In the following Laplace transform, how do you get rid of the cos (2at) * sin (at) ?

L{sin^3 (at) }

L{ sin (at) * [1/2 (1 - cos (2at) ]}

1/2 L{sin (at) } - 1/2 L{cos (2at) * sin (at) }

1/2 * (a^2) / (s^2 + a^2) - 1/2 L{cos (2at) * sin (at) }

Question

Larsen

31 January, 07:59

+2

Answers (1)

Know the Answer?

Kasen Trevino · Answer 1

sin A cos B=1/2[sin (A-B) + sin (A+B) ]

sin (at) * cos (2at) = 1/2[sin (3at) - sin (at) ]

In the following Laplace transform, how do you get rid of the cos (2at) * sin (at) ?L{sin^3 (at) }L{ sin (at) * [1/2 (1 - cos (2at) ]}1/2 L{sin (at) } - 1/2 L{cos (2at) * sin (at) }1/2 * (a^2) / (s^2 + a^2) - 1/2 L{cos (2at) * sin (at) }