The Laplace transform of cos^4t is equal to 3/(8s) + s/[2(s2+4)] + s/[8(s2+16)]. So the Laplace formula of cos4t, that is, L{cos4t} is given as follows.
L{cos4t} = $\dfrac{3}{8s} + \dfrac{s}{2s^2+8} + \dfrac{s}{8s^2+128}$.
Let us now prove the above Laplace transform formula of the fourth power of cost.
Table of Contents
What is the Laplace of cos4t
| Answer: The Laplace of cos4t is equal to 3/(8s) + s/[2(s2+4)] + s/[8(s2+16)]. |
Proof:
Step 1:
By using the trigonometric identity 2cos2θ = 1+cos2θ, let us first simplify the function cos4t. So we get that
cos4t = cos2t × cos2t.
This implies that
cos4t = $\big( \dfrac{1+\cos 2t}{2}\big)^2$
⇒ cos4t = $\dfrac{1+2\cos 2t+\cos^2 2t}{4}$
⇒ cos4t = $\dfrac{1+2\cos 2t+ \frac{1+\cos 4t}{2}}{4}$
⇒ cos4t = $\dfrac{2+4\cos 2t+ 1+\cos 4t}{8}$
⇒ cos4t = $\dfrac{3+4\cos 2t+\cos 4t}{8}$.
Step 2:
Taking Laplace transforms on both sides and using the property of Laplace transforms, we get that
L{cos4t} = $\dfrac{3}{8} L\{1\} + \dfrac{1}{2} L\{\cos 2t\}+ \dfrac{1}{8} L\{\cos 4t\}$
⇒ L{cos4t} = $\dfrac{3}{8} \cdot \dfrac{1}{s} + \dfrac{s}{2(s^2+2^2)}+ \dfrac{s}{8(s^2+4^2)}$, by the Laplace formulae L{1} = 1/s and L{cosat} = s/(s2+a2).
⇒ L{cos4t} = $\dfrac{3}{8s} + \dfrac{s}{2(s^2+4)}+ \dfrac{s}{8(s^2+16)}$
So the Laplace transform of cos4t is equal to 3/(8s) + s/[2(s2+4)] + s/[8(s2+16)]. That is, we have:
| L{cos4t} = 3/(8s) + s/[2(s2+4)] + s/[8(s2+16)] |
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FAQs
Answer: The Laplace transform of cos4t, that is, L{cos4t} is equal to 3/(8s) + s/(2s2+8) + s/(8s2+128).
