What is laplacetransform sin(t-8) ?

The answer to laplacetransform sin(t-8) is (-sin(8)s+cos(8))/(s^2+1)

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laplacetransform sin(t-8)

laplace transform

sin (t - 8)

\frac{- sin ( 8 ) s + cos ( 8 )}{s ^{2} + 1}

Solution steps

L {sin (t - 8)}

Use Laplace Transform table:

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

L {sin (t - 8)} = \frac{s sin ( - 8 ) + 1 \cdot cos ( - 8 )}{s ^{2} + 1 ^{2}}

= \frac{s sin ( - 8 ) + 1 \cdot cos ( - 8 )}{s ^{2} + 1 ^{2}}

= \frac{- sin ( 8 ) s + cos ( 8 )}{s ^{2} + 1}

\frac{\partial}{\partial x} (y e^{x z})

\frac{d}{d x} (6 x^{2} sin (x) tan (x))

\int 1 + (e^{x})^{2} d x

\int 3 \cdot e^{- 3 y + 6} d y

\frac{d}{d x} (\frac{1 - 2 ln ( x )}{x ^{3}})

What is laplacetransform sin(t-8) ?
The answer to laplacetransform sin(t-8) is (-sin(8)s+cos(8))/(s^2+1)