The Laplace transform of cos^4t is equal to 3/(8s) + s/[2(s^{2}+4)] + s/[8(s^{2}+16)]. So the Laplace formula of cos^{4}t, that is, L{cos^{4}t} is given as follows.

L{cos^{4}t} = $\dfrac{3}{8s} + \dfrac{s}{2s^2+8} + \dfrac{s}{8s^2+128}$.

Let us now prove the above Laplace transform formula of the fourth power of cost.

Table of Contents

## What is the Laplace of cos^{4}t

Answer: The Laplace of cos^{4}t is equal to 3/(8s) + s/[2(s^{2}+4)] + s/[8(s^{2}+16)]. |

**Proof:**

**Step 1:**

By using the trigonometric identity 2cos^{2}θ = 1+cos2θ, let us first simplify the function cos^{4}t. So we get that

cos^{4}t = cos^{2}t × cos^{2}t.

This implies that

cos^{4}t = $\big( \dfrac{1+\cos 2t}{2}\big)^2$

⇒ cos^{4}t = $\dfrac{1+2\cos 2t+\cos^2 2t}{4}$

⇒ cos^{4}t = $\dfrac{1+2\cos 2t+ \frac{1+\cos 4t}{2}}{4}$

⇒ cos^{4}t = $\dfrac{2+4\cos 2t+ 1+\cos 4t}{8}$

⇒ cos^{4}t = $\dfrac{3+4\cos 2t+\cos 4t}{8}$.

**Step 2:**

Taking Laplace transforms on both sides and using the property of Laplace transforms, we get that

L{cos^{4}t} = $\dfrac{3}{8} L\{1\} + \dfrac{1}{2} L\{\cos 2t\}+ \dfrac{1}{8} L\{\cos 4t\}$

⇒ L{cos^{4}t} = $\dfrac{3}{8} \cdot \dfrac{1}{s} + \dfrac{s}{2(s^2+2^2)}+ \dfrac{s}{8(s^2+4^2)}$, by the Laplace formulae L{1} = 1/s and L{cosat} = s/(s^{2}+a^{2}).

⇒ L{cos^{4}t} = $\dfrac{3}{8s} + \dfrac{s}{2(s^2+4)}+ \dfrac{s}{8(s^2+16)}$

So the Laplace transform of cos^{4}t is equal to 3/(8s) + s/[2(s^{2}+4)] + s/[8(s^{2}+16)]. That is, we have:

L{cos^{4}t} = 3/(8s) + s/[2(s^{2}+4)] + s/[8(s^{2}+16)] |

Also Read:

Laplace transform of t^{2}u(t-1)

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

## FAQs

**Q1: What is the Laplace transform of cos**

^{4}t?Answer: The Laplace transform of cos^{4}t, that is, L{cos^{4}t} is equal to 3/(8s) + s/(2s^{2}+8) + s/(8s^{2}+128).