What is laplacetransform 6cos(4t-1) ?

The answer to laplacetransform 6cos(4t-1) is (6(cos(1)s+4sin(1)))/(s^2+16)

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laplacetransform 6cos(4t-1)

laplace transform

6 cos (4 t - 1)

\frac{6 ( cos ( 1 ) s + 4 sin ( 1 ) )}{s ^{2} + 16}

Solution steps

L {6 cos (4 t - 1)}

Use the constant multiplication property of Laplace Transform:

For function

f (t)

and constant

a : L {a \cdot f (t)} = a \cdot L {f (t)}

= 6 L {cos (4 t - 1)}

L {cos (4 t - 1)} : \frac{cos ( 1 ) s + 4 sin ( 1 )}{s ^{2} + 16}

= 6 \cdot \frac{cos ( 1 ) s + 4 sin ( 1 )}{s ^{2} + 16}

Refine

6 \frac{cos ( 1 ) s + 4 sin ( 1 )}{s ^{2} + 16} : \frac{6 ( cos ( 1 ) s + 4 sin ( 1 ) )}{s ^{2} + 16}

= \frac{6 ( cos ( 1 ) s + 4 sin ( 1 ) )}{s ^{2} + 16}

What is laplacetransform 6cos(4t-1) ?
The answer to laplacetransform 6cos(4t-1) is (6(cos(1)s+4sin(1)))/(s^2+16)