\dfrac{1}{4}$(3sint – sin3t).
⇒ sin3t = $\dfrac{3}{4}$ sint – $\dfrac{1}{4}$ sin3t
Taking Laplace transforms on both sides and using the linearity property of Laplace transforms, we obtain that
L{sin3t} = $\dfrac{3}{4}$ L{sint} – $\dfrac{1}{4}$ L{sin3t}.
As L{sin at} = a/(s2+a2), it follows that
L{sin3t} = $\dfrac{3}{4} \dfrac{1}{s^2+1^2}$ – $\dfrac{1}{4} \dfrac{3}{s^2+3^2}$
= $\dfrac{3}{4} \Big[\dfrac{1}{s^2+1} – \dfrac{1}{s^2+9} \Big]$
= $\dfrac{3}{4} \dfrac{s^2+9-s^2-1}{(s^2+1)(s^2+9)}$
= $\dfrac{3}{4} \dfrac{8}{(s^2+1)(s^2+9)}$
= $\dfrac{6}{(s^2+1)(s^2+9)}$
Final Answer:
So the Laplace transform of sin^3t is equal to 6/[(s2+1)(s2+9)].
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FAQs
Answer: The Laplace transform of sin^3t is equal to 6/[(s2+1)(s2+9)]. It is obtained by applying the Laplace transform on the formula of sin cube t.
Answer: L{sin3t} = 6/[(s2+1)(s2+9)].
