What is laplacetransform sin(4t-5) ?

The answer to laplacetransform sin(4t-5) is (-sin(5)s+4cos(5))/(s^2+16)

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laplacetransform sin(4t-5)

laplace transform

sin (4 t - 5)

\frac{- sin ( 5 ) s + 4 cos ( 5 )}{s ^{2} + 16}

Solution steps

L {sin (4 t - 5)}

Use Laplace Transform table:

L {sin (a t + b)} = \frac{s \cdot sin ( b ) + a \cdot cos ( b )}{s ^{2} + a ^{2}}

L {sin (4 t - 5)} = \frac{s sin ( - 5 ) + 4 cos ( - 5 )}{s ^{2} + 4 ^{2}}

= \frac{s sin ( - 5 ) + 4 cos ( - 5 )}{s ^{2} + 4 ^{2}}

= \frac{- sin ( 5 ) s + 4 cos ( 5 )}{s ^{2} + 16}

\frac{\partial}{\partial x} (x y (1 - x^{2} - y^{2}))

\frac{d}{d x} (\frac{2}{x - 4})

\int \frac{x}{4 x + 1} d x

(6 x y - 2 y^{2} + 1) d x + (3 x^{2} - 4 x y) d y = 0

\int \frac{5}{6 x + 5} d x

What is laplacetransform sin(4t-5) ?
The answer to laplacetransform sin(4t-5) is (-sin(5)s+4cos(5))/(s^2+16)