The Laplace transform of sin2t is equal to L{sin2t} = 2/(s2+4) and the Laplace of cos2t is L{cos2t} = s/(s2+4). Because, the Laplace of sinat is a/(s2+a2) and the Laplace of cosat is s/(s2+a2). In this post, we will find the Laplace transform of sin2t and cos2t.
Laplace of sin2t and cos2t
To find the Laplace of sin2t and cos2t, we will use the Laplace transform formula of eat which is as follows.
L{eat} = $\dfrac{1}{s-a}$.
So we have that
L{eiat} = $\dfrac{1}{s-ia}$.
Rationalising the denominator of the right hand side, that is, multiplying both the numerator and the denominator by (s-ia) we obtain that
L{eiat} = $\dfrac{s+ia}{(s-ia)(s+ia)}$.
⇒ L{eiat} = $\dfrac{s+ia}{s^2-i^2a^2}$
⇒ L{cos at + i sin at} = $\dfrac{s}{s^2+a^2}+i\dfrac{a}{s^2+a^2}$
Now using the linearity property of Laplace transform, it follows that
L{cos at} + i L{sin at} = $\dfrac{s}{s^2+a^2}+i\dfrac{a}{s^2+a^2}$
We now compare the real part and the imaginary part of both sides. As a result, we deduce that
L{sin 2t} = 2/(s2+4) and L{cos 2t} = s/(s2+4).
So the Laplace transform of sin2t is 2/(s2+4) and the Laplace transform of cos2t is s/(s2+4).
You can read:
Laplace transform of sint by definition
Laplace transform of unit step function
Laplace transform of Dirac delta function
FAQs
Q1: What is the Laplace transform of sin2t?
Answer: The Laplace transform of sin2t is 2/(s2+4), that is, L{sin2t} = 2/(s2+4).
Q2: What is the Laplace transform of cos2t?
Answer: The Laplace transform of cos2t is s/(s2+4), that is, L{cos2t} = s/(s2+4).