The Laplace transform of cost cos2t is equal to s(s2+5)/(s2+1)(s2+9). Here we learn how to find the Laplace of cost cos2t.
The Laplace transform formula of cost cos2t is given below:
$\boxed{\mathcal{L} \{\cos t \cos 2t \} = \dfrac{s(s^2+5)}{(s^2+1)(s^2+9)} }$
Table of Contents
Laplace of cost cos2t
| Answer: The Laplace of cost cos2t is s(s2+5) / (s4+10s2+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/(s2+a2).
= $\dfrac{s}{2} \dfrac{s^2+1+s^2+9}{(s^2+1)(s^2+9)}$
= $\dfrac{s(s^2+5)}{s^4+10s^2+9}$.
So the Laplace of transform of cost cos2t is s(s2+5) / (s4+10s2+9).
More Laplace Transforms:
- Laplace Transform of sin2t sin3t
- Laplace Transform of sint sin2t sin3t
- Find L{sin2t}
- Find L{t cost}
- Find the Laplace Transform of t sin2t
- Find Laplace of (1-cost)/t
- Laplace Transform of (1-et)/t
FAQs
Answer: The Laplace transform of cost cos2t is equal to s(s2+5)/(s2+1)(s2+9).
Answer: L{cost cos2t} = s(s2+5)/(s2+1)(s2+9).
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