Laplace Transform of cos^4t | Find L{cos^4t}

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.

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)]

Also Read:

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Laplace transform of (1-e^t)/t

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