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

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

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laplacetransform cos(t-pi/6)

laplace transform

cos (t - \frac{π}{6})

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

Solution steps

L {cos (t - \frac{π}{6})}

Use Laplace Transform table:

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

L {cos (t - \frac{π}{6})} = \frac{s cos ( - \frac{π}{6} ) - 1 \cdot sin ( - \frac{π}{6} )}{s ^{2} + 1 ^{2}}

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

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

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