Question

Find \( \mathscr{L}\{f(t)\} \) by first using a trigonometric identity. (Write your answer as a function of s.) \[ f(t)=\sin (4 t+3) \] \[ \mathscr{L}\{f(t)\}= \]

A) 4 sin³/s²+16

B) sin³s+ 4 cos³s²+16

C) sin³/s - 4 cos³/s²+16

D) sin³/s

Answer 1

The Laplace transform of f(t) = sin(4t + 3) is 4 / (s² + 16), which is not reflected in the provided answer choices, suggesting a typo in the question or choices.

To find the Laplace transform of f(t) = sin(4t + 3), we first recognize that there is no direct trigonometric identity for sin(4t + 3). Instead, we proceed with the transform directly since the input to the function is a linear transformation of t. The Laplace transform of a sine function is given by:

ℒ{sin(at)} = ℒ{a/(s² + a²)}

In this case, a = 4, so the Laplace transform of f(t) is:

ℒ{sin(4t)} = 4 / (s² + 4²)

The +3 inside the sine does not affect the frequency or the ℒ-transform, only the phase, which is not captured by the transform. So the ℒ-transform of f(t) = sin(4t + 3) does not change.

So our answer is: ℒ{f(t)} = 4 / (s² + 16)

The option matching the result is therefore not listed in the provided choices, indicating a potential typo in the question or answer choices.