Answered: "First using a trigonometric identity, find L{f(t)} f(t)=sin⁡2tcos⁡2t"

Trent Carpenter

Answered question

2021-09-09

First using a trigonometric
identity, find

L {f (t)}

f (t) = \sin 2 t \cos 2 t

Arham Warner

Skilled2021-09-10Added 102 answers

Linearity Property: suppose

f_{1} (p) and f_{2} (p)

are Laplace transformations of

F_{1} (t) and F_{2} (t)

respectively. Then:

L {c_{1} F_{1} (t) + c_{2} F_{2} (t)} = c_{1} L {F_{1} (t)} + c_{2} L {F_{2} (t)} = c_{1} f_{1} (p) + c_{2} f_{2} (p)

Also,

L {\sin a t} = \frac{a}{p^{2} + a^{2}}

Given,

f (t) = \sin 2 t \cos 2 t

We have to find

L {f (t)}

i.e. the Laplace Transformation of giveb

f (t)

f (t) = \sin 2 t \cos 2 t =

= \frac{1}{2} (2 \sin 2 t \cos 2 t) =

= \frac{1}{2} \sin 4 t

Taking Laplace transformation on both sides,

L {f (t)} = L {\frac{1}{2} \sin 4 t} =

= \frac{1}{2} L {\sin 4 t} =

= \frac{1}{2} \frac{4}{p^{2} + 4^{2}} =

= \frac{2}{p^{2} + 4^{2}}

The required solution

L {f (t)} = \frac{2}{p^{2} + 4^{2}}

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