The Laplace transform of sin2t is equal to L{sin2t} = 2/(s^{2}+4) and the Laplace of cos2t is L{cos2t} = s/(s^{2}+4). Because, the Laplace of sinat is a/(s^{2}+a^{2}) and the Laplace of cosat is s/(s^{2}+a^{2}). 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 e^{at} which is as follows.

L{e^{at}} = $\dfrac{1}{s-a}$.

So we have that

L{e^{iat}} = $\dfrac{1}{s-ia}$.

Rationalising the denominator of the right hand side, that is, multiplying both the numerator and the denominator by (s-ia) we obtain that

L{e^{iat}} = $\dfrac{s+ia}{(s-ia)(s+ia)}$.

⇒ L{e^{iat}} = $\dfrac{s+ia}{s^2-i^2a^2}$

⇒ 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/(s^{2}+4) and L{cos 2t} = s/(s^{2}+4).

So the Laplace transform of sin2t is 2/(s^{2}+4) and the Laplace transform of cos2t is s/(s^{2}+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/(s^{2}+4), that is, L{sin2t} = 2/(s^{2}+4).

### Q2: What is the Laplace transform of cos2t?

Answer: The Laplace transform of cos2t is s/(s^{2}+4), that is, L{cos2t} = s/(s^{2}+4).