The Laplace transform of sin^4t is denoted by L{sin4t}, and it is equal to 3/(8s) – s/[2(s2+4)] + s/[8(s2+16)]. So the Laplace formula of sin4t is equal to
L{sin4t} = $\dfrac{3}{8s} – \dfrac{s}{2(s^2+4)} + \dfrac{s}{8(s^2+16)}$.
Let us now prove the above Laplace transform formula of the fourth power of sint.
Table of Contents
Laplace of sin4t
Question: What is the Laplace of sin4t?
Answer: L{sin4t} = 3/(8s) – s/[2(s2+4)] + s/[8(s2+16)]. |
Proof:
To find the Laplace transform of sin4t, we will follow the below steps.
Step 1:
First, we simplify the function sin4t using the trigonometric identity: 2sin2t = 1-cos2t. Therefore,
sin4t = sin2t × sin2t
⇒ sin4t = $\big( \dfrac{1-\cos 2t}{2}\big)^2$
⇒ sin4t = $\dfrac{1-2\cos 2t+\cos^2 2t}{4}$
⇒ sin4t = $\dfrac{1-2\cos 2t+ \frac{1+\cos 4t}{2}}{4}$ by the identity 2cos2θ = 1+cos2θ.
⇒ sin4t = $\dfrac{2-4\cos 2t+ 1+\cos 4t}{8}$
⇒ sin4t = $\dfrac{3-4\cos 2t+\cos 4t}{8}$.
Step 2:
So the Laplace of sin4t will be equal to
L{sin4t} = L$\big\{ \dfrac{3-4\cos 2t+\cos 4t}{8} \big\}$
= $\dfrac{1}{8} \big[ 3L\{1\} – 4L\{\cos 2t\}+ L\{\cos 4t\} \big]$, by the property of Laplace transforms.
= $\dfrac{1}{8} \big[ \dfrac{3}{s} – 4 \dfrac{s}{s^2+2^2}+ \dfrac{s}{s^2+4^2} \big]$ as we know that L{1} = 1/s and L{cosat} = s/(s2+a2).
= $\dfrac{3}{8s} – \dfrac{s}{2(s^2+4)}+ \dfrac{s}{8(s^2+16)}$.
Thus, the Laplace transform of sin4t is equal to L{sin4t} = 3/(8s) – s/[2(s2+4)] + s/[8(s2+16)].
Also Read:
FAQs
Answer: The Laplace transform of sin4t is L{sin4t} = 3/(8s) – s/(2s2+8) + s/(8s2+128).
This article is written by Dr. T, an expert in Mathematics (PhD). On Mathstoon.com you will find Maths from very basic level to advanced level. Thanks for visiting.