The Laplace transform of the function sin2 2t isa)(1/2s)-s/[2(s2+16)]b...
Explanation:
The Laplace transform of a function f(t) is defined as:
F(s) = L{f(t)} = ∫[0,∞] e^(-st) f(t) dt
where s is the complex variable and t is the time variable.
In this case, we need to find the Laplace transform of the function sin^2(2t).
Step 1: Rewrite the function using trigonometric identities.
sin^2(2t) = (1 - cos(4t))/2
Step 2: Apply the Laplace transform to each term separately.
L{sin^2(2t)} = L{(1 - cos(4t))/2}
Using the linearity property of the Laplace transform, we can split the expression into two separate transforms.
L{sin^2(2t)} = (1/2) L{1} - (1/2) L{cos(4t)}
The Laplace transform of a constant function is given by:
L{1} = 1/s
And the Laplace transform of cos(at) is given by:
L{cos(at)} = s/(s^2 + a^2)
Applying these formulas, we get:
L{sin^2(2t)} = (1/2) * (1/s) - (1/2) * (s/(s^2 + 4^2))
Simplifying further, we get:
L{sin^2(2t)} = 1/(2s) - s/(2(s^2 + 16))
Step 3: Compare the result with the given options.
The Laplace transform of sin^2(2t) is given by option A: (1/2s) - s/(2(s^2 + 16))
Therefore, the correct answer is option A.