What is laplacetransform 10cos(t-pi/6) ?

The answer to laplacetransform 10cos(t-pi/6) is (5(sqrt(3)s+1))/(s^2+1)

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laplacetransform 10cos(t-pi/6)

laplace transform

10 cos (t - \frac{π}{6})

\frac{5 ( 3 s + 1 )}{s ^{2} + 1}

Solution steps

L {10 cos (t - \frac{π}{6})}

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)}

= 10 L {cos (t - \frac{π}{6})}

L {cos (t - \frac{π}{6})} : \frac{3 s + 1}{2 ( s ^{2} + 1 )}

= 10 \cdot \frac{3 s + 1}{2 ( s ^{2} + 1 )}

Refine

10 \frac{3 s + 1}{2 ( s ^{2} + 1 )} : \frac{5 ( 3 s + 1 )}{s ^{2} + 1}

= \frac{5 ( 3 s + 1 )}{s ^{2} + 1}

What is laplacetransform 10cos(t-pi/6) ?
The answer to laplacetransform 10cos(t-pi/6) is (5(sqrt(3)s+1))/(s^2+1)