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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
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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)}

User Vadim Chekry
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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-example-1
User Gnanam
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