Question

Find the laplace transform of f(t)=cos^3(2t )

Answer 1

Step-by-step explanation:

Let f(t)=cos3(2t)f(t)=cos3⁡(2t).

We want to manipulate this a bit and re-write it to something simpler.

Using the identity cos2(x)=1+cos(2x)2cos2⁡(x)=1+cos⁡(2x)2,

it follows that cos3(x)=1+cos(2x)2⋅cos(x)cos3⁡(x)=1+cos⁡(2x)2⋅cos⁡(x).

In our case, the x=2tx=2t, so we can re-write our function as:

f(t)=1+cos(4t)2⋅cos(2t)⇒f(t)=1+cos⁡(4t)2⋅cos⁡(2t)⇒

⇒f(t)=cos(2t)2+cos(4t)cos(2t)2⇒f(t)=cos⁡(2t)2+cos⁡(4t)cos⁡(2t)2

Now, using the identity: cos(a)⋅cos(b)=12(cos(a−b)+cos(a+b))cos⁡(a)⋅cos⁡(b)=12(cos⁡(a−b)+cos⁡(a+b)), where a=4ta=4t and b=2tb=2t, we obtain:

1

Answer 2

Given,

$\[{\cos ^3}\left( {2t} \right)\]$

Solution,

Know that,

$\[\cos 3x = 4{\cos ^3}x - 3\cos x\]$

$\[\begin{array}{l}\cos 6t = 4{\cos ^3}\left( {2t} \right) - 3\cos 2t\\ \Rightarrow {\cos ^3}\left( {2t} \right) = \frac{1}{4}\cos 6t + \frac{3}{4}\cos 2t\end{array}\]$

Calculate the Laplace transform.

$\[\begin{array}{l} = L\left( {{{\cos }^3}\left( {2t} \right)} \right)\\ = \frac{1}{4}L\left( {\cos 6t} \right) + \frac{3}{4}L\left( {\cos 2t} \right)\end{array}\]$

$\[ = \frac{1}{4} \times \frac{s}{{{s^2} + {6^2}}} + \frac{3}{4} \times \frac{s}{{{s^2} + {2^2}}}\]$

$\[ = \frac{s}{{4\left( {{s^2} + 36} \right)}} + \frac{{3s}}{{4\left( {{s^2} + 4} \right)}}\]$

Hence the Laplace transform of $\[{\cos ^3}\left( {2t} \right)\]$ is $\[\frac{s}{{4\left( {{s^2} + 36} \right)}} + \frac{{3s}}{{4\left( {{s^2} + 4} \right)}}\]$