The Laplace transform of cosine function cos(at) is s/(s^{2}+a^{2}) and the Laplace transform of cos(t) is s/(s^{2}+1). Here, we will discuss how to find out the Laplace transform of cosine functions.

Recall that the Laplace transform of a function f(t)is denoted and defined as follows:

F(s) = L{f(t)} = $\int_0^\infty$ e^{-st} f(t) dt **…(I)**

## Laplace Transform of Cosine Function

**Theorem:** The Laplace transform of cos at is

L{cos at} = s/(s^{2}+a^{2}).

*Proof:*

In the above definition of Laplace transforms, we put f(t) = cos at. Thus, we have

L{cos at} = $\int_0^\infty$ e^{-st} cos at dt

We use an application of integration by parts formula. Then we have

L{cos at} = $[\dfrac{e^{-st}}{s^2+a^2}(-s \cos at +a \sin at)]_0^\infty$

= lim_{T→∞} $[\dfrac{e^{-st}}{s^2+a^2}(-s \cos at +a \sin at)]_0^T$

= lim_{T→∞} $[\dfrac {e^{-s T} (-s \cos a T + a \sin a T) } {s^2 + a^2}$ $-\dfrac {-s \cos 0 + a \sin 0} {s^2 + a^2}]$

= $0-\dfrac {-s \cdot 1 + a \cdot 0} {s^2 + a^2}$ as sin0=0 and cos 0=1.

= $\dfrac {s} {s^2 + a^2}$

Thus, the Laplace transform of cos(at) is s/(s^{2}+a^{2}).

Find the Laplace transform of cos(at).Summary:L{cos at} = s/(s ^{2}+a^{2}) |

**Read Also:**

**Laplace transform of sin(at)**: The Laplace transform of sinat is a/(s^{2}+a^{2}).

**Laplace transform of 1**: The Laplace transform of 1 is 1/s.

**Inverse Laplace transform of 1**: The inverse Laplace transform of 1 is the Dirac delta function δ(t).

## Laplace Transform of cos t

Note that we have proven above that the Laplace transform of cos(at) is s/(s^{2}+a^{2}). Thus, putting a=1, we get the Laplace transform of cos(t) which is

L{cos t} = s/(s^{2}+1) |

*Proof:*

To find the Laplace transform of cost, we will now use the Laplace transform formula of exponential functions. It is known that

L{e^{it}} = $\dfrac{1}{s-i}$

= $\dfrac{s+i}{(s-i)(s+i)}$, multiplying both numerator and denominator by (s-i).

Therefore, L{e^{it}} = $\dfrac{s+i}{s^2-i^2}$

⇒ L{cos t + sin t} = $\dfrac{s}{s^2+1}+i\dfrac{1}{s^2+1}$

⇒ L{cos t} + L{sin t} = $\dfrac{s}{s^2+1}+i\dfrac{1}{s^2+1}$ by the linearity property of Laplace transform.

Now, comparing the real part and the imaginary part of both sides, we obtain that

L{cos t} = s/(s^{2}+1).

So the Laplace transform of cost is s/(s^{2}+1).

**Alternative Proof:** Let us now use the Laplace transform of second derivatives formula to find the Laplace transform of cost. We have

L{$f^{\prime\prime}(t)$} = s^{2} L{f(t)}-sf(0)-$f^\prime(0)$.

Put f(t) = cos t.

∴ f$^{\prime}$(t) = -sin t, f$^{\prime\prime}$(t)= -cos t, f(0)=cos0 =1, f$^{\prime}$(0)=-sin0=0. Thus, we obtain that

L{- cos t} = s^{2}L{cos t}-s⋅1

⇒ – L{cos t} = s^{2} L{cos t}-s

⇒ s = (s^{2}+1) L{cos t}

⇒ L{cos t} = s/(s^{2}+1).

Hence, the Laplace transform of cos(t) is s/(s^{2}+1).

## FAQs

**Q1: What is the Laplace transform formula of cos t?**

Answer: The Laplace transform formula of cos t is s/(s^{2}+1).

**Q2: cos(at) Laplace transform formula?**

Answer: The Laplace transform of cos at is s/(s^{2}+a^{2}).