Final Answer:
The Laplace transform of f(t) = cos(bt) is F(s) = s / (s^2 + b^2) for s > 0.
Step-by-step explanation:
The Laplace transform of a function f(t) is given by the integral of e^(-st) * f(t) from 0 to ∞, where s is a complex number. To find the Laplace transform of f(t) = cos(bt), we use the formula for the Laplace transform. The Laplace transform of cos(bt) is ∫[0 to ∞] e^(-st) * cos(bt) dt.
Using Euler's formula, cos(bt) can be expressed as Re(e^(ibt)), where Re denotes the real part of the complex number. Therefore, the Laplace transform of cos(bt) is the real part of the Laplace transform of e^(ibt). The Laplace transform of e^(ibt) is 1 / (s - ib), derived from the Laplace transform of e^(at) = 1 / (s - a).
However, we're interested in the real part of the Laplace transform, so we manipulate the expression to make it suitable for taking the real part. Multiplying the numerator and denominator by (s + ib), we get the Laplace transform of cos(bt) as s / (s^2 + b^2). Hence, the final expression for the Laplace transform of f(t) = cos(bt) is F(s) = s / (s^2 + b^2) for s > 0. This result represents the Laplace transform of the cosine function with respect to time.c