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

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

Popular >

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}

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

(1.9) (4.1)

\int (- x^{\frac{π}{3}} - 3 x^{3} + 5) d x

\int \frac{8}{16 - ( x + 4 ) ^{2}} d x

x \to 0 lim (\frac{x o t - 1}{x ^{2}})

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