The Laplace transform of cos^2t is equal to 1/2s + s/(2s2+8). In this article, we will learn how to find the Laplace of cos square t. The formula of the Laplace of cos2t is given below:
L{cos2t} = 1/2s + s/(2s2+8).
Table of Contents
Find the Laplace Transform of cos2t
Answer: The Laplace of cos square t is 1/2s + s/(2s2+8).
Proof:
From the trigonometric formulas, we know that 1+cos2t=2cos2t. Thus, it follows that
cos2t = $\frac{1}{2}$(1+cos2t)
Taking Laplace transforms on both sides, we get that L{cos2t} = L{$\frac{1}{2}$(1+cos2t)}. Now, by the linearity property of Laplace transforms it follows that
L{cos2t} = $\frac{1}{2}$ (L{1} + L{cos2t}).
As the Laplace Transform of 1 is 1/s and the Laplace transform of cosat is s/(s2+a2), we deduce that
L{cos2t} = $\dfrac{1}{2} \left(\dfrac{1}{s}+\dfrac{s}{s^2+4} \right)$
Simplifying the above, we get that
L{cos2t} = $\dfrac{1}{2s}+\dfrac{s}{2(s^2+4)}$
So the Laplace transform of cos^2t is equal to 1/2s + s/(2s2+8).
Main Article: Laplace Transform: Definition, Table, Formulas, Properties
| Laplace transform of sin at |
| Laplace transform of sint/t |
| Laplace transform of sin2t/t |
| Laplace transform of cos2t/t |
| Laplace transform of t sint |
FAQs
Answer: The Laplace transform of cos^2t is equal to 1/2s + s/(2s2+8). That is, L{cos2t} = 1/2s + s/(2s2+8).
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