The Laplace transform of cost cos2t is equal to s(s^{2}+5)/(s^{2}+1)(s^{2}+9). Here we learn how to find the Laplace of cost cos2t.

The Laplace transform formula of cost cos2t is given below:

$\boxed{\mathcal{L} \{\cos t \cos 2t \} = \dfrac{s(s^2+5)}{(s^2+1)(s^2+9)} }$

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## Laplace of cost cos2t

Answer: The Laplace of cost cos2t is s(s^{2}+5) / (s^{4}+10s^{2}+9). |

*Proof:*

At first, we will use the formula of cosx cosy in order to simplify cost cos2t. The formula of cosx cosy is provided below:

cosx cosy = $\dfrac{1}{2} [\cos(x+y)+\cos(x-y)]$.

Hence,

cost cos2t = $\dfrac{1}{2} [\cos(t+2t)+\cos(t-2t)]$

⇒ cost cos2t = $\dfrac{1}{2} (\cos 3t+\cos t)$ as cos(-x)=cosx.

Therefore,

L{cost cos2t} = L $\big\{\dfrac{1}{2} (\cos 3t+\cos t) \big\}$

= $\dfrac{1}{2} L\{ \cos 3t+\cos t\}$

= $\dfrac{1}{2} L\{ \cos 3t \}$ + $\dfrac{1}{2} L\{ \cos t \}$ by the linear property of Laplace transforms.

= $\dfrac{1}{2} \dfrac{s}{s^2+9}$ + $\dfrac{1}{2} \dfrac{s}{s^2+1}$, using the Laplace formula L{cos at} = s/(s^{2}+a^{2}).

= $\dfrac{s}{2} \dfrac{s^2+1+s^2+9}{(s^2+1)(s^2+9)}$

= $\dfrac{s(s^2+5)}{s^4+10s^2+9}$.

So the Laplace of transform of cost cos2t is s(s^{2}+5) / (s^{4}+10s^{2}+9).

More Laplace Transforms:

- Laplace Transform of sin2t sin3t
- Laplace Transform of sint sin2t sin3t
- Find L{sin
^{2}t} - Find L{t cost}
- Find the Laplace Transform of t sin2t
- Find Laplace of (1-cost)/t
- Laplace Transform of (1-e
^{t})/t

## FAQs

**Q1: What is the Laplace transform of cost cos2t?**

Answer: The Laplace transform of cost cos2t is equal to s(s^{2}+5)/(s^{2}+1)(s^{2}+9).

**Q2: What is L{cost cos2t}?**

Answer: L{cost cos2t} = s(s^{2}+5)/(s^{2}+1)(s^{2}+9).

This article is written by Dr. T. Mandal, Ph.D in Mathematics. On Mathstoon.com you will find Maths from very basic level to advanced level. Thanks for visiting.