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The Laplace transform of the function—
f(t) = sin at cos bt
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a + b + a – b 
s2 + (a + b)2 s2 + (a – b)2 -
s + a + b + a – b s2 + (a + b)2 s2 + (a – b)2 -
1 a + b + a – b 2 s2 + (a + b)2 s2 + (a – b)2 -
2 a + b + a – b s2 + (a + b)2 s2 + (a – b)2
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Correct Option: A
sin at cos bt can be written as
| = | {sin (a + b)t + sin (a – b)t} | 2 |
By taking Laplace transform, we get L{sin at·cos bt}
| = L | | {sin (a + b)t + sin (a – b)t} | | ||
| 2 |
| = | | + | | |||||
| 2 | s2 + (a + b)2 | s2 + (a – b)2 |
