Question

Find ℒ{f(t)} by first using a trigonometric identity. (Write your answer as a function of s.)

f(t) = 16cos(t−π/6)

ℒ{f(t)} = ?

Answer 1

`L\{f(t)\}=(8(\sqrt3s+1))/(s^2+1)`

Explanation:

Given :
`f(t)=16\cos (t-(\pi)/(6))`

To find : ℒ{f(t)} by first using a trigonometric identity ?

Solution :

First we solve the function,

`f(t)=16\cos (t-(\pi)/(6))`

Applying trigonometric identity,
`\cos (A-B)=\cos A\cos B+\sin A\sin B`

`f(t)=16(\cos t\cos ((\pi)/(6))+\sin t\sin((\pi)/(6))`

`f(t)=16((\sqrt3)/(2)\cos t+(1)/(2)\sin t)`

`f(t)=(16)/(2)(\sqrt3\cos t+\sin t)`

`f(t)=8(\sqrt3\cos t+\sin t)`

We know,
`L(\cos at)=(s)/(s^2+a^2)` and
`L(\sin at)=(a)/(s^2+a^2)`

Applying Laplace in function,

`L\{f(t)\}=8\sqrt3L(\cos t)+8L(\sin t)`

`L\{f(t)\}=8\sqrt3((s)/(s^2+1))+8((1)/(s^2+1))`

`L\{f(t)\}=(8\sqrt3s+8)/(s^2+1)`

`L\{f(t)\}=(8(\sqrt3s+1))/(s^2+1)`

Therefore, The Laplace transformation is
`L\{f(t)\}=(8(\sqrt3s+1))/(s^2+1)`