Question

L{cos(at+b)} solved by Laplace transformation​

Answer 1

Answer:

we know that cos(m+n)=cos(m) cos(n)- sin (m) sin(n)

L{cos(at+b)}=L{cos(at) cos(b) - sin(at) sin(b)}

=cos(b) L{(cosat)} - sin(b) L{sin(at)}

. =cos(b) s/(s^2+a^2) -

sin(b) a/(s^2+a^2)

In the above steps denominators are same hence we get

= {s cos(b) - a sin (b)}/ (a^2+s^2)

Answer 2

$\dfrac{s(cosb)-a(sinb)}{s^{2} + a^{2} }$ is the Laplace Tansformation of $L[cos(at+b)]$.

Step-by-step explanation:

Given: $L[cos(at+b)]$

To Find: Laplace Transformation

Solution:

• Laplace Tansformation of $L[cos(at+b)]$

We have $L[cos(at+b)]$ such that we can write,

$\Rightarrow L[cos(at+b)] = L[cos(at) cos(b) - sin(at) sin(b)]$

$\Rightarrow L[cos(at+b)] = cos(b) L[cos(at)] - sin(b) L[sin(at)]$

Since the Laplace Transformation of $L[cosat] = \frac{s}{s^{2} + a^{2} }$ and $L[sinat] = \frac{a}{s^{2} + a^{2} }$, substituting them in the above expression, we get,

$\Rightarrow L[cos(at+b)] = cos(b) \Big( \dfrac{s}{s^{2} + a^{2} } \Big) - sin(b) \Big( \dfrac{a}{s^{2} + a^{2} } \Big)$

$\Rightarrow L[cos(at+b)] = \dfrac{s cos(b) - a sin(b) }{s^{2} + a^{2} }$

Hence, $\dfrac{s(cosb)-a(sinb)}{s^{2} + a^{2} }$ is the Laplace Tansformation of $L[cos(at+b)]$.